# Maximum size square Sub-Matrix with sum less than or equals to K

Given a Matrix arr[][] of size M x N having positive integers and a number K, the task is to find the size of the largest square sub-matrix whose sum of elements is less than or equals to K

Example:

```Input:
arr[][] = { { 1, 1, 3, 2, 4, 3, 2 },
{ 1, 1, 3, 2, 4, 3, 2 },
{ 1, 1, 3, 2, 4, 3, 2 } },
K = 4
Output:
2
Explanation:
Maximum size square Sub-Matrix
with sum less than or equals to 4
1 1
1 1
Size is 2.

Input:
arr[][] = { { 1, 1, 3, 2, 4, 3, 2 },
{ 1, 1, 3, 2, 4, 3, 2 },
{ 1, 1, 3, 2, 4, 3, 2 } },
K = 22
Output:
3
Explanation:
Maximum size square Sub-Matrix
with sum less than or equals to 22
1 1 3
1 1 3
1 1 3
Size is 3. ```

Approach:

1. For the given matrix arr[][] create a prefix sum matrix(say sum[][]) such that sum[i][j] stores the sum of all the elements of the matrix of size i x j.
2. For each row in prefix sum matrix sum[][] using Binary Search do the following:
• Perform Binary search with the lower limit as 0 end the upper limit as to maximum size of square matrix.
• Find the middle index (say mid).
• If the sum of elements of all possible square matrix of size mid is less than or equals to K, then update the lower limit as mid + 1 to find the maximum sum with size greater than mid.
• Else Update the upper limit as mid – 1 to find the maximum sum with size less than mid.
3. Keep updating the maximum size of square matrix in each iteration for the given valid condition above.

Below is the implementation of the above approach:

## Java

 `// Java program for the above approach` `import` `java.util.*;`   `class` `GFG {`   `    ``// Function to find the maximum size` `    ``// of matrix with sum <= K` `    ``static` `void` `findMaxMatrixSize(``int``[][] arr, ``int` `K)` `    ``{` `        ``int` `i, j;`   `        ``// N size of rows and M size of cols` `        ``int` `n = arr.length;` `        ``int` `m = arr[``0``].length;`   `        ``// To store the prefix sum of matrix` `        ``int``[][] sum = ``new` `int``[n + ``1``][m + ``1``];`   `        ``// Create prefix sum` `        ``for` `(i = ``0``; i <= n; i++) {`   `            ``// Traverse each rows` `            ``for` `(j = ``0``; j <= m; j++) {` `                ``if` `(i == ``0` `|| j == ``0``) {` `                    ``sum[i][j] = ``0``;` `                    ``continue``;` `                ``}`   `                ``// Update the prefix sum` `                ``// till index i x j` `                ``sum[i][j] = arr[i - ``1``][j - ``1``]` `                            ``+ sum[i - ``1``][j] + sum[i][j - ``1``]` `                            ``- sum[i - ``1``][j - ``1``];` `            ``}` `        ``}`   `        ``// To store the maximum size of` `        ``// matrix with sum <= K` `        ``int` `ans = ``0``;`   `        ``// Traverse the sum matrix` `        ``for` `(i = ``1``; i <= n; i++) {` `            ``for` `(j = ``1``; j <= m; j++) {`   `                ``// Index out of bound` `                ``if` `(i + ans - ``1` `> n || j + ans - ``1` `> m)` `                    ``break``;`   `                ``int` `mid, lo = ans;`   `                ``// Maximum possible size` `                ``// of matrix` `                ``int` `hi = Math.min(n - i + ``1``, m - j + ``1``);`   `                ``// Binary Search` `                ``while` `(lo < hi) {`   `                    ``// Find middle index` `                    ``mid = (hi + lo + ``1``) / ``2``;`   `                    ``// Check whether sum <= K` `                    ``// or not` `                    ``// If Yes check for other` `                    ``// half of the search` `                    ``if` `(sum[i + mid - ``1``][j + mid - ``1``]` `                            ``+ sum[i - ``1``][j - ``1``]` `                            ``- sum[i + mid - ``1``][j - ``1``]` `                            ``- sum[i - ``1``][j + mid - ``1``]` `                        ``<= K) {` `                        ``lo = mid;` `                    ``}`   `                    ``// Else check it in first` `                    ``// half` `                    ``else` `{` `                        ``hi = mid - ``1``;` `                    ``}` `                ``}`   `                ``// Update the maximum size matrix` `                ``ans = Math.max(ans, lo);` `            ``}` `        ``}`   `        ``// Print the final answer` `        ``System.out.print(ans + ``"\n"``);` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int``[][] arr = { { ``1``, ``1``, ``3``, ``2``, ``4``, ``3``, ``2` `},` `                        ``{ ``1``, ``1``, ``3``, ``2``, ``4``, ``3``, ``2` `},` `                        ``{ ``1``, ``1``, ``3``, ``2``, ``4``, ``3``, ``2` `} };`   `        ``// Given target sum` `        ``int` `K = ``4``;`   `        ``// Function Call` `        ``findMaxMatrixSize(arr, K);` `    ``}` `}`   `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 program for the above approach`   `# Function to find the maximum size` `# of matrix with sum <= K`     `def` `findMaxMatrixSize(arr, K):`   `    ``# N size of rows and M size of cols` `    ``n ``=` `len``(arr)` `    ``m ``=` `len``(arr[``0``])`   `    ``# To store the prefix sum of matrix` `    ``sum` `=` `[[``0` `for` `i ``in` `range``(m ``+` `1``)] ``for` `j ``in` `range``(n ``+` `1``)]`   `    ``# Create Prefix Sum` `    ``for` `i ``in` `range``(n ``+` `1``):`   `        ``# Traverse each rows` `        ``for` `j ``in` `range``(m``+``1``):` `            ``if` `(i ``=``=` `0` `or` `j ``=``=` `0``):` `                ``sum``[i][j] ``=` `0` `                ``continue`   `            ``# Update the prefix sum` `            ``# till index i x j` `            ``sum``[i][j] ``=` `arr[i ``-` `1``][j ``-` `1``] ``+` `sum``[i ``-` `1``][j] ``+` `\` `                ``sum``[i][j ``-` `1``]``-``sum``[i ``-` `1``][j ``-` `1``]`   `    ``# To store the maximum size of` `    ``# matrix with sum <= K` `    ``ans ``=` `0`   `    ``# Traverse the sum matrix` `    ``for` `i ``in` `range``(``1``, n ``+` `1``):` `        ``for` `j ``in` `range``(``1``, m ``+` `1``):`   `            ``# Index out of bound` `            ``if` `(i ``+` `ans ``-` `1` `> n ``or` `j ``+` `ans ``-` `1` `> m):` `                ``break`   `            ``mid ``=` `ans` `            ``lo ``=` `ans`   `            ``# Maximum possible size` `            ``# of matrix` `            ``hi ``=` `min``(n ``-` `i ``+` `1``, m ``-` `j ``+` `1``)`   `            ``# Binary Search` `            ``while` `(lo < hi):`   `                ``# Find middle index` `                ``mid ``=` `(hi ``+` `lo ``+` `1``) ``/``/` `2`   `                ``# Check whether sum <= K` `                ``# or not` `                ``# If Yes check for other` `                ``# half of the search` `                ``if` `(``sum``[i ``+` `mid ``-` `1``][j ``+` `mid ``-` `1``] ``+` `                    ``sum``[i ``-` `1``][j ``-` `1``] ``-` `                    ``sum``[i ``+` `mid ``-` `1``][j ``-` `1``] ``-` `                        ``sum``[i ``-` `1``][j ``+` `mid ``-` `1``] <``=` `K):` `                    ``lo ``=` `mid`   `                ``# Else check it in first` `                ``# half` `                ``else``:` `                    ``hi ``=` `mid ``-` `1`   `            ``# Update the maximum size matrix` `            ``ans ``=` `max``(ans, lo)`   `    ``# Print the final answer` `    ``print``(ans)`     `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `    ``arr ``=` `[[``1``, ``1``, ``3``, ``2``, ``4``, ``3``, ``2``],` `           ``[``1``, ``1``, ``3``, ``2``, ``4``, ``3``, ``2``],` `           ``[``1``, ``1``, ``3``, ``2``, ``4``, ``3``, ``2``]]`   `    ``# Given target sum` `    ``K ``=` `4`   `    ``# Function Call` `    ``findMaxMatrixSize(arr, K)`   `# This code is contributed by Surendra_Gangwar`

## C#

 `// C# program for the above approach` `using` `System;`   `class` `GFG {`   `    ``// Function to find the maximum size` `    ``// of matrix with sum <= K` `    ``static` `void` `findMaxMatrixSize(``int``[, ] arr, ``int` `K)` `    ``{` `        ``int` `i, j;`   `        ``// N size of rows and M size of cols` `        ``int` `n = arr.GetLength(0);` `        ``int` `m = arr.GetLength(1);`   `        ``// To store the prefix sum of matrix` `        ``int``[, ] sum = ``new` `int``[n + 1, m + 1];`   `        ``// Create prefix sum` `        ``for` `(i = 0; i <= n; i++) {`   `            ``// Traverse each rows` `            ``for` `(j = 0; j <= m; j++) {` `                ``if` `(i == 0 || j == 0) {` `                    ``sum[i, j] = 0;` `                    ``continue``;` `                ``}`   `                ``// Update the prefix sum` `                ``// till index i x j` `                ``sum[i, j] = arr[i - 1, j - 1]` `                            ``+ sum[i - 1, j] + sum[i, j - 1]` `                            ``- sum[i - 1, j - 1];` `            ``}` `        ``}`   `        ``// To store the maximum size` `        ``// of matrix with sum <= K` `        ``int` `ans = 0;`   `        ``// Traverse the sum matrix` `        ``for` `(i = 1; i <= n; i++) {` `            ``for` `(j = 1; j <= m; j++) {`   `                ``// Index out of bound` `                ``if` `(i + ans - 1 > n || j + ans - 1 > m)` `                    ``break``;`   `                ``int` `mid, lo = ans;`   `                ``// Maximum possible size` `                ``// of matrix` `                ``int` `hi = Math.Min(n - i + 1, m - j + 1);`   `                ``// Binary Search` `                ``while` `(lo < hi) {`   `                    ``// Find middle index` `                    ``mid = (hi + lo + 1) / 2;`   `                    ``// Check whether sum <= K` `                    ``// or not` `                    ``// If Yes check for other` `                    ``// half of the search` `                    ``if` `(sum[i + mid - 1, j + mid - 1]` `                            ``+ sum[i - 1, j - 1]` `                            ``- sum[i + mid - 1, j - 1]` `                            ``- sum[i - 1, j + mid - 1]` `                        ``<= K) {` `                        ``lo = mid;` `                    ``}`   `                    ``// Else check it in first` `                    ``// half` `                    ``else` `{` `                        ``hi = mid - 1;` `                    ``}` `                ``}`   `                ``// Update the maximum size matrix` `                ``ans = Math.Max(ans, lo);` `            ``}` `        ``}`   `        ``// Print the readonly answer` `        ``Console.Write(ans + ``"\n"``);` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `Main(String[] args)` `    ``{` `        ``int``[, ] arr = { { 1, 1, 3, 2, 4, 3, 2 },` `                        ``{ 1, 1, 3, 2, 4, 3, 2 },` `                        ``{ 1, 1, 3, 2, 4, 3, 2 } };`   `        ``// Given target sum` `        ``int` `K = 4;`   `        ``// Function Call` `        ``findMaxMatrixSize(arr, K);` `    ``}` `}`   `// This code is contributed by Amit Katiyar`

## Javascript

 ``

## C++

 `// C++ program for the above approach`   `#include ` `using` `namespace` `std;`   `// Function to find the maximum size` `// of matrix with sum <= K` `void` `findMaxMatrixSize(vector > arr, ``int` `K)` `{`   `    ``int` `i, j;`   `    ``// N size of rows and M size of cols` `    ``int` `n = arr.size();` `    ``int` `m = arr[0].size();`   `    ``// To store the prefix sum of matrix` `    ``int` `sum[n + 1][m + 1];`   `    ``// Create Prefix Sum` `    ``for` `(``int` `i = 0; i <= n; i++) {`   `        ``// Traverse each rows` `        ``for` `(``int` `j = 0; j <= m; j++) {`   `            ``if` `(i == 0 || j == 0) {` `                ``sum[i][j] = 0;` `                ``continue``;` `            ``}`   `            ``// Update the prefix sum` `            ``// till index i x j` `            ``sum[i][j] = arr[i - 1][j - 1] + sum[i - 1][j]` `                        ``+ sum[i][j - 1] - sum[i - 1][j - 1];` `        ``}` `    ``}`   `    ``// To store the maximum size of` `    ``// matrix with sum <= K` `    ``int` `ans = 0;`   `    ``// Traverse the sum matrix` `    ``for` `(i = 1; i <= n; i++) {`   `        ``for` `(j = 1; j <= m; j++) {`   `            ``// Index out of bound` `            ``if` `(i + ans - 1 > n || j + ans - 1 > m)` `                ``break``;`   `            ``int` `mid, lo = ans;`   `            ``// Maximum possible size` `            ``// of matrix` `            ``int` `hi = min(n - i + 1, m - j + 1);`   `            ``// Binary Search` `            ``while` `(lo < hi) {`   `                ``// Find middle index` `                ``mid = (hi + lo + 1) / 2;`   `                ``// Check whether sum <= K` `                ``// or not` `                ``// If Yes check for other` `                ``// half of the search` `                ``if` `(sum[i + mid - 1][j + mid - 1]` `                        ``+ sum[i - 1][j - 1]` `                        ``- sum[i + mid - 1][j - 1]` `                        ``- sum[i - 1][j + mid - 1]` `                    ``<= K) {` `                    ``lo = mid;` `                ``}`   `                ``// Else check it in first` `                ``// half` `                ``else` `{` `                    ``hi = mid - 1;` `                ``}` `            ``}`   `            ``// Update the maximum size matrix` `            ``ans = max(ans, lo);` `        ``}` `    ``}`   `    ``// Print the final answer` `    ``cout << ans << endl;` `}`   `// Driver Code` `int` `main()` `{` `    ``vector > arr;`   `    ``arr = { { 1, 1, 3, 2, 4, 3, 2 },` `            ``{ 1, 1, 3, 2, 4, 3, 2 },` `            ``{ 1, 1, 3, 2, 4, 3, 2 } };`   `    ``// Given target sum` `    ``int` `K = 4;`   `    ``// Function Call` `    ``findMaxMatrixSize(arr, K);` `    ``return` `0;` `}`

Output

`2`

Time Complexity: O(N*N*log(N))
Auxiliary Space: O(M*N)

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