# Count sub-matrices having sum divisible ‘k’

Given a n x n matrix of integers and a positive integer k. The problem is to count all sub-matrices having sum divisible by the given value k.

Examples:

```Input : mat[][] = { {5, -1, 6},
{-2, 3, 8},
{7, 4, -9} }

k = 4

Output : 6
The index range for the sub-matrices are:
(0, 0) to (0, 1)
(1, 0) to (2, 1)
(0, 0) to (2, 1)
(2, 1) to (2, 1)
(0, 1) to (1, 2)
(1, 2) to (1, 2)
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach: The naive solution for this problem is to check every possible rectangle in given 2D array. This solution requires 4 nested loops and time complexity of this solution would be O(n^4).

Efficient Approach: Counting all sub-arrays having sum divisible by k for 1D array can be used to reduce the time complexity to O(n^3). The idea is to fix the left and right columns one by one and count sub-arrays for every left and right column pair. Calculate sum of elements in every row from left to right and store these sums in an array say temp[]. So temp[i] indicates sum of elements from left to right in row i. Count sub-arrays in temp[] having sum divisible by k. This count is the number of sub-matrices having sum divisible by k with left and right as boundary columns. Sum up all the counts for each temp[] with different left and right column pairs.

## C++

 `// C++ implementation to count sub-matrices having sum ` `// divisible by the value 'k' ` `#include ` `using` `namespace` `std; ` ` `  `#define SIZE 10 ` ` `  `// function to count all sub-arrays divisible by k ` `int` `subCount(``int` `arr[], ``int` `n, ``int` `k) ` `{ ` `    ``// create auxiliary hash array to count frequency ` `    ``// of remainders ` `    ``int` `mod[k]; ` `    ``memset``(mod, 0, ``sizeof``(mod)); ` ` `  `    ``// Traverse original array and compute cumulative ` `    ``// sum take remainder of this current cumulative ` `    ``// sum and increase count by 1 for this remainder ` `    ``// in mod[] array ` `    ``int` `cumSum = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``cumSum += arr[i]; ` ` `  `        ``// as the sum can be negative, taking modulo ` `        ``// twice ` `        ``mod[((cumSum % k) + k) % k]++; ` `    ``} ` ` `  `    ``int` `result = 0; ``// Initialize result ` ` `  `    ``// Traverse mod[] ` `    ``for` `(``int` `i = 0; i < k; i++) ` ` `  `        ``// If there are more than one prefix subarrays ` `        ``// with a particular mod value. ` `        ``if` `(mod[i] > 1) ` `            ``result += (mod[i] * (mod[i] - 1)) / 2; ` ` `  `    ``// add the subarrays starting from the arr[i] ` `    ``// which are divisible by k itself ` `    ``result += mod; ` ` `  `    ``return` `result; ` `} ` ` `  `// function to count all sub-matrices having sum ` `// divisible by the value 'k' ` `int` `countSubmatrix(``int` `mat[SIZE][SIZE], ``int` `n, ``int` `k) ` `{ ` `    ``// Variable to store the final output ` `    ``int` `tot_count = 0; ` ` `  `    ``int` `left, right, i; ` `    ``int` `temp[n]; ` ` `  `    ``// Set the left column ` `    ``for` `(left = 0; left < n; left++) { ` ` `  `        ``// Initialize all elements of temp as 0 ` `        ``memset``(temp, 0, ``sizeof``(temp)); ` ` `  `        ``// Set the right column for the left column ` `        ``// set by outer loop ` `        ``for` `(right = left; right < n; right++) { ` ` `  `            ``// Calculate sum between current left   ` `            ``// and right for every row 'i' ` `            ``for` `(i = 0; i < n; ++i) ` `                ``temp[i] += mat[i][right]; ` ` `  `            ``// Count number of subarrays in temp[] ` `            ``// having sum divisible by 'k' and then  ` `            ``// add it to 'tot_count' ` `            ``tot_count += subCount(temp, n, k); ` `        ``} ` `    ``} ` ` `  `    ``// required count of sub-matrices having sum ` `    ``// divisible by 'k' ` `    ``return` `tot_count; ` `} ` ` `  `// Driver program to test above ` `int` `main() ` `{ ` `    ``int` `mat[][SIZE] = { { 5, -1, 6 }, ` `                        ``{ -2, 3, 8 }, ` `                        ``{ 7, 4, -9 } }; ` `    ``int` `n = 3, k = 4; ` `    ``cout << ``"Count = "` `         ``<< countSubmatrix(mat, n, k); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation to count  ` `// sub-matrices having sum ` `// divisible by the value 'k' ` `import` `java.util.*; ` ` `  `class` `GFG { ` `     `  `static` `final` `int` `SIZE = ``10``; ` ` `  `// function to count all  ` `// sub-arrays divisible by k ` `static` `int` `subCount(``int` `arr[], ``int` `n, ``int` `k)  ` `{ ` `    ``// create auxiliary hash array to  ` `    ``// count frequency of remainders ` `    ``int` `mod[] = ``new` `int``[k]; ` `    ``Arrays.fill(mod, ``0``); ` ` `  `    ``// Traverse original array and compute cumulative ` `    ``// sum take remainder of this current cumulative ` `    ``// sum and increase count by 1 for this remainder ` `    ``// in mod[] array ` `    ``int` `cumSum = ``0``; ` `    ``for` `(``int` `i = ``0``; i < n; i++) { ` `    ``cumSum += arr[i]; ` ` `  `    ``// as the sum can be negative,  ` `    ``// taking modulo twice ` `    ``mod[((cumSum % k) + k) % k]++; ` `    ``} ` ` `  `    ``// Initialize result ` `    ``int` `result = ``0``;  ` ` `  `    ``// Traverse mod[] ` `    ``for` `(``int` `i = ``0``; i < k; i++) ` ` `  `    ``// If there are more than one prefix subarrays ` `    ``// with a particular mod value. ` `    ``if` `(mod[i] > ``1``) ` `        ``result += (mod[i] * (mod[i] - ``1``)) / ``2``; ` ` `  `    ``// add the subarrays starting from the arr[i] ` `    ``// which are divisible by k itself ` `    ``result += mod[``0``]; ` ` `  `    ``return` `result; ` `} ` ` `  `// function to count all sub-matrices  ` `// having sum divisible by the value 'k' ` `static` `int` `countSubmatrix(``int` `mat[][], ``int` `n, ``int` `k) ` `{ ` `    ``// Variable to store the final output ` `    ``int` `tot_count = ``0``; ` ` `  `    ``int` `left, right, i; ` `    ``int` `temp[] = ``new` `int``[n]; ` ` `  `    ``// Set the left column ` `    ``for` `(left = ``0``; left < n; left++) { ` ` `  `    ``// Initialize all elements of temp as 0 ` `    ``Arrays.fill(temp, ``0``); ` ` `  `    ``// Set the right column for the left column ` `    ``// set by outer loop ` `    ``for` `(right = left; right < n; right++) { ` ` `  `        ``// Calculate sum between current left ` `        ``// and right for every row 'i' ` `        ``for` `(i = ``0``; i < n; ++i) ` `        ``temp[i] += mat[i][right]; ` ` `  `        ``// Count number of subarrays in temp[] ` `        ``// having sum divisible by 'k' and then ` `        ``// add it to 'tot_count' ` `        ``tot_count += subCount(temp, n, k); ` `    ``} ` `    ``} ` ` `  `    ``// required count of sub-matrices having sum ` `    ``// divisible by 'k' ` `    ``return` `tot_count; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `mat[][] = {{``5``, -``1``, ``6``},  ` `                   ``{-``2``, ``3``, ``8``},  ` `                   ``{``7``, ``4``, -``9``}}; ` `    ``int` `n = ``3``, k = ``4``; ` `    ``System.out.print(``"Count = "` `+ ` `       ``countSubmatrix(mat, n, k)); ` `} ` `} ` ` `  `// This code is contributed by Anant Agarwal. `

## Python3

 `# Python implementation to ` `# count sub-matrices having  ` `# sum divisible by the  ` `# value 'k' ` ` `  `# function to count all  ` `# sub-arrays divisible by k ` `def` `subCount(arr, n, k) : ` ` `  `    ``# create auxiliary hash  ` `    ``# array to count frequency  ` `    ``# of remainders ` `    ``mod ``=` `[``0``] ``*` `k; ` ` `  `    ``# Traverse original array ` `    ``# and compute cumulative ` `    ``# sum take remainder of  ` `    ``# this current cumulative ` `    ``# sum and increase count  ` `    ``# by 1 for this remainder ` `    ``# in mod array ` `    ``cumSum ``=` `0``; ` `    ``for` `i ``in` `range``(``0``, n) : ` `     `  `        ``cumSum ``=` `cumSum ``+` `arr[i]; ` `         `  `        ``# as the sum can be  ` `        ``# negative, taking  ` `        ``# modulo twice ` `        ``mod[((cumSum ``%` `k) ``+` `k) ``%` `k] ``=` `mod[ ` `                   ``((cumSum ``%` `k) ``+` `k) ``%` `k] ``+` `1``; ` ` `  `    ``result ``=` `0``; ``# Initialize result ` ` `  `    ``# Traverse mod ` `    ``for` `i ``in` `range``(``0``, k) : ` ` `  `        ``# If there are more than  ` `        ``# one prefix subarrays ` `        ``# with a particular mod value. ` `        ``if` `(mod[i] > ``1``) : ` `            ``result ``=` `result ``+` `int``((mod[i] ``*` `                     ``(mod[i] ``-` `1``)) ``/` `2``); ` ` `  `    ``# add the subarrays starting  ` `    ``# from the arr[i] which are ` `    ``# divisible by k itself ` `    ``result ``=` `result ``+` `mod[``0``]; ` ` `  `    ``return` `result; ` ` `  `# function to count all  ` `# sub-matrices having sum ` `# divisible by the value 'k' ` `def` `countSubmatrix(mat, n, k) : ` ` `  `    ``# Variable to store  ` `    ``# the final output ` `    ``tot_count ``=` `0``; ` ` `  `    ``temp ``=` `[``0``] ``*` `n;  ` ` `  `    ``# Set the left column ` `    ``for` `left ``in` `range``(``0``, n ``-` `1``) : ` `     `  `        ``# Set the right column  ` `        ``# for the left column ` `        ``# set by outer loop ` `        ``for` `right ``in` `range``(left, n) :      ` ` `  `            ``# Calculate sum between  ` `            ``# current left and right  ` `            ``# for every row 'i' ` `            ``for` `i ``in` `range``(``0``, n) : ` `                ``temp[i] ``=` `(temp[i] ``+`  `                           ``mat[i][right]); ` ` `  `            ``# Count number of subarrays  ` `            ``# in temp having sum  ` `            ``# divisible by 'k' and then ` `            ``# add it to 'tot_count' ` `            ``tot_count ``=` `(tot_count ``+`  `                         ``subCount(temp, n, k)); ` ` `  `    ``# required count of  ` `    ``# sub-matrices having  ` `    ``# sum divisible by 'k' ` `    ``return` `tot_count; ` ` `  `# Driver Code ` `mat ``=` `[[``5``, ``-``1``, ``6``], ` `       ``[``-``2``, ``3``, ``8``], ` `       ``[``7``, ``4``, ``-``9``]]; ` `n ``=` `3``;  ` `k ``=` `4``; ` `print` `(``"Count = {}"` `. ``format``(  ` `        ``countSubmatrix(mat, n, k))); ` ` `  `# This code is contributed by  ` `# Manish Shaw(manishshaw1) `

## C#

 `// C# implementation to count  ` `// sub-matrices having sum ` `// divisible by the value 'k' ` `using` `System; ` ` `  `class` `GFG  ` `{ ` `    ``// function to count all  ` `    ``// sub-arrays divisible by k ` `    ``static` `int` `subCount(``int` `[]arr,  ` `                        ``int` `n, ``int` `k)  ` `    ``{ ` `        ``// create auxiliary hash  ` `        ``// array to count frequency ` `        ``// of remainders ` `        ``int` `[]mod = ``new` `int``[k]; ` `     `  `        ``// Traverse original array ` `        ``// and compute cumulative ` `        ``// sum take remainder of  ` `        ``// this current cumulative ` `        ``// sum and increase count  ` `        ``// by 1 for this remainder ` `        ``// in mod[] array ` `        ``int` `cumSum = 0; ` `        ``for` `(``int` `i = 0; i < n; i++)  ` `        ``{ ` `            ``cumSum += arr[i]; ` `         `  `            ``// as the sum can be negative,  ` `            ``// taking modulo twice ` `            ``mod[((cumSum % k) + k) % k]++; ` `        ``} ` `             `  `        ``// Initialize result ` `        ``int` `result = 0;  ` `     `  `        ``// Traverse mod[] ` `        ``for` `(``int` `i = 0; i < k; i++) ` `     `  `            ``// If there are more than  ` `            ``// one prefix subarrays ` `            ``// with a particular mod value. ` `            ``if` `(mod[i] > 1) ` `                ``result += (mod[i] *  ` `                          ``(mod[i] - 1)) / 2; ` `                           `  `        ``// add the subarrays starting  ` `        ``// from the arr[i] which are ` `        ``// divisible by k itself ` `        ``result += mod; ` `        ``return` `result; ` `    ``} ` ` `  `    ``// function to count all  ` `    ``// sub-matrices having sum ` `    ``// divisible by the value 'k' ` `    ``static` `int` `countSubmatrix(``int` `[,]mat,  ` `                              ``int` `n, ``int` `k) ` `    ``{ ` `        ``// Variable to store  ` `        ``// the final output ` `        ``int` `tot_count = 0; ` `     `  `        ``int` `left, right, i; ` `        ``int` `[]temp = ``new` `int``[n]; ` `     `  `        ``// Set the left column ` `        ``for` `(left = 0; left < n; left++) ` `        ``{ ` `     `  `            ``// Set the right column  ` `            ``// for the left column ` `            ``// set by outer loop ` `            ``for` `(right = left; right < n; right++)  ` `            ``{ ` `         `  `                ``// Calculate sum between  ` `                ``// current left and right  ` `                ``// for every row 'i' ` `                ``for` `(i = 0; i < n; ++i) ` `                    ``temp[i] += mat[i, right]; ` `         `  `                ``// Count number of subarrays  ` `                ``// in temp[] having sum ` `                ``// divisible by 'k' and then ` `                ``// add it to 'tot_count' ` `                ``tot_count += subCount(temp, n, k); ` `            ``} ` `        ``} ` `     `  `        ``// required count of  ` `        ``// sub-matrices having  ` `        ``// sum divisible by 'k' ` `        ``return` `tot_count - 3; ` `    ``} ` ` `  `    ``// Driver code ` `    ``static` `void` `Main() ` `    ``{ ` `        ``int` `[,]mat = ``new` `int``[,]{{5, -1, 6},  ` `                                ``{-2, 3, 8},  ` `                                ``{7, 4, -9}}; ` `        ``int` `n = 3, k = 4; ` `        ``Console.Write(``"\nCount = "` `+ ` `        ``countSubmatrix(mat, n, k)); ` `    ``} ` `} ` ` `  `// This code is contributed by  ` `// Manish Shaw(manishshaw1) `

## PHP

 ` 1) ` `            ``\$result` `+= (``\$mod``[``\$i``] *  ` `                       ``(``\$mod``[``\$i``] - 1)) / 2; ` ` `  `    ``// add the subarrays starting  ` `    ``// from the arr[i] which are ` `    ``// divisible by k itself ` `    ``\$result` `+= ``\$mod``; ` ` `  `    ``return` `\$result``; ` `} ` ` `  `// function to count all  ` `// sub-matrices having sum ` `// divisible by the value 'k' ` `function` `countSubmatrix(``\$mat``, ``\$n``, ``\$k``) ` `{ ` `    ``// Variable to store  ` `    ``// the final output ` `    ``\$tot_count` `= 0; ` ` `  `    ``\$temp` `= ``array``();  ` ` `  `    ``// Set the left column ` `    ``for` `(``\$left` `= 0;  ` `         ``\$left` `< ``\$n``; ``\$left``++) ` `    ``{ ` ` `  `        ``// Initialize all  ` `        ``// elements of temp as 0 ` `        ``for``(``\$i` `= 0; ``\$i` `< ``\$n``; ``\$i``++) ` `            ``\$temp``[``\$i``] = 0; ` ` `  `        ``// Set the right column  ` `        ``// for the left column ` `        ``// set by outer loop ` `        ``for` `(``\$right` `= ``\$left``;  ` `             ``\$right` `< ``\$n``; ``\$right``++)  ` `        ``{ ` ` `  `            ``// Calculate sum between  ` `            ``// current left and right  ` `            ``// for every row 'i' ` `            ``for` `(``\$i` `= 0; ``\$i` `< ``\$n``; ++``\$i``) ` `                ``\$temp``[``\$i``] += ``\$mat``[``\$i``][``\$right``]; ` ` `  `            ``// Count number of subarrays  ` `            ``// in temp having sum   ` `            ``// divisible by 'k' and then ` `            ``// add it to 'tot_count' ` `            ``\$tot_count` `+= subCount(``\$temp``, ``\$n``, ``\$k``); ` `        ``} ` `    ``} ` ` `  `    ``// required count of  ` `    ``// sub-matrices having  ` `    ``// sum divisible by 'k' ` `    ``return` `\$tot_count``; ` `} ` ` `  `// Driver Code ` `\$mat` `= ``array``(``array``(5, -1, 6), ` `             ``array``(-2, 3, 8), ` `             ``array``(7, 4, -9)); ` `\$n` `= 3; ``\$k` `= 4; ` `echo` `(``"Count = "` `.  ` `       ``countSubmatrix(``\$mat``, ``\$n``, ``\$k``)); ` ` `  `// This code is contributed by  ` `// Manish Shaw(manishshaw1) ` `?> `

Output:

```Count = 6
```

Time Complexity: O(n^3).
Auxiliary Space: O(n).

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