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I was recently given this interview question and I'm curious what a good solution to it would be.
Say I'm given a 2d array where all the
numbers in the array are in increasing
order from left to right and top to
bottom.
What is the best way to search and
determine if a target number is in the
array?
Now, my first inclination is to utilize a binary search since my data is sorted. I can determine if a number is in a single row in O(log N) time. However, it is the 2 directions that throw me off.
Another solution I thought may work is to start somewhere in the middle. If the middle value is less than my target, then I can be sure it is in the left square portion of the matrix from the middle. I then move diagonally and check again, reducing the size of the square that the target could potentially be in until I have honed in on the target number.
Does anyone have any good ideas on solving this problem?
Example array:
Sorted left to right, top to bottom.
1 2 4 5 6
2 3 5 7 8
4 6 8 9 10
5 8 9 10 11
Here's a simple approach:
Start at the bottom-left corner.
If the target is less than that value, it must be above us, so move up one.
Otherwise we know that the target can't be in that column, so move right one.
Goto 2.
For an NxM array, this runs in O(N+M). I think it would be difficult to do better. :)
Edit: Lots of good discussion. I was talking about the general case above; clearly, if N or M are small, you could use a binary search approach to do this in something approaching logarithmic time.
Here are some details, for those who are curious:
History
This simple algorithm is called a Saddleback Search. It's been around for a while, and it is optimal when N == M. Some references:
David Gries, The Science of Programming. Springer-Verlag, 1989.
Edsgar Dijkstra, The Saddleback Search. Note EWD-934, 1985.
However, when N < M, intuition suggests that binary search should be able to do better than O(N+M): For example, when N == 1, a pure binary search will run in logarithmic rather than linear time.
Worst-case bound
Richard Bird examined this intuition that binary search could improve the Saddleback algorithm in a 2006 paper:
Richard S. Bird, Improving Saddleback Search: A Lesson in Algorithm Design, in Mathematics of Program Construction, pp. 82--89, volume 4014, 2006.
Using a rather unusual conversational technique, Bird shows us that for N <= M, this problem has a lower bound of Ω(N * log(M/N)). This bound make sense, as it gives us linear performance when N == M and logarithmic performance when N == 1.
Algorithms for rectangular arrays
One approach that uses a row-by-row binary search looks like this:
Start with a rectangular array where N < M. Let's say N is rows and M is columns.
Do a binary search on the middle row for value. If we find it, we're done.
Otherwise we've found an adjacent pair of numbers s and g, where s < value < g.
The rectangle of numbers above and to the left of s is less than value, so we can eliminate it.
The rectangle below and to the right of g is greater than value, so we can eliminate it.
Go to step (2) for each of the two remaining rectangles.
In terms of worst-case complexity, this algorithm does log(M) work to eliminate half the possible solutions, and then recursively calls itself twice on two smaller problems. We do have to repeat a smaller version of that log(M) work for every row, but if the number of rows is small compared to the number of columns, then being able to eliminate all of those columns in logarithmic time starts to become worthwhile.
This gives the algorithm a complexity of T(N,M) = log(M) + 2 * T(M/2, N/2), which Bird shows to be O(N * log(M/N)).
Another approach posted by Craig Gidney describes an algorithm similar the approach above: it examines a row at a time using a step size of M/N. His analysis shows that this results in O(N * log(M/N)) performance as well.
Performance Comparison
Big-O analysis is all well and good, but how well do these approaches work in practice? The chart below examines four algorithms for increasingly "square" arrays:
(The "naive" algorithm simply searches every element of the array. The "recursive" algorithm is described above. The "hybrid" algorithm is an implementation of Gidney's algorithm. For each array size, performance was measured by timing each algorithm over fixed set of 1,000,000 randomly-generated arrays.)
Some notable points:
As expected, the "binary search" algorithms offer the best performance on rectangular arrays and the Saddleback algorithm works the best on square arrays.
The Saddleback algorithm performs worse than the "naive" algorithm for 1-d arrays, presumably because it does multiple comparisons on each item.
The performance hit that the "binary search" algorithms take on square arrays is presumably due to the overhead of running repeated binary searches.
Summary
Clever use of binary search can provide O(N * log(M/N) performance for both rectangular and square arrays. The O(N + M) "saddleback" algorithm is much simpler, but suffers from performance degradation as arrays become increasingly rectangular.
This problem takes Θ(b lg(t)) time, where b = min(w,h) and t=b/max(w,h). I discuss the solution in this blog post.
Lower bound
An adversary can force an algorithm to make Ω(b lg(t)) queries, by restricting itself to the main diagonal:
Legend: white cells are smaller items, gray cells are larger items, yellow cells are smaller-or-equal items and orange cells are larger-or-equal items. The adversary forces the solution to be whichever yellow or orange cell the algorithm queries last.
Notice that there are b independent sorted lists of size t, requiring Ω(b lg(t)) queries to completely eliminate.
Algorithm
(Assume without loss of generality that w >= h)
Compare the target item against the cell t to the left of the top right corner of the valid area
If the cell's item matches, return the current position.
If the cell's item is less than the target item, eliminate the remaining t cells in the row with a binary search. If a matching item is found while doing this, return with its position.
Otherwise the cell's item is more than the target item, eliminating t short columns.
If there's no valid area left, return failure
Goto step 2
Finding an item:
Determining an item doesn't exist:
Legend: white cells are smaller items, gray cells are larger items, and the green cell is an equal item.
Analysis
There are b*t short columns to eliminate. There are b long rows to eliminate. Eliminating a long row costs O(lg(t)) time. Eliminating t short columns costs O(1) time.
In the worst case we'll have to eliminate every column and every row, taking time O(lg(t)*b + b*t*1/t) = O(b lg(t)).
Note that I'm assuming lg clamps to a result above 1 (i.e. lg(x) = log_2(max(2,x))). That's why when w=h, meaning t=1, we get the expected bound of O(b lg(1)) = O(b) = O(w+h).
Code
public static Tuple<int, int> TryFindItemInSortedMatrix<T>(this IReadOnlyList<IReadOnlyList<T>> grid, T item, IComparer<T> comparer = null) {
if (grid == null) throw new ArgumentNullException("grid");
comparer = comparer ?? Comparer<T>.Default;
// check size
var width = grid.Count;
if (width == 0) return null;
var height = grid[0].Count;
if (height < width) {
var result = grid.LazyTranspose().TryFindItemInSortedMatrix(item, comparer);
if (result == null) return null;
return Tuple.Create(result.Item2, result.Item1);
}
// search
var minCol = 0;
var maxRow = height - 1;
var t = height / width;
while (minCol < width && maxRow >= 0) {
// query the item in the minimum column, t above the maximum row
var luckyRow = Math.Max(maxRow - t, 0);
var cmpItemVsLucky = comparer.Compare(item, grid[minCol][luckyRow]);
if (cmpItemVsLucky == 0) return Tuple.Create(minCol, luckyRow);
// did we eliminate t rows from the bottom?
if (cmpItemVsLucky < 0) {
maxRow = luckyRow - 1;
continue;
}
// we eliminated most of the current minimum column
// spend lg(t) time eliminating rest of column
var minRowInCol = luckyRow + 1;
var maxRowInCol = maxRow;
while (minRowInCol <= maxRowInCol) {
var mid = minRowInCol + (maxRowInCol - minRowInCol + 1) / 2;
var cmpItemVsMid = comparer.Compare(item, grid[minCol][mid]);
if (cmpItemVsMid == 0) return Tuple.Create(minCol, mid);
if (cmpItemVsMid > 0) {
minRowInCol = mid + 1;
} else {
maxRowInCol = mid - 1;
maxRow = mid - 1;
}
}
minCol += 1;
}
return null;
}
I would use the divide-and-conquer strategy for this problem, similar to what you suggested, but the details are a bit different.
This will be a recursive search on subranges of the matrix.
At each step, pick an element in the middle of the range. If the value found is what you are seeking, then you're done.
Otherwise, if the value found is less than the value that you are seeking, then you know that it is not in the quadrant above and to the left of your current position. So recursively search the two subranges: everything (exclusively) below the current position, and everything (exclusively) to the right that is at or above the current position.
Otherwise, (the value found is greater than the value that you are seeking) you know that it is not in the quadrant below and to the right of your current position. So recursively search the two subranges: everything (exclusively) to the left of the current position, and everything (exclusively) above the current position that is on the current column or a column to the right.
And ba-da-bing, you found it.
Note that each recursive call only deals with the current subrange only, not (for example) ALL rows above the current position. Just those in the current subrange.
Here's some pseudocode for you:
bool numberSearch(int[][] arr, int value, int minX, int maxX, int minY, int maxY)
if (minX == maxX and minY == maxY and arr[minX,minY] != value)
return false
if (arr[minX,minY] > value) return false; // Early exits if the value can't be in
if (arr[maxX,maxY] < value) return false; // this subrange at all.
int nextX = (minX + maxX) / 2
int nextY = (minY + maxY) / 2
if (arr[nextX,nextY] == value)
{
print nextX,nextY
return true
}
else if (arr[nextX,nextY] < value)
{
if (numberSearch(arr, value, minX, maxX, nextY + 1, maxY))
return true
return numberSearch(arr, value, nextX + 1, maxX, minY, nextY)
}
else
{
if (numberSearch(arr, value, minX, nextX - 1, minY, maxY))
return true
reutrn numberSearch(arr, value, nextX, maxX, minY, nextY)
}
The two main answers give so far seem to be the arguably O(log N) "ZigZag method" and the O(N+M) Binary Search method. I thought I'd do some testing comparing the two methods with some various setups. Here are the details:
The array is N x N square in every test, with N varying from 125 to 8000 (the largest my JVM heap could handle). For each array size, I picked a random place in the array to put a single 2. I then put a 3 everywhere possible (to the right and below of the 2) and then filled the rest of the array with 1. Some of the earlier commenters seemed to think this type of setup would yield worst case run time for both algorithms. For each array size, I picked 100 different random locations for the 2 (search target) and ran the test. I recorded avg run time and worst case run time for each algorithm. Because it was happening too fast to get good ms readings in Java, and because I don't trust Java's nanoTime(), I repeated each test 1000 times just to add a uniform bias factor to all the times. Here are the results:
ZigZag beat binary in every test for both avg and worst case times, however, they are all within an order of magnitude of each other more or less.
Here is the Java code:
public class SearchSortedArray2D {
static boolean findZigZag(int[][] a, int t) {
int i = 0;
int j = a.length - 1;
while (i <= a.length - 1 && j >= 0) {
if (a[i][j] == t) return true;
else if (a[i][j] < t) i++;
else j--;
}
return false;
}
static boolean findBinarySearch(int[][] a, int t) {
return findBinarySearch(a, t, 0, 0, a.length - 1, a.length - 1);
}
static boolean findBinarySearch(int[][] a, int t,
int r1, int c1, int r2, int c2) {
if (r1 > r2 || c1 > c2) return false;
if (r1 == r2 && c1 == c2 && a[r1][c1] != t) return false;
if (a[r1][c1] > t) return false;
if (a[r2][c2] < t) return false;
int rm = (r1 + r2) / 2;
int cm = (c1 + c2) / 2;
if (a[rm][cm] == t) return true;
else if (a[rm][cm] > t) {
boolean b1 = findBinarySearch(a, t, r1, c1, r2, cm - 1);
boolean b2 = findBinarySearch(a, t, r1, cm, rm - 1, c2);
return (b1 || b2);
} else {
boolean b1 = findBinarySearch(a, t, r1, cm + 1, rm, c2);
boolean b2 = findBinarySearch(a, t, rm + 1, c1, r2, c2);
return (b1 || b2);
}
}
static void randomizeArray(int[][] a, int N) {
int ri = (int) (Math.random() * N);
int rj = (int) (Math.random() * N);
a[ri][rj] = 2;
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
if (i == ri && j == rj) continue;
else if (i > ri || j > rj) a[i][j] = 3;
else a[i][j] = 1;
}
}
}
public static void main(String[] args) {
int N = 8000;
int[][] a = new int[N][N];
int randoms = 100;
int repeats = 1000;
long start, end, duration;
long zigMin = Integer.MAX_VALUE, zigMax = Integer.MIN_VALUE;
long binMin = Integer.MAX_VALUE, binMax = Integer.MIN_VALUE;
long zigSum = 0, zigAvg;
long binSum = 0, binAvg;
for (int k = 0; k < randoms; k++) {
randomizeArray(a, N);
start = System.currentTimeMillis();
for (int i = 0; i < repeats; i++) findZigZag(a, 2);
end = System.currentTimeMillis();
duration = end - start;
zigSum += duration;
zigMin = Math.min(zigMin, duration);
zigMax = Math.max(zigMax, duration);
start = System.currentTimeMillis();
for (int i = 0; i < repeats; i++) findBinarySearch(a, 2);
end = System.currentTimeMillis();
duration = end - start;
binSum += duration;
binMin = Math.min(binMin, duration);
binMax = Math.max(binMax, duration);
}
zigAvg = zigSum / randoms;
binAvg = binSum / randoms;
System.out.println(findZigZag(a, 2) ?
"Found via zigzag method. " : "ERROR. ");
//System.out.println("min search time: " + zigMin + "ms");
System.out.println("max search time: " + zigMax + "ms");
System.out.println("avg search time: " + zigAvg + "ms");
System.out.println();
System.out.println(findBinarySearch(a, 2) ?
"Found via binary search method. " : "ERROR. ");
//System.out.println("min search time: " + binMin + "ms");
System.out.println("max search time: " + binMax + "ms");
System.out.println("avg search time: " + binAvg + "ms");
}
}
This is a short proof of the lower bound on the problem.
You cannot do it better than linear time (in terms of array dimensions, not the number of elements). In the array below, each of the elements marked as * can be either 5 or 6 (independently of other ones). So if your target value is 6 (or 5) the algorithm needs to examine all of them.
1 2 3 4 *
2 3 4 * 7
3 4 * 7 8
4 * 7 8 9
* 7 8 9 10
Of course this expands to bigger arrays as well. This means that this answer is optimal.
Update: As pointed out by Jeffrey L Whitledge, it is only optimal as the asymptotic lower bound on running time vs input data size (treated as a single variable). Running time treated as two-variable function on both array dimensions can be improved.
I think Here is the answer and it works for any kind of sorted matrix
bool findNum(int arr[][ARR_MAX],int xmin, int xmax, int ymin,int ymax,int key)
{
if (xmin > xmax || ymin > ymax || xmax < xmin || ymax < ymin) return false;
if ((xmin == xmax) && (ymin == ymax) && (arr[xmin][ymin] != key)) return false;
if (arr[xmin][ymin] > key || arr[xmax][ymax] < key) return false;
if (arr[xmin][ymin] == key || arr[xmax][ymax] == key) return true;
int xnew = (xmin + xmax)/2;
int ynew = (ymin + ymax)/2;
if (arr[xnew][ynew] == key) return true;
if (arr[xnew][ynew] < key)
{
if (findNum(arr,xnew+1,xmax,ymin,ymax,key))
return true;
return (findNum(arr,xmin,xmax,ynew+1,ymax,key));
} else {
if (findNum(arr,xmin,xnew-1,ymin,ymax,key))
return true;
return (findNum(arr,xmin,xmax,ymin,ynew-1,key));
}
}
Interesting question. Consider this idea - create one boundary where all the numbers are greater than your target and another where all the numbers are less than your target. If anything is left in between the two, that's your target.
If I'm looking for 3 in your example, I read across the first row until I hit 4, then look for the smallest adjacent number (including diagonals) greater than 3:
1 2 4 5 6
2 3 5 7 8
4 6 8 9 10
5 8 9 10 11
Now I do the same for those numbers less than 3:
1 2 4 5 6
2 3 5 7 8
4 6 8 9 10
5 8 9 10 11
Now I ask, is anything inside the two boundaries? If yes, it must be 3. If no, then there is no 3. Sort of indirect since I don't actually find the number, I just deduce that it must be there. This has the added bonus of counting ALL the 3's.
I tried this on some examples and it seems to work OK.
Binary search through the diagonal of the array is the best option.
We can find out whether the element is less than or equal to the elements in the diagonal.
I've been asking this question in interviews for the better part of a decade and I think there's only been one person who has been able to come up with an optimal algorithm.
My solution has always been:
Binary search the middle diagonal, which is the diagonal running down and right, containing the item at (rows.count/2, columns.count/2).
If the target number is found, return true.
Otherwise, two numbers (u and v) will have been found such that u is smaller than the target, v is larger than the target, and v is one right and one down from u.
Recursively search the sub-matrix to the right of u and top of v and the one to the bottom of u and left of v.
I believe this is a strict improvement over the algorithm given by Nate here, since searching the diagonal often allows a reduction of over half the search space (if the matrix is close to square), whereas searching a row or column always results in an elimination of exactly half.
Here's the code in (probably not terribly Swifty) Swift:
import Cocoa
class Solution {
func searchMatrix(_ matrix: [[Int]], _ target: Int) -> Bool {
if (matrix.isEmpty || matrix[0].isEmpty) {
return false
}
return _searchMatrix(matrix, 0..<matrix.count, 0..<matrix[0].count, target)
}
func _searchMatrix(_ matrix: [[Int]], _ rows: Range<Int>, _ columns: Range<Int>, _ target: Int) -> Bool {
if (rows.count == 0 || columns.count == 0) {
return false
}
if (rows.count == 1) {
return _binarySearch(matrix, rows.lowerBound, columns, target, true)
}
if (columns.count == 1) {
return _binarySearch(matrix, columns.lowerBound, rows, target, false)
}
var lowerInflection = (-1, -1)
var upperInflection = (Int.max, Int.max)
var currentRows = rows
var currentColumns = columns
while (currentRows.count > 0 && currentColumns.count > 0 && upperInflection.0 > lowerInflection.0+1) {
let rowMidpoint = (currentRows.upperBound + currentRows.lowerBound) / 2
let columnMidpoint = (currentColumns.upperBound + currentColumns.lowerBound) / 2
let value = matrix[rowMidpoint][columnMidpoint]
if (value == target) {
return true
}
if (value > target) {
upperInflection = (rowMidpoint, columnMidpoint)
currentRows = currentRows.lowerBound..<rowMidpoint
currentColumns = currentColumns.lowerBound..<columnMidpoint
} else {
lowerInflection = (rowMidpoint, columnMidpoint)
currentRows = rowMidpoint+1..<currentRows.upperBound
currentColumns = columnMidpoint+1..<currentColumns.upperBound
}
}
if (lowerInflection.0 == -1) {
lowerInflection = (upperInflection.0-1, upperInflection.1-1)
} else if (upperInflection.0 == Int.max) {
upperInflection = (lowerInflection.0+1, lowerInflection.1+1)
}
return _searchMatrix(matrix, rows.lowerBound..<lowerInflection.0+1, upperInflection.1..<columns.upperBound, target) || _searchMatrix(matrix, upperInflection.0..<rows.upperBound, columns.lowerBound..<lowerInflection.1+1, target)
}
func _binarySearch(_ matrix: [[Int]], _ rowOrColumn: Int, _ range: Range<Int>, _ target: Int, _ searchRow : Bool) -> Bool {
if (range.isEmpty) {
return false
}
let midpoint = (range.upperBound + range.lowerBound) / 2
let value = (searchRow ? matrix[rowOrColumn][midpoint] : matrix[midpoint][rowOrColumn])
if (value == target) {
return true
}
if (value > target) {
return _binarySearch(matrix, rowOrColumn, range.lowerBound..<midpoint, target, searchRow)
} else {
return _binarySearch(matrix, rowOrColumn, midpoint+1..<range.upperBound, target, searchRow)
}
}
}
A. Do a binary search on those lines where the target number might be on.
B. Make it a graph : Look for the number by taking always the smallest unvisited neighbour node and backtracking when a too big number is found
Binary search would be the best approach, imo. Starting at 1/2 x, 1/2 y will cut it in half. IE a 5x5 square would be something like x == 2 / y == 3 . I rounded one value down and one value up to better zone in on the direction of the targeted value.
For clarity the next iteration would give you something like x == 1 / y == 2 OR x == 3 / y == 5
Well, to begin with, let us assume we are using a square.
1 2 3
2 3 4
3 4 5
1. Searching a square
I would use a binary search on the diagonal. The goal is the locate the smaller number that is not strictly lower than the target number.
Say I am looking for 4 for example, then I would end up locating 5 at (2,2).
Then, I am assured that if 4 is in the table, it is at a position either (x,2) or (2,x) with x in [0,2]. Well, that's just 2 binary searches.
The complexity is not daunting: O(log(N)) (3 binary searches on ranges of length N)
2. Searching a rectangle, naive approach
Of course, it gets a bit more complicated when N and M differ (with a rectangle), consider this degenerate case:
1 2 3 4 5 6 7 8
2 3 4 5 6 7 8 9
10 11 12 13 14 15 16 17
And let's say I am looking for 9... The diagonal approach is still good, but the definition of diagonal changes. Here my diagonal is [1, (5 or 6), 17]. Let's say I picked up [1,5,17], then I know that if 9 is in the table it is either in the subpart:
5 6 7 8
6 7 8 9
10 11 12 13 14 15 16
This gives us 2 rectangles:
5 6 7 8 10 11 12 13 14 15 16
6 7 8 9
So we can recurse! probably beginning by the one with less elements (though in this case it kills us).
I should point that if one of the dimensions is less than 3, we cannot apply the diagonal methods and must use a binary search. Here it would mean:
Apply binary search on 10 11 12 13 14 15 16, not found
Apply binary search on 5 6 7 8, not found
Apply binary search on 6 7 8 9, not found
It's tricky because to get good performance you might want to differentiate between several cases, depending on the general shape....
3. Searching a rectangle, brutal approach
It would be much easier if we dealt with a square... so let's just square things up.
1 2 3 4 5 6 7 8
2 3 4 5 6 7 8 9
10 11 12 13 14 15 16 17
17 . . . . . . 17
. .
. .
. .
17 . . . . . . 17
We now have a square.
Of course, we will probably NOT actually create those rows, we could simply emulate them.
def get(x,y):
if x < N and y < M: return table[x][y]
else: return table[N-1][M-1] # the max
so it behaves like a square without occupying more memory (at the cost of speed, probably, depending on cache... oh well :p)
EDIT:
I misunderstood the question. As the comments point out this only works in the more restricted case.
In a language like C that stores data in row-major order, simply treat it as a 1D array of size n * m and use a binary search.
I have a recursive Divide & Conquer Solution.
Basic Idea for one step is: We know that the Left-Upper(LU) is smallest and the right-bottom(RB) is the largest no., so the given No(N) must: N>=LU and N<=RB
IF N==LU and N==RB::::Element Found and Abort returning the position/Index
If N>=LU and N<=RB = FALSE, No is not there and abort.
If N>=LU and N<=RB = TRUE, Divide the 2D array in 4 equal parts of 2D array each in logical manner..
And then apply the same algo step to all four sub-array.
My Algo is Correct I have implemented on my friends PC.
Complexity: each 4 comparisons can b used to deduce the total no of elements to one-fourth at its worst case..
So My complexity comes to be 1 + 4 x lg(n) + 4
But really expected this to be working on O(n)
I think something is wrong somewhere in my calculation of Complexity, please correct if so..
The optimal solution is to start at the top-left corner, that has minimal value. Move diagonally downwards to the right until you hit an element whose value >= value of the given element. If the element's value is equal to that of the given element, return found as true.
Otherwise, from here we can proceed in two ways.
Strategy 1:
Move up in the column and search for the given element until we reach the end. If found, return found as true
Move left in the row and search for the given element until we reach the end. If found, return found as true
return found as false
Strategy 2:
Let i denote the row index and j denote the column index of the diagonal element we have stopped at. (Here, we have i = j, BTW). Let k = 1.
Repeat the below steps until i-k >= 0
Search if a[i-k][j] is equal to the given element. if yes, return found as true.
Search if a[i][j-k] is equal to the given element. if yes, return found as true.
Increment k
1 2 4 5 6
2 3 5 7 8
4 6 8 9 10
5 8 9 10 11
public boolean searchSortedMatrix(int arr[][] , int key , int minX , int maxX , int minY , int maxY){
// base case for recursion
if(minX > maxX || minY > maxY)
return false ;
// early fails
// array not properly intialized
if(arr==null || arr.length==0)
return false ;
// arr[0][0]> key return false
if(arr[minX][minY]>key)
return false ;
// arr[maxX][maxY]<key return false
if(arr[maxX][maxY]<key)
return false ;
//int temp1 = minX ;
//int temp2 = minY ;
int midX = (minX+maxX)/2 ;
//if(temp1==midX){midX+=1 ;}
int midY = (minY+maxY)/2 ;
//if(temp2==midY){midY+=1 ;}
// arr[midX][midY] = key ? then value found
if(arr[midX][midY] == key)
return true ;
// alas ! i have to keep looking
// arr[midX][midY] < key ? search right quad and bottom matrix ;
if(arr[midX][midY] < key){
if( searchSortedMatrix(arr ,key , minX,maxX , midY+1 , maxY))
return true ;
// search bottom half of matrix
if( searchSortedMatrix(arr ,key , midX+1,maxX , minY , maxY))
return true ;
}
// arr[midX][midY] > key ? search left quad matrix ;
else {
return(searchSortedMatrix(arr , key , minX,midX-1,minY,midY-1));
}
return false ;
}
I suggest, store all characters in a 2D list. then find index of required element if it exists in list.
If not present print appropriate message else print row and column as:
row = (index/total_columns) and column = (index%total_columns -1)
This will incur only the binary search time in a list.
Please suggest any corrections. :)
If O(M log(N)) solution is ok for an MxN array -
template <size_t n>
struct MN * get(int a[][n], int k, int M, int N){
struct MN *result = new MN;
result->m = -1;
result->n = -1;
/* Do a binary search on each row since rows (and columns too) are sorted. */
for(int i = 0; i < M; i++){
int lo = 0; int hi = N - 1;
while(lo <= hi){
int mid = lo + (hi-lo)/2;
if(k < a[i][mid]) hi = mid - 1;
else if (k > a[i][mid]) lo = mid + 1;
else{
result->m = i;
result->n = mid;
return result;
}
}
}
return result;
}
Working C++ demo.
Please do let me know if this wouldn't work or if there is a bug it it.
class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
if(matrix == null)
return false;
int i=0;
int j=0;
int m = matrix.length;
int n = matrix[0].length;
boolean found = false;
while(i<m && !found){
while(j<n && !found){
if(matrix[i][j] == target)
found = true;
if(matrix[i][j] < target)
j++;
else
break;
}
i++;
j=0;
}
return found;
}}
129 / 129 test cases passed.
Status: Accepted
Runtime: 39 ms
Memory Usage: 55 MB
Given a square matrix as follows:
[ a b c ]
[ d e f ]
[ i j k ]
We know that a < c, d < f, i < k. What we don't know is whether d < c or d > c, etc. We have guarantees only in 1-dimension.
Looking at the end elements (c,f,k), we can do a sort of filter: is N < c ? search() : next(). Thus, we have n iterations over the rows, with each row taking either O( log( n ) ) for binary search or O( 1 ) if filtered out.
Let me given an EXAMPLE where N = j,
1) Check row 1. j < c? (no, go next)
2) Check row 2. j < f? (yes, bin search gets nothing)
3) Check row 3. j < k? (yes, bin search finds it)
Try again with N = q,
1) Check row 1. q < c? (no, go next)
2) Check row 2. q < f? (no, go next)
3) Check row 3. q < k? (no, go next)
There is probably a better solution out there but this is easy to explain.. :)
As this is an interview question, it would seem to lead towards a discussion of Parallel programming and Map-reduce algorithms.
See http://code.google.com/intl/de/edu/parallel/mapreduce-tutorial.html
I need an alternate sequence like 1, -1, 1, -1... asf. First I used if-statements for it, but it's dumb. Then I tried something like:
int n = 1;
...
do{
n = 0 + ( n * (-1));
} while(blabla)
It's ok, but I have to store n value from iteration to iteration. This isn't so pretty. How to compute that sequence from a control variable like frameCount?
Sorry, I am learning not only to code, but English too.
It's not very readable, but if you're looking for something purely "elegant," I suppose you could do something like:
int n = 1 - ((frameCount % 2) * 2);
If you're on an even frame you'll be subtracting (1 - 0), if you're on odd frame you'll be subtracting (1 - 2).
For readability, I recommend:
int n = (frameCount & 1) == 1 ? 1 : -1;
or
int n = -1;
if ((frameCount & 1) == 1) {
n = 1;
}
(Note that x & 1 extracts the lowest bit from x just like x % 2 does.)
Why don't you just do this
float n = 1 ;
void setup() {
....
}
void draw() {
....
n =-n;
....
}
I recently saw another fun way to create such a sequence using XOR magic.
int n = -1, t = 1 ^ -1;
...
do{
n ^= t;
}while(blabla);
Found in Algorithms for programmers, page 5.
One possibility is
((framecount & 1)<<1) -1;
On even frames, framecount & 1 will yield 0, and the result will then be -1. On odd frames, framecount & 1 will yield 1, and the result will be 1.
Or, less immediate but possibly much faster on some architectures,
frameCount-(frameCount^1)
When frameCount is odd, frameCount^1 will be equal to frameCount-1, and the result will then be 1. If frameCount is even, frameCount^1 will be equal to frameCount+1, giving -1.
I need help solving problem N from this earlier competition:
Problem N: Digit Sums
Given 3 positive integers A, B and C,
find how many positive integers less
than or equal to A, when expressed in
base B, have digits which sum to C.
Input will consist of a series of
lines, each containing three integers,
A, B and C, 2 ≤ B ≤ 100, 1 ≤ A, C ≤
1,000,000,000. The numbers A, B and C
are given in base 10 and are separated
by one or more blanks. The input is
terminated by a line containing three
zeros.
Output will be the number of numbers,
for each input line (it must be given
in base 10).
Sample input
100 10 9
100 10 1
750000 2 2
1000000000 10 40
100000000 100 200
0 0 0
Sample output
10
3
189
45433800
666303
The relevant rules:
Read all input from the keyboard, i.e. use stdin, System.in, cin or equivalent. Input will be redirected from a file to form the input to your submission.
Write all output to the screen, i.e. use stdout, System.out, cout or equivalent. Do not write to stderr. Do NOT use, or even include, any module that allows direct manipulation of the screen, such as conio, Crt or anything similar. Output from your program is redirected to a file for later checking. Use of direct I/O means that such output is not redirected and hence cannot be checked. This could mean that a correct program is rejected!
Unless otherwise stated, all integers in the input will fit into a standard 32-bit computer word. Adjacent integers on a line will be separated by one or more spaces.
Of course, it's fair to say that I should learn more before trying to solve this, but i'd really appreciate it if someone here told me how it's done.
Thanks in advance, John.
Other people pointed out trivial solution: iterate over all numbers from 1 to A. But this problem, actually, can be solved in nearly constant time: O(length of A), which is O(log(A)).
Code provided is for base 10. Adapting it for arbitrary base is trivial.
To reach above estimate for time, you need to add memorization to recursion. Let me know if you have questions about that part.
Now, recursive function itself. Written in Java, but everything should work in C#/C++ without any changes. It's big, but mostly because of comments where I try to clarify algorithm.
// returns amount of numbers strictly less than 'num' with sum of digits 'sum'
// pay attention to word 'strictly'
int count(int num, int sum) {
// no numbers with negative sum of digits
if (sum < 0) {
return 0;
}
int result = 0;
// imagine, 'num' == 1234
// let's check numbers 1233, 1232, 1231, 1230 manually
while (num % 10 > 0) {
--num;
// check if current number is good
if (sumOfDigits(num) == sum) {
// one more result
++result;
}
}
if (num == 0) {
// zero reached, no more numbers to check
return result;
}
num /= 10;
// Using example above (1234), now we're left with numbers
// strictly less than 1230 to check (1..1229)
// It means, any number less than 123 with arbitrary digit appended to the right
// E.g., if this digit in the right (last digit) is 3,
// then sum of the other digits must be "sum - 3"
// and we need to add to result 'count(123, sum - 3)'
// let's iterate over all possible values of last digit
for (int digit = 0; digit < 10; ++digit) {
result += count(num, sum - digit);
}
return result;
}
Helper function
// returns sum of digits, plain and simple
int sumOfDigits(int x) {
int result = 0;
while (x > 0) {
result += x % 10;
x /= 10;
}
return result;
}
Now, let's write a little tester
int A = 12345;
int C = 13;
// recursive solution
System.out.println(count(A + 1, C));
// brute-force solution
int total = 0;
for (int i = 1; i <= A; ++i) {
if (sumOfDigits(i) == C) {
++total;
}
}
System.out.println(total);
You can write more comprehensive tester checking all values of A, but overall solution seems to be correct. (I tried several random A's and C's.)
Don't forget, you can't test solution for A == 1000000000 without memorization: it'll run too long. But with memorization, you can test it even for A == 10^1000.
edit
Just to prove a concept, poor man's memorization. (in Java, in other languages hashtables are declared differently) But if you want to learn something, it might be better to try to do it yourself.
// hold values here
private Map<String, Integer> mem;
int count(int num, int sum) {
// no numbers with negative sum of digits
if (sum < 0) {
return 0;
}
String key = num + " " + sum;
if (mem.containsKey(key)) {
return mem.get(key);
}
// ...
// continue as above...
// ...
mem.put(key, result);
return result;
}
Here's the same memoized recursive solution that Rybak posted, but with a simpler implementation, in my humble opinion:
HashMap<String, Integer> cache = new HashMap<String, Integer>();
int count(int bound, int base, int sum) {
// No negative digit sums.
if (sum < 0)
return 0;
// Handle one digit case.
if (bound < base)
return (sum <= bound) ? 1 : 0;
String key = bound + " " + sum;
if (cache.containsKey(key))
return cache.get(key);
int count = 0;
for (int digit = 0; digit < base; digit++)
count += count((bound - digit) / base, base, sum - digit);
cache.put(key, count);
return count;
}
This is not the complete solution (no input parsing). To get the number in base B, repeatedly take the modulo B, and then divide by B until the result is 0. This effectively computes the base-B digit from the right, and then shifts the number right.
int A,B,C; // from input
for (int x=1; x<A; x++)
{
int sumDigits = 0;
int v = x;
while (v!=0) {
sumDigits += (v % B);
v /= B;
}
if (sumDigits==C)
cout << x;
}
This is a brute force approach. It may be possible to compute this quicker by determining which sets of base B digits add up to C, arranging these in all permutations that are less than A, and then working backwards from that to create the original number.
Yum.
Try this:
int number, digitSum, resultCounter = 0;
for(int i=1; i<=A, i++)
{
number = i; //to avoid screwing up our counter
digitSum = 0;
while(number > 1)
{
//this is the next "digit" of the number as it would be in base B;
//works with any base including 10.
digitSum += (number % B);
//remove this digit from the number, square the base, rinse, repeat
number /= B;
}
digitSum += number;
//Does the sum match?
if(digitSum == C)
resultCounter++;
}
That's your basic algorithm for one line. Now you wrap this in another For loop for each input line you received, preceded by the input collection phase itself. This process can be simplified, but I don't feel like coding your entire answer to see if my algorithm works, and this looks right whereas the simpler tricks are harder to pass by inspection.
The way this works is by modulo dividing by powers of the base. Simple example, 1234 in base 10:
1234 % 10 = 4
1234 / 10 = 123 //integer division truncates any fraction
123 % 10 = 3 //sum is 7
123 / 10 = 12
12 % 10 = 2 //sum is 9
12 / 10 = 1 //end condition, add this and the sum is 10
A harder example to figure out by inspection would be the same number in base 12:
1234 % 12 = 10 //you can call it "A" like in hex, but we need a sum anyway
1234 / 12 = 102
102 % 12 = 6 // sum 16
102/12 = 8
8 % 12 = 8 //sum 24
8 / 12 = 0 //end condition, sum still 24.
So 1234 in base 12 would be written 86A. Check the math:
8*12^2 + 6*12 + 10 = 1152 + 72 + 10 = 1234
Have fun wrapping the rest of the code around this.
I was recently given this interview question and I'm curious what a good solution to it would be.
Say I'm given a 2d array where all the
numbers in the array are in increasing
order from left to right and top to
bottom.
What is the best way to search and
determine if a target number is in the
array?
Now, my first inclination is to utilize a binary search since my data is sorted. I can determine if a number is in a single row in O(log N) time. However, it is the 2 directions that throw me off.
Another solution I thought may work is to start somewhere in the middle. If the middle value is less than my target, then I can be sure it is in the left square portion of the matrix from the middle. I then move diagonally and check again, reducing the size of the square that the target could potentially be in until I have honed in on the target number.
Does anyone have any good ideas on solving this problem?
Example array:
Sorted left to right, top to bottom.
1 2 4 5 6
2 3 5 7 8
4 6 8 9 10
5 8 9 10 11
Here's a simple approach:
Start at the bottom-left corner.
If the target is less than that value, it must be above us, so move up one.
Otherwise we know that the target can't be in that column, so move right one.
Goto 2.
For an NxM array, this runs in O(N+M). I think it would be difficult to do better. :)
Edit: Lots of good discussion. I was talking about the general case above; clearly, if N or M are small, you could use a binary search approach to do this in something approaching logarithmic time.
Here are some details, for those who are curious:
History
This simple algorithm is called a Saddleback Search. It's been around for a while, and it is optimal when N == M. Some references:
David Gries, The Science of Programming. Springer-Verlag, 1989.
Edsgar Dijkstra, The Saddleback Search. Note EWD-934, 1985.
However, when N < M, intuition suggests that binary search should be able to do better than O(N+M): For example, when N == 1, a pure binary search will run in logarithmic rather than linear time.
Worst-case bound
Richard Bird examined this intuition that binary search could improve the Saddleback algorithm in a 2006 paper:
Richard S. Bird, Improving Saddleback Search: A Lesson in Algorithm Design, in Mathematics of Program Construction, pp. 82--89, volume 4014, 2006.
Using a rather unusual conversational technique, Bird shows us that for N <= M, this problem has a lower bound of Ω(N * log(M/N)). This bound make sense, as it gives us linear performance when N == M and logarithmic performance when N == 1.
Algorithms for rectangular arrays
One approach that uses a row-by-row binary search looks like this:
Start with a rectangular array where N < M. Let's say N is rows and M is columns.
Do a binary search on the middle row for value. If we find it, we're done.
Otherwise we've found an adjacent pair of numbers s and g, where s < value < g.
The rectangle of numbers above and to the left of s is less than value, so we can eliminate it.
The rectangle below and to the right of g is greater than value, so we can eliminate it.
Go to step (2) for each of the two remaining rectangles.
In terms of worst-case complexity, this algorithm does log(M) work to eliminate half the possible solutions, and then recursively calls itself twice on two smaller problems. We do have to repeat a smaller version of that log(M) work for every row, but if the number of rows is small compared to the number of columns, then being able to eliminate all of those columns in logarithmic time starts to become worthwhile.
This gives the algorithm a complexity of T(N,M) = log(M) + 2 * T(M/2, N/2), which Bird shows to be O(N * log(M/N)).
Another approach posted by Craig Gidney describes an algorithm similar the approach above: it examines a row at a time using a step size of M/N. His analysis shows that this results in O(N * log(M/N)) performance as well.
Performance Comparison
Big-O analysis is all well and good, but how well do these approaches work in practice? The chart below examines four algorithms for increasingly "square" arrays:
(The "naive" algorithm simply searches every element of the array. The "recursive" algorithm is described above. The "hybrid" algorithm is an implementation of Gidney's algorithm. For each array size, performance was measured by timing each algorithm over fixed set of 1,000,000 randomly-generated arrays.)
Some notable points:
As expected, the "binary search" algorithms offer the best performance on rectangular arrays and the Saddleback algorithm works the best on square arrays.
The Saddleback algorithm performs worse than the "naive" algorithm for 1-d arrays, presumably because it does multiple comparisons on each item.
The performance hit that the "binary search" algorithms take on square arrays is presumably due to the overhead of running repeated binary searches.
Summary
Clever use of binary search can provide O(N * log(M/N) performance for both rectangular and square arrays. The O(N + M) "saddleback" algorithm is much simpler, but suffers from performance degradation as arrays become increasingly rectangular.
This problem takes Θ(b lg(t)) time, where b = min(w,h) and t=b/max(w,h). I discuss the solution in this blog post.
Lower bound
An adversary can force an algorithm to make Ω(b lg(t)) queries, by restricting itself to the main diagonal:
Legend: white cells are smaller items, gray cells are larger items, yellow cells are smaller-or-equal items and orange cells are larger-or-equal items. The adversary forces the solution to be whichever yellow or orange cell the algorithm queries last.
Notice that there are b independent sorted lists of size t, requiring Ω(b lg(t)) queries to completely eliminate.
Algorithm
(Assume without loss of generality that w >= h)
Compare the target item against the cell t to the left of the top right corner of the valid area
If the cell's item matches, return the current position.
If the cell's item is less than the target item, eliminate the remaining t cells in the row with a binary search. If a matching item is found while doing this, return with its position.
Otherwise the cell's item is more than the target item, eliminating t short columns.
If there's no valid area left, return failure
Goto step 2
Finding an item:
Determining an item doesn't exist:
Legend: white cells are smaller items, gray cells are larger items, and the green cell is an equal item.
Analysis
There are b*t short columns to eliminate. There are b long rows to eliminate. Eliminating a long row costs O(lg(t)) time. Eliminating t short columns costs O(1) time.
In the worst case we'll have to eliminate every column and every row, taking time O(lg(t)*b + b*t*1/t) = O(b lg(t)).
Note that I'm assuming lg clamps to a result above 1 (i.e. lg(x) = log_2(max(2,x))). That's why when w=h, meaning t=1, we get the expected bound of O(b lg(1)) = O(b) = O(w+h).
Code
public static Tuple<int, int> TryFindItemInSortedMatrix<T>(this IReadOnlyList<IReadOnlyList<T>> grid, T item, IComparer<T> comparer = null) {
if (grid == null) throw new ArgumentNullException("grid");
comparer = comparer ?? Comparer<T>.Default;
// check size
var width = grid.Count;
if (width == 0) return null;
var height = grid[0].Count;
if (height < width) {
var result = grid.LazyTranspose().TryFindItemInSortedMatrix(item, comparer);
if (result == null) return null;
return Tuple.Create(result.Item2, result.Item1);
}
// search
var minCol = 0;
var maxRow = height - 1;
var t = height / width;
while (minCol < width && maxRow >= 0) {
// query the item in the minimum column, t above the maximum row
var luckyRow = Math.Max(maxRow - t, 0);
var cmpItemVsLucky = comparer.Compare(item, grid[minCol][luckyRow]);
if (cmpItemVsLucky == 0) return Tuple.Create(minCol, luckyRow);
// did we eliminate t rows from the bottom?
if (cmpItemVsLucky < 0) {
maxRow = luckyRow - 1;
continue;
}
// we eliminated most of the current minimum column
// spend lg(t) time eliminating rest of column
var minRowInCol = luckyRow + 1;
var maxRowInCol = maxRow;
while (minRowInCol <= maxRowInCol) {
var mid = minRowInCol + (maxRowInCol - minRowInCol + 1) / 2;
var cmpItemVsMid = comparer.Compare(item, grid[minCol][mid]);
if (cmpItemVsMid == 0) return Tuple.Create(minCol, mid);
if (cmpItemVsMid > 0) {
minRowInCol = mid + 1;
} else {
maxRowInCol = mid - 1;
maxRow = mid - 1;
}
}
minCol += 1;
}
return null;
}
I would use the divide-and-conquer strategy for this problem, similar to what you suggested, but the details are a bit different.
This will be a recursive search on subranges of the matrix.
At each step, pick an element in the middle of the range. If the value found is what you are seeking, then you're done.
Otherwise, if the value found is less than the value that you are seeking, then you know that it is not in the quadrant above and to the left of your current position. So recursively search the two subranges: everything (exclusively) below the current position, and everything (exclusively) to the right that is at or above the current position.
Otherwise, (the value found is greater than the value that you are seeking) you know that it is not in the quadrant below and to the right of your current position. So recursively search the two subranges: everything (exclusively) to the left of the current position, and everything (exclusively) above the current position that is on the current column or a column to the right.
And ba-da-bing, you found it.
Note that each recursive call only deals with the current subrange only, not (for example) ALL rows above the current position. Just those in the current subrange.
Here's some pseudocode for you:
bool numberSearch(int[][] arr, int value, int minX, int maxX, int minY, int maxY)
if (minX == maxX and minY == maxY and arr[minX,minY] != value)
return false
if (arr[minX,minY] > value) return false; // Early exits if the value can't be in
if (arr[maxX,maxY] < value) return false; // this subrange at all.
int nextX = (minX + maxX) / 2
int nextY = (minY + maxY) / 2
if (arr[nextX,nextY] == value)
{
print nextX,nextY
return true
}
else if (arr[nextX,nextY] < value)
{
if (numberSearch(arr, value, minX, maxX, nextY + 1, maxY))
return true
return numberSearch(arr, value, nextX + 1, maxX, minY, nextY)
}
else
{
if (numberSearch(arr, value, minX, nextX - 1, minY, maxY))
return true
reutrn numberSearch(arr, value, nextX, maxX, minY, nextY)
}
The two main answers give so far seem to be the arguably O(log N) "ZigZag method" and the O(N+M) Binary Search method. I thought I'd do some testing comparing the two methods with some various setups. Here are the details:
The array is N x N square in every test, with N varying from 125 to 8000 (the largest my JVM heap could handle). For each array size, I picked a random place in the array to put a single 2. I then put a 3 everywhere possible (to the right and below of the 2) and then filled the rest of the array with 1. Some of the earlier commenters seemed to think this type of setup would yield worst case run time for both algorithms. For each array size, I picked 100 different random locations for the 2 (search target) and ran the test. I recorded avg run time and worst case run time for each algorithm. Because it was happening too fast to get good ms readings in Java, and because I don't trust Java's nanoTime(), I repeated each test 1000 times just to add a uniform bias factor to all the times. Here are the results:
ZigZag beat binary in every test for both avg and worst case times, however, they are all within an order of magnitude of each other more or less.
Here is the Java code:
public class SearchSortedArray2D {
static boolean findZigZag(int[][] a, int t) {
int i = 0;
int j = a.length - 1;
while (i <= a.length - 1 && j >= 0) {
if (a[i][j] == t) return true;
else if (a[i][j] < t) i++;
else j--;
}
return false;
}
static boolean findBinarySearch(int[][] a, int t) {
return findBinarySearch(a, t, 0, 0, a.length - 1, a.length - 1);
}
static boolean findBinarySearch(int[][] a, int t,
int r1, int c1, int r2, int c2) {
if (r1 > r2 || c1 > c2) return false;
if (r1 == r2 && c1 == c2 && a[r1][c1] != t) return false;
if (a[r1][c1] > t) return false;
if (a[r2][c2] < t) return false;
int rm = (r1 + r2) / 2;
int cm = (c1 + c2) / 2;
if (a[rm][cm] == t) return true;
else if (a[rm][cm] > t) {
boolean b1 = findBinarySearch(a, t, r1, c1, r2, cm - 1);
boolean b2 = findBinarySearch(a, t, r1, cm, rm - 1, c2);
return (b1 || b2);
} else {
boolean b1 = findBinarySearch(a, t, r1, cm + 1, rm, c2);
boolean b2 = findBinarySearch(a, t, rm + 1, c1, r2, c2);
return (b1 || b2);
}
}
static void randomizeArray(int[][] a, int N) {
int ri = (int) (Math.random() * N);
int rj = (int) (Math.random() * N);
a[ri][rj] = 2;
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
if (i == ri && j == rj) continue;
else if (i > ri || j > rj) a[i][j] = 3;
else a[i][j] = 1;
}
}
}
public static void main(String[] args) {
int N = 8000;
int[][] a = new int[N][N];
int randoms = 100;
int repeats = 1000;
long start, end, duration;
long zigMin = Integer.MAX_VALUE, zigMax = Integer.MIN_VALUE;
long binMin = Integer.MAX_VALUE, binMax = Integer.MIN_VALUE;
long zigSum = 0, zigAvg;
long binSum = 0, binAvg;
for (int k = 0; k < randoms; k++) {
randomizeArray(a, N);
start = System.currentTimeMillis();
for (int i = 0; i < repeats; i++) findZigZag(a, 2);
end = System.currentTimeMillis();
duration = end - start;
zigSum += duration;
zigMin = Math.min(zigMin, duration);
zigMax = Math.max(zigMax, duration);
start = System.currentTimeMillis();
for (int i = 0; i < repeats; i++) findBinarySearch(a, 2);
end = System.currentTimeMillis();
duration = end - start;
binSum += duration;
binMin = Math.min(binMin, duration);
binMax = Math.max(binMax, duration);
}
zigAvg = zigSum / randoms;
binAvg = binSum / randoms;
System.out.println(findZigZag(a, 2) ?
"Found via zigzag method. " : "ERROR. ");
//System.out.println("min search time: " + zigMin + "ms");
System.out.println("max search time: " + zigMax + "ms");
System.out.println("avg search time: " + zigAvg + "ms");
System.out.println();
System.out.println(findBinarySearch(a, 2) ?
"Found via binary search method. " : "ERROR. ");
//System.out.println("min search time: " + binMin + "ms");
System.out.println("max search time: " + binMax + "ms");
System.out.println("avg search time: " + binAvg + "ms");
}
}
This is a short proof of the lower bound on the problem.
You cannot do it better than linear time (in terms of array dimensions, not the number of elements). In the array below, each of the elements marked as * can be either 5 or 6 (independently of other ones). So if your target value is 6 (or 5) the algorithm needs to examine all of them.
1 2 3 4 *
2 3 4 * 7
3 4 * 7 8
4 * 7 8 9
* 7 8 9 10
Of course this expands to bigger arrays as well. This means that this answer is optimal.
Update: As pointed out by Jeffrey L Whitledge, it is only optimal as the asymptotic lower bound on running time vs input data size (treated as a single variable). Running time treated as two-variable function on both array dimensions can be improved.
I think Here is the answer and it works for any kind of sorted matrix
bool findNum(int arr[][ARR_MAX],int xmin, int xmax, int ymin,int ymax,int key)
{
if (xmin > xmax || ymin > ymax || xmax < xmin || ymax < ymin) return false;
if ((xmin == xmax) && (ymin == ymax) && (arr[xmin][ymin] != key)) return false;
if (arr[xmin][ymin] > key || arr[xmax][ymax] < key) return false;
if (arr[xmin][ymin] == key || arr[xmax][ymax] == key) return true;
int xnew = (xmin + xmax)/2;
int ynew = (ymin + ymax)/2;
if (arr[xnew][ynew] == key) return true;
if (arr[xnew][ynew] < key)
{
if (findNum(arr,xnew+1,xmax,ymin,ymax,key))
return true;
return (findNum(arr,xmin,xmax,ynew+1,ymax,key));
} else {
if (findNum(arr,xmin,xnew-1,ymin,ymax,key))
return true;
return (findNum(arr,xmin,xmax,ymin,ynew-1,key));
}
}
Interesting question. Consider this idea - create one boundary where all the numbers are greater than your target and another where all the numbers are less than your target. If anything is left in between the two, that's your target.
If I'm looking for 3 in your example, I read across the first row until I hit 4, then look for the smallest adjacent number (including diagonals) greater than 3:
1 2 4 5 6
2 3 5 7 8
4 6 8 9 10
5 8 9 10 11
Now I do the same for those numbers less than 3:
1 2 4 5 6
2 3 5 7 8
4 6 8 9 10
5 8 9 10 11
Now I ask, is anything inside the two boundaries? If yes, it must be 3. If no, then there is no 3. Sort of indirect since I don't actually find the number, I just deduce that it must be there. This has the added bonus of counting ALL the 3's.
I tried this on some examples and it seems to work OK.
Binary search through the diagonal of the array is the best option.
We can find out whether the element is less than or equal to the elements in the diagonal.
I've been asking this question in interviews for the better part of a decade and I think there's only been one person who has been able to come up with an optimal algorithm.
My solution has always been:
Binary search the middle diagonal, which is the diagonal running down and right, containing the item at (rows.count/2, columns.count/2).
If the target number is found, return true.
Otherwise, two numbers (u and v) will have been found such that u is smaller than the target, v is larger than the target, and v is one right and one down from u.
Recursively search the sub-matrix to the right of u and top of v and the one to the bottom of u and left of v.
I believe this is a strict improvement over the algorithm given by Nate here, since searching the diagonal often allows a reduction of over half the search space (if the matrix is close to square), whereas searching a row or column always results in an elimination of exactly half.
Here's the code in (probably not terribly Swifty) Swift:
import Cocoa
class Solution {
func searchMatrix(_ matrix: [[Int]], _ target: Int) -> Bool {
if (matrix.isEmpty || matrix[0].isEmpty) {
return false
}
return _searchMatrix(matrix, 0..<matrix.count, 0..<matrix[0].count, target)
}
func _searchMatrix(_ matrix: [[Int]], _ rows: Range<Int>, _ columns: Range<Int>, _ target: Int) -> Bool {
if (rows.count == 0 || columns.count == 0) {
return false
}
if (rows.count == 1) {
return _binarySearch(matrix, rows.lowerBound, columns, target, true)
}
if (columns.count == 1) {
return _binarySearch(matrix, columns.lowerBound, rows, target, false)
}
var lowerInflection = (-1, -1)
var upperInflection = (Int.max, Int.max)
var currentRows = rows
var currentColumns = columns
while (currentRows.count > 0 && currentColumns.count > 0 && upperInflection.0 > lowerInflection.0+1) {
let rowMidpoint = (currentRows.upperBound + currentRows.lowerBound) / 2
let columnMidpoint = (currentColumns.upperBound + currentColumns.lowerBound) / 2
let value = matrix[rowMidpoint][columnMidpoint]
if (value == target) {
return true
}
if (value > target) {
upperInflection = (rowMidpoint, columnMidpoint)
currentRows = currentRows.lowerBound..<rowMidpoint
currentColumns = currentColumns.lowerBound..<columnMidpoint
} else {
lowerInflection = (rowMidpoint, columnMidpoint)
currentRows = rowMidpoint+1..<currentRows.upperBound
currentColumns = columnMidpoint+1..<currentColumns.upperBound
}
}
if (lowerInflection.0 == -1) {
lowerInflection = (upperInflection.0-1, upperInflection.1-1)
} else if (upperInflection.0 == Int.max) {
upperInflection = (lowerInflection.0+1, lowerInflection.1+1)
}
return _searchMatrix(matrix, rows.lowerBound..<lowerInflection.0+1, upperInflection.1..<columns.upperBound, target) || _searchMatrix(matrix, upperInflection.0..<rows.upperBound, columns.lowerBound..<lowerInflection.1+1, target)
}
func _binarySearch(_ matrix: [[Int]], _ rowOrColumn: Int, _ range: Range<Int>, _ target: Int, _ searchRow : Bool) -> Bool {
if (range.isEmpty) {
return false
}
let midpoint = (range.upperBound + range.lowerBound) / 2
let value = (searchRow ? matrix[rowOrColumn][midpoint] : matrix[midpoint][rowOrColumn])
if (value == target) {
return true
}
if (value > target) {
return _binarySearch(matrix, rowOrColumn, range.lowerBound..<midpoint, target, searchRow)
} else {
return _binarySearch(matrix, rowOrColumn, midpoint+1..<range.upperBound, target, searchRow)
}
}
}
A. Do a binary search on those lines where the target number might be on.
B. Make it a graph : Look for the number by taking always the smallest unvisited neighbour node and backtracking when a too big number is found
Binary search would be the best approach, imo. Starting at 1/2 x, 1/2 y will cut it in half. IE a 5x5 square would be something like x == 2 / y == 3 . I rounded one value down and one value up to better zone in on the direction of the targeted value.
For clarity the next iteration would give you something like x == 1 / y == 2 OR x == 3 / y == 5
Well, to begin with, let us assume we are using a square.
1 2 3
2 3 4
3 4 5
1. Searching a square
I would use a binary search on the diagonal. The goal is the locate the smaller number that is not strictly lower than the target number.
Say I am looking for 4 for example, then I would end up locating 5 at (2,2).
Then, I am assured that if 4 is in the table, it is at a position either (x,2) or (2,x) with x in [0,2]. Well, that's just 2 binary searches.
The complexity is not daunting: O(log(N)) (3 binary searches on ranges of length N)
2. Searching a rectangle, naive approach
Of course, it gets a bit more complicated when N and M differ (with a rectangle), consider this degenerate case:
1 2 3 4 5 6 7 8
2 3 4 5 6 7 8 9
10 11 12 13 14 15 16 17
And let's say I am looking for 9... The diagonal approach is still good, but the definition of diagonal changes. Here my diagonal is [1, (5 or 6), 17]. Let's say I picked up [1,5,17], then I know that if 9 is in the table it is either in the subpart:
5 6 7 8
6 7 8 9
10 11 12 13 14 15 16
This gives us 2 rectangles:
5 6 7 8 10 11 12 13 14 15 16
6 7 8 9
So we can recurse! probably beginning by the one with less elements (though in this case it kills us).
I should point that if one of the dimensions is less than 3, we cannot apply the diagonal methods and must use a binary search. Here it would mean:
Apply binary search on 10 11 12 13 14 15 16, not found
Apply binary search on 5 6 7 8, not found
Apply binary search on 6 7 8 9, not found
It's tricky because to get good performance you might want to differentiate between several cases, depending on the general shape....
3. Searching a rectangle, brutal approach
It would be much easier if we dealt with a square... so let's just square things up.
1 2 3 4 5 6 7 8
2 3 4 5 6 7 8 9
10 11 12 13 14 15 16 17
17 . . . . . . 17
. .
. .
. .
17 . . . . . . 17
We now have a square.
Of course, we will probably NOT actually create those rows, we could simply emulate them.
def get(x,y):
if x < N and y < M: return table[x][y]
else: return table[N-1][M-1] # the max
so it behaves like a square without occupying more memory (at the cost of speed, probably, depending on cache... oh well :p)
EDIT:
I misunderstood the question. As the comments point out this only works in the more restricted case.
In a language like C that stores data in row-major order, simply treat it as a 1D array of size n * m and use a binary search.
I have a recursive Divide & Conquer Solution.
Basic Idea for one step is: We know that the Left-Upper(LU) is smallest and the right-bottom(RB) is the largest no., so the given No(N) must: N>=LU and N<=RB
IF N==LU and N==RB::::Element Found and Abort returning the position/Index
If N>=LU and N<=RB = FALSE, No is not there and abort.
If N>=LU and N<=RB = TRUE, Divide the 2D array in 4 equal parts of 2D array each in logical manner..
And then apply the same algo step to all four sub-array.
My Algo is Correct I have implemented on my friends PC.
Complexity: each 4 comparisons can b used to deduce the total no of elements to one-fourth at its worst case..
So My complexity comes to be 1 + 4 x lg(n) + 4
But really expected this to be working on O(n)
I think something is wrong somewhere in my calculation of Complexity, please correct if so..
The optimal solution is to start at the top-left corner, that has minimal value. Move diagonally downwards to the right until you hit an element whose value >= value of the given element. If the element's value is equal to that of the given element, return found as true.
Otherwise, from here we can proceed in two ways.
Strategy 1:
Move up in the column and search for the given element until we reach the end. If found, return found as true
Move left in the row and search for the given element until we reach the end. If found, return found as true
return found as false
Strategy 2:
Let i denote the row index and j denote the column index of the diagonal element we have stopped at. (Here, we have i = j, BTW). Let k = 1.
Repeat the below steps until i-k >= 0
Search if a[i-k][j] is equal to the given element. if yes, return found as true.
Search if a[i][j-k] is equal to the given element. if yes, return found as true.
Increment k
1 2 4 5 6
2 3 5 7 8
4 6 8 9 10
5 8 9 10 11
public boolean searchSortedMatrix(int arr[][] , int key , int minX , int maxX , int minY , int maxY){
// base case for recursion
if(minX > maxX || minY > maxY)
return false ;
// early fails
// array not properly intialized
if(arr==null || arr.length==0)
return false ;
// arr[0][0]> key return false
if(arr[minX][minY]>key)
return false ;
// arr[maxX][maxY]<key return false
if(arr[maxX][maxY]<key)
return false ;
//int temp1 = minX ;
//int temp2 = minY ;
int midX = (minX+maxX)/2 ;
//if(temp1==midX){midX+=1 ;}
int midY = (minY+maxY)/2 ;
//if(temp2==midY){midY+=1 ;}
// arr[midX][midY] = key ? then value found
if(arr[midX][midY] == key)
return true ;
// alas ! i have to keep looking
// arr[midX][midY] < key ? search right quad and bottom matrix ;
if(arr[midX][midY] < key){
if( searchSortedMatrix(arr ,key , minX,maxX , midY+1 , maxY))
return true ;
// search bottom half of matrix
if( searchSortedMatrix(arr ,key , midX+1,maxX , minY , maxY))
return true ;
}
// arr[midX][midY] > key ? search left quad matrix ;
else {
return(searchSortedMatrix(arr , key , minX,midX-1,minY,midY-1));
}
return false ;
}
I suggest, store all characters in a 2D list. then find index of required element if it exists in list.
If not present print appropriate message else print row and column as:
row = (index/total_columns) and column = (index%total_columns -1)
This will incur only the binary search time in a list.
Please suggest any corrections. :)
If O(M log(N)) solution is ok for an MxN array -
template <size_t n>
struct MN * get(int a[][n], int k, int M, int N){
struct MN *result = new MN;
result->m = -1;
result->n = -1;
/* Do a binary search on each row since rows (and columns too) are sorted. */
for(int i = 0; i < M; i++){
int lo = 0; int hi = N - 1;
while(lo <= hi){
int mid = lo + (hi-lo)/2;
if(k < a[i][mid]) hi = mid - 1;
else if (k > a[i][mid]) lo = mid + 1;
else{
result->m = i;
result->n = mid;
return result;
}
}
}
return result;
}
Working C++ demo.
Please do let me know if this wouldn't work or if there is a bug it it.
class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
if(matrix == null)
return false;
int i=0;
int j=0;
int m = matrix.length;
int n = matrix[0].length;
boolean found = false;
while(i<m && !found){
while(j<n && !found){
if(matrix[i][j] == target)
found = true;
if(matrix[i][j] < target)
j++;
else
break;
}
i++;
j=0;
}
return found;
}}
129 / 129 test cases passed.
Status: Accepted
Runtime: 39 ms
Memory Usage: 55 MB
Given a square matrix as follows:
[ a b c ]
[ d e f ]
[ i j k ]
We know that a < c, d < f, i < k. What we don't know is whether d < c or d > c, etc. We have guarantees only in 1-dimension.
Looking at the end elements (c,f,k), we can do a sort of filter: is N < c ? search() : next(). Thus, we have n iterations over the rows, with each row taking either O( log( n ) ) for binary search or O( 1 ) if filtered out.
Let me given an EXAMPLE where N = j,
1) Check row 1. j < c? (no, go next)
2) Check row 2. j < f? (yes, bin search gets nothing)
3) Check row 3. j < k? (yes, bin search finds it)
Try again with N = q,
1) Check row 1. q < c? (no, go next)
2) Check row 2. q < f? (no, go next)
3) Check row 3. q < k? (no, go next)
There is probably a better solution out there but this is easy to explain.. :)
As this is an interview question, it would seem to lead towards a discussion of Parallel programming and Map-reduce algorithms.
See http://code.google.com/intl/de/edu/parallel/mapreduce-tutorial.html
Is there any nice algorithm to find the nearest prime number to a given real number? I only need to search within the first 100 primes or so.
At present, I've a bunch of prime numbers stored in an array and I'm checking the difference one number at a time (O(n)?).
Rather than a sorted list of primes, given the relatively small range targetted, have an array indexed by all the odd numbers in the range (you know there are no even primes except the special case of 2) and containing the closest prime. Finding the solution becomes O(1) time-wise.
I think the 100th prime is circa 541. an array of 270 [small] ints is all that is needed.
This approach is particularly valid, given the relative high density of primes (in particular relative to odd numbers), in the range below 1,000. (As this affects the size of a binary tree)
If you only need to search in the first 100 primes or so, just create a sorted table of those primes, and do a binary search. This will either get you to one prime number, or a spot between two, and you check which of those is closer.
Edit: Given the distribution of primes in that range, you could probably speed things up (a tiny bit) by using an interpolation search -- instead of always starting at the middle of the table, use linear interpolation to guess at a more accurate starting point. The 100th prime number should be somewhere around 250 or so (at a guess -- I haven't checked), so if (for example) you wanted the one closest to 50, you'd start about 1/5th of the way into the array instead of halfway. You can pretty much treat the primes as starting at 1, so just divide the number you want by the largest in your range to get a guess at the starting point.
Answers so far are rather complicated, given the task in hand. The first hundred primes are all less then 600. I would create an array of size 600 and place in each the value of the nearest prime to that number. Then, given a number to test, I would round it both up and down using the floor and ceil functions to get one or two candidate answers. A simple comparison with the distances to these numbers will give you a very fast answer.
The simplest approach would be to store the primes in a sorted list and modify your algorithm to do a binary search.
The standard binary search algorithm would return null for a miss, but it should be straight-forward to modify it for your purposes.
The fastest algorithm? Create a lookup table with p[100]=541 elements and return the result for floor(x), with special logic for x on [2,3]. That would be O(1).
You should sort your number in array then you can use binary search. This algorithm is O(log n) performance in worst case.
public static boolean p(int n){
for(int i=3;i*i<=n;i+=2) {
if(n%i==0)
return false;
}
return n%2==0? false: true; }
public static void main(String args[]){
String n="0";
int x = Integer.parseInt(n);
int z=x;
int a=0;
int i=1;
while(!p(x)){
a = i*(int)Math.pow(-1, i);
i++;
x+=a;
}
System.out.println( (int) Math.abs(x-z));}
this is for n>=2.
In python:
>>> def nearest_prime(n):
incr = -1
multiplier = -1
count = 1
while True:
if prime(n):
return n
else:
n = n + incr
multiplier = multiplier * -1
count = count + 1
incr = multiplier * count
>>> nearest_prime(3)
3
>>> nearest_prime(4)
3
>>> nearest_prime(5)
5
>>> nearest_prime(6)
5
>>> nearest_prime(7)
7
>>> nearest_prime(8)
7
>>> nearest_prime(9)
7
>>> nearest_prime(10)
11
<?php
$N1Diff = null;
$N2Diff = null;
$n1 = null;
$n2 = null;
$number = 16;
function isPrime($x) {
for ($i = 2; $i < $x; $i++) {
if ($x % $i == 0) {
return false;
}
}
return true;
}
for ($j = $number; ; $j--) {
if( isPrime($j) ){
$N1Diff = abs($number - $j);
$n1 = $j;
break;
}
}
for ($j = $number; ; $j++) {
if( isPrime($j) ){
$N2Diff = abs($number - $j);
$n2 = $j;
break;
}
}
if($N1Diff < $N2Diff) {
echo $n1;
} else if ($N1Diff2 < $N1Diff ){
echo $n2;
}
If you want to write an algorithm, A Wikipedia search for prime number led me to another article on the Sieve of Eratosthenes. The algorithm looks a bit simple and I'm thinking a recursive function would suit it well imo. (I could be wrong about that.)
If the array solution isn't a valid solution for you (it is the best one for your scenario), you can try the code below. After the "2 or 3" case, it will check every odd number away from the starting value until it finds a prime.
static int NearestPrime(double original)
{
int above = (int)Math.Ceiling(original);
int below = (int)Math.Floor(original);
if (above <= 2)
{
return 2;
}
if (below == 2)
{
return (original - 2 < 0.5) ? 2 : 3;
}
if (below % 2 == 0) below -= 1;
if (above % 2 == 0) above += 1;
double diffBelow = double.MaxValue, diffAbove = double.MaxValue;
for (; ; above += 2, below -= 2)
{
if (IsPrime(below))
{
diffBelow = original - below;
}
if (IsPrime(above))
{
diffAbove = above - original;
}
if (diffAbove != double.MaxValue || diffBelow != double.MaxValue)
{
break;
}
}
//edit to your liking for midpoint cases (4.0, 6.0, 9.0, etc)
return (int) (diffAbove < diffBelow ? above : below);
}
static bool IsPrime(int p) //intentionally incomplete due to checks in NearestPrime
{
for (int i = 3; i < Math.Sqrt(p); i += 2)
{
if (p % i == 0)
return false;
}
return true;
}
Lookup table whit size of 100 bytes; (unsigned chars)
Round real number and use lookup table.
Maybe we can find the left and right nearest prime numbers, and then compare to get the nearest one. (I've assumed that the next prime number shows up within next 10 occurrences)
def leftnearestprimeno(n):
n1 = n-1
while(n1 >= 0):
if isprime(n1):
return n1
else:
n1 -= 1
return -1
def rightnearestprimeno(n):
n1 = n+1
while(n1 < (n+10)):
if isprime(n1):
return n1
else:
n1 += 1
return -1
n = int(input())
a = leftnearestprimeno(n)
b = rightnearestprimeno(n)
if (n - a) < (b - n):
print("nearest: ", a)
elif (n - a) > (b - n):
print("nearest: ", b)
else:
print("nearest: ", a) #in case the difference is equal, choose min
#value
Simplest answer-
Every prime number can be represented in the form (6*x-1 and 6*X +1) (except 2 and 3).
let number is N.divide it with 6.
t=N/6;
now
a=(t-1)*6
b=(t+1)*6
and check which one is closer to N.