Gold mine problem. Following sequence for loop is giving correct result.
//see link for other code
static int getMaxGold(int gold[][], int m, int n) {
//see link for other code
for (int col = n-1; col >= 0; col--) {
for (int row = 0; row < m; row++) {
int right = (col == n-1) ? 0 : goldTable[row][col+1];
int right_up = (row == 0 || col == n-1) ? 0 : goldTable[row-1][col+1];
int right_down = (row == m-1 || col == n-1) ? 0 : goldTable[row+1][col+1];
goldTable[row][col] = gold[row][col] + Math.max(right, Math.max(right_up, right_down));
}
}
}
//see link for other code
While other way round does not give the expected result. For example
for (int col = 0; col < n; col ++) {
for (int row = 0; row < m; row++) {
//same code to calculate right, rightUp and rightDown
}
}
Any explanation for this behaviour?
You don't need to store a whole matrix.
When you build the table, you just need to keep the last layer you processed.
In your recursion, there is diagonally right, or right, so the layer is a column because to compute the value of some cell, you need to know three values (on its right)
You conclude (as spotted by Damien already) that you start from the rightmost column, then to compute the value of every cells of the n-1 column, you only need to know the nth column (which you luckily computed already)
In below example. m_ij refers to i-th line, j-th column. (e.g m_01 == 2, m_10 = 5)
1 2 3 4
5 6 7 8
9 1 2 3
4 5 6 3
The last column is {4,8,3,3}
To compute the max value for m_02 you need to choose between 4 and 8
3 - 4
\
8
m_02 = 3 + 8 = 11
To compute the max value of m_12 you need to choose between 4, 8 and 3
4
/
7 - 8
\
3
m_12 = 7 + 8 = 15
Skipping stuff
m_22 = 2 + 8 = 10
m_32 = 6 + 3 = 9
Now you know the max value for each square of the third column
1 2 11 .
5 6 15 .
9 1 10 .
4 5 9 .
You do the same for m_10, m_11, ...
idem
m_01 = 2 + max(11, 15) = 17
m_11 = 6 + 15
m_21 = 1 + 15
m_31 = 5 + 10
Left to process is thus
1 17
5 21
9 16
4 15
Then
1+21
5+21
9+21
4+16
And finally score = max(22, 26, 30, 20)
As you have noticed you only need to keep track of the last processed column. Not a whole table of computation. And the last processed column must start from the right and always be the rightmost one...
I don't think an implem is relevant to help you understand but in case
const s = `
1 2 3 4
5 6 7 8
9 1 2 3
4 5 6 3`
const m = s.trim().split('\n').map(x => x.trim().split(' ').map(y => parseInt(y)))
let layer = [0, 0, 0, 0]
for (let j = 3; j >= 0; --j) {
const nextLayer = []
for (let i = 0; i < 4; ++i) {
nextLayer[i] = m[i][j] + Math.max(
layer[i-1] || 0, // we default undefined value as 0 supposing s only holds positive coefficient
layer[i],
layer[i+1] || 0
)
}
layer = nextLayer
}
console.log(Math.max(...layer))
Related
I have a matrix in Rcpp (C++ for R) which is stored in column order in memory. Ie, it looks like:
[,1] [,2] [,3] [,4] [,5]
[1,] 1 6 11 16 21
[2,] 2 7 12 17 22
[3,] 3 8 13 18 23
[4,] 4 9 14 19 24
[5,] 5 10 15 20 25
Now, I have a single for loop that runs from i = 1 to 25 (bear in mind, it is all zero based, but here I am just saying one for convenience).
For every element of the matrix, I want its Moore neighbourhood. This is easy for the elements that are not on the edge. So if our selected index is idx and the size of the square matrix is nrow then we have
leftmid = idx - nrow
lefttop = (idx - nrow) - 1
leftbot = (idx - nrow) + 1
rightmid = idx + nrow
righttop = (idx + nrow) - 1
rightbot = (idx + nrow) + 1
midtop = idx - 1
midbot = idx + 1
But i cant figure out how to deal with the edge cases. For example, if idx = 3, then i want the neighbours:
leftmid = 23
lefttop = 22
leftbot = 24
rightmid = 8
righttop = 7
rightbot = 9
midtop = 2
midbot = 4
It's a little bit more complicated at the corner cases as well. My goal here is to reduce time. I am currently running my program with a double for loop which works, but is slower than reasonable. I want to change it into a single for loop to improve performance.
Edit: I realized the left and right boundaries can be obtained by modulus. So 3 - 5 %% 25 = 23. But I still have the top and bottom edge cases.
It appears you're interested in "cyclic" boundary conditions, where the matrix has a toroidal topology, i.e. the top wraps around to the bottom and the right wraps around to the left.
It might be easier to iterate with four loops, one each over the row and column, and then one each over the row and column of the neighborhood. Something like this should work:
int mooreNeighbors[3][3];
int nRows = 5;
int nCols = 5;
// Loop over the rows and columns of the matrix
for (int i = 0; i < nRows; ++i) {
for (int j = 0; j < nCols; ++j) {
// Loop over the cyclic Moore neighborhood
for (int mnI = 0; mnI < 3; ++mnI) {
for (int mnJ = 0; mnJ < 3; ++mnJ) {
// Sub-matrix indices
int subI = (i - mnI - 1) % nRows;
int subJ = (j - mnJ - 1) % nCols;
// Index into column-dominant matrix
int idx = subI + subJ*nRows;
mooreNeighbors[mnI][mnJ] = matrix[idx];
}
}
}
}
I haven't tried compiling this, but it should be close to correct and clear enough to correct any mistakes. Think of it as pseudo-code.
Also, I'm preferring clarity over optimality. For example, you don't have to do everything in the inner-most loop.
There is a m x n array, I need to output each possible combination Of each line's element. For example, for array{{1,2,3},{4,5,6}}, I need to output{{1,4},{1,5},{1,6},{2,4},{2,5},{2,6},{3,4},{3,5},{3,6}}.
I think there should be a m loop to solve this. For the example above, I wrote the code:
int[,] array = new int[,] {{1, 2, 3}, {4, 5, 6}};
for (var i = 0; i < 3; i++)
{
for (var j = 0; j < 3; j++)
{
Console.WriteLine($"{{{array[0, i]},{array[1, j]}}}");
}
}
With m changes, the number of for loop also changes. But m is unknown when I write the code. How can I solve it?
Maintain a list of active combinations c. Initialize c to be the array's first row. Then iterate every additional row and update c. Basically, you augment each of the current combinations with each of the row's items. Here is some pseudo code:
c = array[0]; //read: the first row of the array
for(var i = 1; i < m; ++i) { //iterate all rows
var c_modified = [];
for(var j = 0; j < n; ++j) { //iterate all row entries
for(var k = 0; k < c.length; ++k) { //iterate all entries of c
add c[k].array[i, j] to c_modified; // . represents concatenation
}
}
c = c_modified;
}
This combination of elements (n^m sets) is called Cartesian product. There are ready-to-use functions for its generation in some language libraries
I believe that the simplest code is recursive one.
type
TArr2D = TArray<TArray<Integer>>;
procedure CartesianProduct(const A: TArr2D; Line: Integer; Reslt: TArray<Integer>);
var
i: integer;
begin
if Line > High(A) then
Memo1.Lines.Add(ArrayToString(Reslt)) // output m-element set
else
for i in A[Line] do
CartesianProduct(A, Line + 1, Reslt + [i]); // include element in set
end;
var
A: TArr2D;
n, m, i, j: Integer;
begin
m := 3;
n := 3;
SetLength(A, m, n);
for j := 0 to m - 1 do
for i := 0 to n - 1 do
A[j, i] := j * n + i;
//0 1 2
//3 4 5
//6 7 8
CartesianProduct(A, 0, []);
gives
0 3 6
0 3 7
0 3 8
0 4 6
0 4 7
0 4 8
0 5 6
0 5 7
0 5 8
1 3 6
1 3 7
1 3 8
1 4 6
1 4 7
1 4 8
1 5 6
1 5 7
1 5 8
2 3 6
2 3 7
2 3 8
2 4 6
2 4 7
2 4 8
2 5 6
2 5 7
2 5 8
Add all the values between 100 and 4000000 inclusively that are divisable by 3 or 5 but not both 3 and 5
Can't figure out how to implement second part of that stipulation. Here's what I have so far:
var sum = 0;
for (var i = 100; i < 4000001; i++) {
if (i % 3 || i % 5 === 0) {
sum = sum + i;
}
}
You can compute the sum without any loop, using the formula for the sum of an arithmetic progression: We have
3 + 5 + 6 + 9 + 10 + 12 + 18 + 20 + ...
= 3 + 6 + 9 + 12 + 15 + 18 + ...
+ 5 + 10 + 15 + 20 + ...
- 2*(15 + 30 + 45 + ...)
Note that we add all the multiples of 3 and 5 but then subtract the multiples of 15 twice, because they were counted twice as multiples of both 3 and 5.
Let g(n) be the sum of integers from 1 to n. We have g(n) = n*(n+1)/2.
Let f(n) be the sum of integers between 1 and n that are divisible by 3 or 5, but not both. Then we have
f(n) = 3*g(floor(n / 3)) + 5*g(floor(n/5)) - 30*g(floor(n/15))
And the sum of integers between m and n that are divisible by 3 or 5, but not both is then just f(n) - f(m - 1). This can be computed in O(1).
You simply need to escape only those part which involves division by 15, and other higher numbers(multiple of 15) will be avoided further automatically.
Note that checking divisibility by 15 should be at the top, which on being true will continue further iteration without executing the below codes of divisibility by 3 and 5. If false, then a number can only be divisible by 3 or 5 or none, but not both.
for (var i = 100; i < 4000001; i++) {
if(i % 15 == 0 )
continue;
if (i % 3 == 0) {
sum = sum + i;
}
if (i % 5 == 0) {
sum = sum + i;
}
}
Also, note that you have used === operator which I don't think is a valid operator, probably you want ==. BTW, I am not sure whether any language supports ===, I think Javascript supports that. So, be careful at that step.
You can use != instead of || since this is exactly what you want. Only divisible by 3 or 5 but not by both.
var sum = 0;
for (var i = 100; i < 4000001; i++) {
if ((i % 3 == 0) != (i % 5 == 0)) {
sum = sum + i;
}
}
var sum = 0;
for (var i = 100; i < 4000001; i++) {
if (i % 3 === 0 ^ i % 5 === 0) {
sum = sum + i;
}
}
use the exclusive OR , XOR ^ returns true only when one of the conditions not both is true.
A sequence of integers is called zigzag sequence if each of its elements is either strictly less or strictly greater than its neighbors.
Example : The sequence 4 2 3 1 5 3 forms a zigzag, but 7 3 5 5 2 and 3 8 6 4 5 don't.
For a given array of integers we need to find the length of its largest (contiguous) sub-array that forms a zigzag sequence.
Can this be done in O(N) ?
Currently my solution is O(N^2) which is just simply taking every two points and checking each possible sub-array if it satisfies the condition or not.
I claim that the length of overlapping sequence of any 2 zigzag sub-sequences is a most 1
Proof by contradiction:
Assume a_i .. a_j is the longest zigzag sub-sequence, and there is another zigzag sub-sequence b_m...b_n overlapping it.
without losing of generality, let's say the overlapping part is
a_i ... a_k...a_j
--------b_m...b_k'...b_n
a_k = b_m, a_k+1 = b_m+1....a_j = b_k' where k'-m = j-k > 0 (at least 2 elements are overlapping)
Then they can merge to form a longer zig-zag sequence, contradiction.
This means the only case they can be overlapping each other is like
3 5 3 2 3 2 3
3 5 3 and 3 2 3 2 3 is overlapping at 1 element
This can still be solved in O(N) I believe, like just greedily increase the zig-zag length whenever possible. If fails, move iterator 1 element back and treat it as a new zig-zag starting point
Keep record the latest and longest zig-zag length you have found
Walk along the array and see if the current item belongs to (fits a definition of) a zigzag. Remember the las zigzag start, which is either the array's start or the first zigzag element after the most recent non-zigzag element. This and the current item define some zigzag subarray. When it appears longer than the previously found, store the new longest zigzag length. Proceed till the end of array and you should complete the task in O(N).
Sorry I use perl to write this.
#!/usr/bin/perl
#a = ( 5, 4, 2, 3, 1, 5, 3, 7, 3, 5, 5, 2, 3, 8, 6, 4, 5 );
$n = scalar #a;
$best_start = 0;
$best_end = 1;
$best_length = 2;
$start = 0;
$end = 1;
$direction = ($a[0] > $a[1]) ? 1 : ($a[0] < $a[1]) ? -1 : 0;
for($i=2; $i<$n; $i++) {
// a trick here, same value make $new_direction = $direction
$new_direction = ($a[$i-1] > $a[$i]) ? 1 : ($a[$i-1] < $a[$i]) ? -1 : $direction;
print "$a[$i-1] > $a[$i] : direction $new_direction Vs $direction\n";
if ($direction != $new_direction) {
$end = $i;
} else {
$this_length = $end - $start + 1;
if ($this_length > $best_length) {
$best_start = $start;
$best_end = $end;
$best_length = $this_length;
}
$start = $i-1;
$end = $i;
}
$direction = $new_direction;
}
$this_length = $end - $start + 1;
if ($this_length > $best_length) {
$best_start = $start;
$best_end = $end;
$best_length = $this_length;
}
print "BEST $best_start to $best_end length $best_length\n";
for ($i=$best_start; $i <= $best_end; $i++) {
print $a[$i], " ";
}
print "\n";
For each index i, you can find the smallest j such that the subarray with index j,j+1,...,i-1,i is a zigzag. This can be done in two phases:
Find the longest "increasing" zig zag (starts with a[1]>a[0]):
start = 0
increasing[0] = 0
sign = true
for (int i = 1; i < n; i ++)
if ((arr[i] > arr[i-1] && sign) || )arr[i] < arr[i-1] && !sign)) {
increasing[i] = start
sign = !sign
} else if (arr[i-1] < arr[i]) { //increasing and started last element
start = i-1
sign = false
increasing[i] = i-1
} else { //started this element
start = i
sign = true
increasing[i] = i
}
}
Do similarly for "decreasing" zig-zag, and you can find for each index the "earliest" possible start for a zig-zag subarray.
From there, finding the maximal possible zig-zag is easy.
Since all oporations are done in O(n), and you basically do one after the other, this is your complexity.
You can combine the both "increasing" and "decreasing" to one go:
start = 0
maxZigZagStart[0] = 0
sign = true
for (int i = 1; i < n; i ++)
if ((arr[i] > arr[i-1] && sign) || )arr[i] < arr[i-1] && !sign)) {
maxZigZagStart[i] = start
sign = !sign
} else if (arr[i-1] > arr[i]) { //decreasing:
start = i-1
sign = false
maxZigZagStart[i] = i-1
} else if (arr[i-1] < arr[i]) { //increasing:
start = i-1
sign = true
maxZigZagStart[i] = i-1
} else { //equality
start = i
//guess it is increasing, if it is not - will be taken care of next iteration
sign = true
maxZigZagStart[i] = i
}
}
You can see that you can actually even let go of maxZigZagStart aux array and stored local maximal length instead.
A sketch of simple one-pass algorithm. Cmp compares neighbour elements, returning -1, 0, 1 for less, equal and greater cases.
Zigzag ends for cases of Cmp transitions:
0 0
-1 0
1 0
Zigzag ends and new series starts:
0 -1
0 1
-1 -1
1 1
Zigzag series continues for transitions
-1 1
1 -1
Algo:
Start = 0
LastCmp = - Compare(A[i], A[i - 1]) //prepare to use the first element individually
MaxLen = 0
for i = 1 to N - 1 do
Cmp = Compare(A[i], A[i - 1]) //returns -1, 0, 1 for less, equal and greater cases
if Abs(Cmp - LastCmp) <> 2 then
//zigzag condition is violated, series ends, new series starts
MaxLen = Max(MaxLen, i - 1 - Start)
Start = i
//else series continues, nothing to do
LastCmp = Cmp
//check for ending zigzag
if LastCmp <> 0 then
MaxLen = Max(MaxLen, N - Start)
examples of output:
2 6 7 1 7 0 7 3 1 1 7 4
5 (7 1 7 0 7)
8 0 0 3 5 8
1
0 0 7 0
2
1 2 0 7 9
3
8 3 5 2
4
1 3 7 1 6 6
2
1 4 0 6 6 3 4 3 8 0 9 9
5
Lets consider sequence 5 9 3 4 5 4 2 3 6 5 2 1 3 as an example. You have a condition which every internal element of subsequence should satisfy (element is strictly less or strictly greater than its neighbors). Lets compute this condition for every element of the whole sequence:
5 9 3 6 5 7 2 3 6 5 2 1 3
0 1 1 1 1 1 1 0 1 0 0 1 0
The condition is undefined for outermost elements because they have only one neighbor each. But I defined it as 0 for convenience.
The longest subsequence of 1's (9 3 6 5 7 2) is the internal part of the longest zigzag subsequence (5 9 3 6 5 7 2 3). So the algorithm is:
Find the longest subsequence of elements satisfying condition.
Add to it one element to each side.
The first step can be done in O(n) by the following algorithm:
max_length = 0
current_length = 0
for i from 2 to len(a) - 1:
if a[i - 1] < a[i] > a[i + 1] or a[i - 1] > a[i] < a[i + 1]:
current_length += 1
else:
max_length = max(max_length, current_length)
current_length = 0
max_length = max(max_length, current_length)
The only special case is if the sequence total length is 0 or 1. Then the whole sequence would be the longest zigzag subsequence.
#include "iostream"
using namespace std ;
int main(){
int t ; scanf("%d",&t) ;
while(t--){
int n ; scanf("%d",&n) ;
int size1 = 1 , size2 = 1 , seq1 , seq2 , x ;
bool flag1 = true , flag2 = true ;
for(int i=1 ; i<=n ; i++){
scanf("%d",&x) ;
if( i== 1 )seq1 = seq2 = x ;
else {
if( flag1 ){
if( x>seq1){
size1++ ;
seq1 = x ;
flag1 = !flag1 ;
}
else if( x < seq1 )
seq1 = x ;
}
else{
if( x<seq1){
size1++ ;
seq1=x ;
flag1 = !flag1 ;
}
else if( x > seq1 )
seq1 = x ;
}
if( flag2 ){
if( x < seq2 ){
size2++ ;
seq2=x ;
flag2 = !flag2 ;
}
else if( x > seq2 )
seq2 = x ;
}
else {
if( x > seq2 ){
size2++ ;
seq2 = x ;
flag2 = !flag2 ;
}
else if( x < seq2 )
seq2 = x ;
}
}
}
printf("%d\n",max(size1,size2)) ;
}
return 0 ;
}
After the you guys helped me out so gracefully last time, here is another tricky array sorter for you.
I have the following array:
a = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16]
I use it for some visual stuff and render it like this:
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Now I want to sort the array to have a "snake" later:
// rearrange the array according to this schema
1 2 3 4
12 13 14 5
11 16 15 6
10 9 8 7
// the original array should look like this
a = [1,2,3,4,12,13,14,5,11,16,15,6,10,9,8,7]
Now I'm looking for a smart formula / smart loop to do that
ticker = 0;
rows = 4; // can be n
cols = 4; // can be n
originalArray = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16];
newArray = [];
while(ticker < originalArray.length)
{
//do the magic here
ticker++;
}
Thanks again for the help.
I was bored, so I made a python version for you with 9 lines of code inside the loop.
ticker = 0
rows = 4
cols = 4
originalArray = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16]
newArray = [None] * (rows * cols)
row = 0
col = 0
dir_x = 1
dir_y = 0
taken = {}
while (ticker < len(originalArray)):
newArray[row * cols + col] = originalArray[ticker]
taken[row * cols + col] = True
if col + dir_x >= cols or row + dir_y >= rows or col + dir_x < 0:
dir_x, dir_y = -dir_y, dir_x
elif ((row + dir_y) * cols + col + dir_x) in taken:
dir_x, dir_y = -dir_y, dir_x
row += dir_y
col += dir_x
ticker += 1
print newArray
You can index into the snake coil directly if you recall that
1 + 2 + 3 + ... + n = n*(n+1)/2
m^2 + m - k = 0 => m - (-1+sqrt(1+4*k))/2
and look at the pattern of the coils. (I'll leave it as a hint for the time being--you could also use that n^2 = (n-1)^2 + (2*n+1) with reverse-indexing, or a variety of other things to solve the problem.)
When translating to code, it's not really any shorter than Tuomas' solution if all you want to do is fill the matrix, however.