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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
How can I write a merge sort but divide to 3?
int merge_sort(int input[], int p, int r)
{
if ( p >= r )
return 0;
int mid = floor((p + r) / 2);
merge_sort(input, p, mid);
**merge_sort(input, mid + 1, r);**
merge(input, p, r);
}
This is probably supposed to be a 3 way merge. You may want to consider using a bottom up merge sort. For either top down or bottom up merge, most of the complexity is going to be in the merge function. As mentioned in the answer linked to by zwergmaster, it's a 3 way merge of runs. Each run needs a current and ending index or pointer. A sequence of if / else statements end up doing two compares to determine which of 3 runs has the smallest element, and then that smallest element is moved to the destination array (or vector or ...) and the next element from that run is retrieved. When the end of one of the 3 runs is reached, the code switches into a 2 way merge. When the end of the next run is reached, the code copies the rest of the remaining run. Then the next set of 3 runs are merged, repeating the process until the end of the array is reached, which could happen within any of the 3 runs, so the last merge near the end of the array may be a merge of 3 or 2 runs, or just a copy of 1 run.
It would be more efficient to have an initial function that allocates a temp array the same size as the array to be sorted, then have it call the merge sort function passing the temp array as a parameter, rather than constantly allocating and freeing small temp arrays during the merge sort process.
So using top down merge sort partial code to help explain this:
merge_sort(int *a, int n)
{
int *b = new int[n];
top_down_merge_sort(a, b, 0, n);
/* ... */
delete[] b;
}
top_down_merge_sort(int *a, int *b, int beg, int end)
{
if(end - beg < 3){
/* sort in place */
return;
}
int run0 = beg;
int run1 = beg + (end-beg)/3;
int run2 = beg + 2*(end-beg)/3;
top_down_merge_sort(a, b, run0, run1);
top_down_merge_sort(a, b, run1, run2);
top_down_merge_sort(a, b, run2, end);
merge_runs(a, b, run0, run1, run2, end);
}
So i have some ranges like these:
2 4
1 9
4 5
4 7
For this the result should be
1 -> 1
2 -> 2
3 -> 2
4 -> 4
5 -> 3
6 -> 2
7 -> 2
8 -> 1
9 -> 1
The naive approach will be to loop through all the ranges but that would be very inefficient and the worst case would take O(n * n)
What would be the efficient approach probably in O(n) or O(log(n))
Here's the solution, in O(n):
The rationale is to add a range [a, b] as a +1 in a, and a -1 after b. Then, after adding all the ranges, then compute the accumulated sums for that array and display it.
If you need to perform queries while adding the values, a better choice would be to use a Binary Indexed Tree, but your question doesn't seem to require this, so I left it out.
#include <iostream>
#define MAX 1000
using namespace std;
int T[MAX];
int main() {
int a, b;
int min_index = 0x1f1f1f1f, max_index = 0;
while(cin >> a >> b) {
T[a] += 1;
T[b+1] -= 1;
min_index = min(min_index, a);
max_index = max(max_index, b);
}
for(int i=min_index; i<=max_index; i++) {
T[i] += T[i-1];
cout << i << " -> " << T[i] << endl;
}
}
UPDATE: Based on the "provocations" (in a good sense) by גלעד ברקן, you can also do this in O(n log n):
#include <iostream>
#include <map>
#define ull unsigned long long
#define miit map<ull, int>::iterator
using namespace std;
map<ull, int> T;
int main() {
ull a, b;
while(cin >> a >> b) {
T[a] += 1;
T[b+1] -= 1;
}
ull last;
int count = 0;
for(miit it = T.begin(); it != T.end(); it++) {
if (count > 0)
for(ull i=last; i<it->first; i++)
cout << i << " " << count << endl;
count += it->second;
last = it->first;
}
}
The advantage of this solution is being able to support ranges with much larger values (as long as the output isn't so large).
The solution would be pretty simple:
generate two lists with the indices of all starting and ending indices of the ranges and sort them.
Generate a counter for the number of ranges that cover the current index. Start at the first item that is at any range and iterate over all numbers to the last element that is in any range. Now if an index is either part of the list of starting-indices, we add 1 to the counter, if it's an element of the ending-indices, we substract 1 from the counter.
Implementation:
vector<int> count(int** ranges , int rangecount , int rangemin , int rangemax)
{
vector<int> res;
set<int> open, close;
for(int** r = ranges ; r < ranges + sizeof(int*) * rangecount ; r++)
{
open.add((*r)[0]);
close.add((*r)[1]);
}
int rc = 0;
for(int i = rangemin ; i < rangemax ; i++)
{
if(open.count(i))
++rc;
res.add(rc);
if(close.count(i))
--rc;
}
return res;
}
Paul's answer still counts from "the first item that is at any range and iterate[s] over all numbers to the last element that is in any range." But what is we could aggregate overlapping counts? For example, if we have three (or say a very large number of) overlapping ranges [(2,6),[1,6],[2,8] the section (2,6) could be dependent only on the number of ranges, if we were to label the overlaps with their counts [(1),3(2,6),(7,8)]).
Using binary search (once for the start and a second time for the end of each interval), we could split the intervals and aggregate the counts in O(n * log m * l) time, where n is our number of given ranges and m is the number of resulting groups in the total range and l varies as the number of disjoint updates required for a particular overlap (the number of groups already within that range). Notice that at any time, we simply have a sorted list grouped as intervals with labeled count.
2 4
1 9
4 5
4 7
=>
(2,4)
(1),2(2,4),(5,9)
(1),2(2,3),3(4),2(5),(6,9)
(1),2(2,3),4(4),3(5),2(6,7),(8,9)
So you want the output to be an array, where the value of each element is the number of input ranges that include it?
Yeah, the obvious solution would be to increment every element in the range by 1, for each range.
I think you can get more efficient if you sort the input ranges by start (primary), end (secondary). So for 32bit start and end, start:end can be a 64bit sort key. Actually, just sorting by start is fine, we need to sort the ends differently anyway.
Then you can see how many ranges you enter for an element, and (with a pqueue of range-ends) see how many you already left.
# pseudo-code with possible bugs.
# TODO: peek or put-back the element from ranges / ends
# that made the condition false.
pqueue ends; // priority queue
int depth = 0; // how many ranges contain this element
for i in output.len {
while (r = ranges.next && r.start <= i) {
ends.push(r.end);
depth++;
}
while (ends.pop < i) {
depth--;
}
output[i] = depth;
}
assert ends.empty();
Actually, we can just sort the starts and ends separately into two separate priority queues. There's no need to build the pqueue on the fly. (Sorting an array of integers is more efficient than sorting an array of structs by one struct member, because you don't have to copy around as much data.)
This is an interview question.
Given a string such as: 123456abcdef consisting of n/2 integers followed by n/2 characters. Reorder the string to contain as 1a2b3c4d5e6f . The algortithm should be in-place.
The solution I gave was trivial - O(n^2). Just shift the characters by n/2 places to the left.
I tried using recursion as -
a. Swap later half of the first half with the previous half of the 2nd part - eg
123 456 abc def
123 abc 456 def
b. Recurse on the two halves.
The pbm I am stuck is that the swapping varies with the number of elements - for eg.
What to do next?
123 abc
12ab 3c
And what to do for : 12345 abcde
123abc 45ab
This is a pretty old question and may be a duplicate. Please let me know.. :)
Another example:
Input: 38726zfgsa
Output: 3z8f7g2s6a
Here's how I would approach the problem:
1) Divide the string into two partitions, number part and letter part
2) Divide each of those partitions into two more (equal sized)
3) Swap the second the third partition (inner number and inner letter)
4) Recurse on the original two partitions (with their newly swapped bits)
5) Stop when the partition has a size of 2
For example:
123456abcdef -> 123456 abcdef -> 123 456 abc def -> 123 abc 456 def
123abc -> 123 abc -> 12 3 ab c -> 12 ab 3 c
12 ab -> 1 2 a b -> 1 a 2 b
... etc
And the same for the other half of the recursion..
All can be done in place with the only gotcha being swapping partitions that aren't the same size (but it'll be off by one, so not difficult to handle).
It is easy to permute an array in place by chasing elements round cycles if you have a bit-map to mark which elements have been moved. We don't have a separate bit-map, but IF your characters are letters (or at least have the high order bit clear) then we can use the top bit of each character to mark this. This produces the following program, which is not recursive and so does not use stack space.
class XX
{
/** new position given old position */
static int newFromOld(int x, int n)
{
if (x < n / 2)
{
return x * 2;
}
return (x - n / 2) * 2 + 1;
}
private static int HIGH_ORDER_BIT = 1 << 15; // 16-bit chars
public static void main(String[] s)
{
// input data - create an array so we can modify
// characters in place
char[] x = s[0].toCharArray();
if ((x.length & 1) != 0)
{
System.err.println("Only works with even length strings");
return;
}
// Character we have read but not yet written, if any
char holding = 0;
// where character in hand was read from
int holdingPos = 0;
// whether picked up a character in our hand
boolean isHolding = false;
int rpos = 0;
while (rpos < x.length)
{ // Here => moved out everything up to rpos
// and put in place with top bit set to mark new occupant
if (!isHolding)
{ // advance read pointer to read new character
char here = x[rpos];
holdingPos = rpos++;
if ((here & HIGH_ORDER_BIT) != 0)
{
// already dealt with
continue;
}
int targetPos = newFromOld(holdingPos, x.length);
// pick up char at target position
holding = x[targetPos];
// place new character, and mark as new
x[targetPos] = (char)(here | HIGH_ORDER_BIT);
// Now holding a character that needs to be put in its
// correct place
isHolding = true;
holdingPos = targetPos;
}
int targetPos = newFromOld(holdingPos, x.length);
char here = x[targetPos];
if ((here & HIGH_ORDER_BIT) != 0)
{ // back to where we picked up a character to hold
isHolding = false;
continue;
}
x[targetPos] = (char)(holding | HIGH_ORDER_BIT);
holding = here;
holdingPos = targetPos;
}
for (int i = 0; i < x.length; i++)
{
x[i] ^= HIGH_ORDER_BIT;
}
System.out.println("Result is " + new String(x));
}
}
These days, if I asked someone that question, what I'm looking for them to write on the whiteboard first is:
assertEquals("1a2b3c4d5e6f",funnySort("123456abcdef"));
...
and then maybe ask for more examples.
(And then, depending, if the task is to interleave numbers & letters, I think you can do it with two walking-pointers, indexLetter and indexDigit, and advance them across swapping as needed til you reach the end.)
In your recursive solution why don't you just make a test if n/2 % 2 == 0 (n%4 ==0 ) and treat the 2 situations differently
As templatetypedef commented your recursion cannot be in-place.
But here is a solution (not in place) using the way you wanted to make your recursion :
def f(s):
n=len(s)
if n==2: #initialisation
return s
elif n%4 == 0 : #if n%4 == 0 it's easy
return f(s[:n/4]+s[n/2:3*n/4])+f(s[n/4:n/2]+s[3*n/4:])
else: #otherwise, n-2 %4 == 0
return s[0]+s[n/2]+f(s[1:n/2]+s[n/2+1:])
Here we go. Recursive, cuts it in half each time, and in-place. Uses the approach outlined by #Chris Mennie. Getting the splitting right was tricky. A lot longer than Python, innit?
/* In-place, divide-and-conquer, recursive riffle-shuffle of strings;
* even length only. No wide characters or Unicode; old school. */
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void testrif(const char *s);
void riffle(char *s);
void rif_recur(char *s, size_t len);
void swap(char *s, size_t midpt, size_t len);
void flip(char *s, size_t len);
void if_odd_quit(const char *s);
int main(void)
{
testrif("");
testrif("a1");
testrif("ab12");
testrif("abc123");
testrif("abcd1234");
testrif("abcde12345");
testrif("abcdef123456");
return 0;
}
void testrif(const char *s)
{
char mutable[20];
strcpy(mutable, s);
printf("'%s'\n", mutable);
riffle(mutable);
printf("'%s'\n\n", mutable);
}
void riffle(char *s)
{
if_odd_quit(s);
rif_recur(s, strlen(s));
}
void rif_recur(char *s, size_t len)
{
/* Turn, e.g., "abcde12345" into "abc123de45", then recurse. */
size_t pivot = len / 2;
size_t half = (pivot + 1) / 2;
size_t twice = half * 2;
if (len < 4)
return;
swap(s + half, pivot - half, pivot);
rif_recur(s, twice);
rif_recur(s + twice, len - twice);
}
void swap(char *s, size_t midpt, size_t len)
{
/* Swap s[0..midpt] with s[midpt..len], in place. Algorithm from
* Programming Pearls, Chapter 2. */
flip(s, midpt);
flip(s + midpt, len - midpt);
flip(s, len);
}
void flip(char *s, size_t len)
{
/* Reverse order of characters in s, in place. */
char *p, *q, tmp;
if (len < 2)
return;
for (p = s, q = s + len - 1; p < q; p++, q--) {
tmp = *p;
*p = *q;
*q = tmp;
}
}
void if_odd_quit(const char *s)
{
if (strlen(s) % 2) {
fputs("String length is odd; aborting.\n", stderr);
exit(1);
}
}
By comparing 123456abcdef and 1a2b3c4d5e6f we can note that only the first and the last characters are in their correct position. We can also note that for each remaining n-2 characters we can compute their correct position directly from their original position. They will get there, and the element that was there surely was not in the correct position, so it will have to replace another one. By doing n-2 such steps all the elements will get to the correct positions:
void funny_sort(char* arr, int n){
int pos = 1; // first unordered element
char aux = arr[pos];
for (int iter = 0; iter < n-2; iter++) { // n-2 unordered elements
pos = (pos < n/2) ? pos*2 : (pos-n/2)*2+1;// correct pos for aux
swap(&aux, arr + pos);
}
}
Score each digit as its numerical value. Score each letter as a = 1.5, b = 2.5 c = 3.5 etc. Run an insertion sort of the string based on the score of each character.
[ETA] Simple scoring won't work so use two pointers and reverse the piece of the string between the two pointers. One pointer starts at the front of the string and advances one step each cycle. The other pointer starts in the middle of the string and advances every second cycle.
123456abcdef
^ ^
1a65432bcdef
^ ^
1a23456bcdef
^ ^
1a2b6543cdef
^ ^
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