# Find Maximum Length Of A Square Submatrix Having Sum Of Elements At-Most K

Given a N x M matrix where N is the number of rows and M is the number of columns in the given matrix and an integer K. The task is to find maximum length of a square submatrix having sum of elements less than or equal to K or print 0 if no such square exits.

Examples:

```Input: r = 4, c = 4 , k = 6
matrix[][] = { {1, 1, 1, 1},
{1, 0, 0, 0},
{1, 0, 0, 0},
{1, 0, 0, 0},
}
Output: 3
Explanation:
Square from (0,0) to (2,2) with
sum 5 is one of the valid answer.

Input: r = 4, c = 4 , k = 1
matrix[][] = { {2, 2, 2},
{2, 2, 2},
{2, 2, 2},
{2, 2, 2},
}
Output: 0
Explanation:
```

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

Approach:
The idea is to calculate prefix sum matrix. Once the prefix sum matrix is calculated, required sub-matrix sum can be calculate in O(1) time complexity. Use sliding window technique to calculate the maximum length square submatrix. For every square of length cur_max+1, where cur_max is the currently found maximum length of a square submatrix, we check whether the sum of elements in the current submatrix, having length of cur_max+1, is less than or equal to K or not. If yes than increase the result by 1 otherwise we continue to check.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach ` `#include ` `using` `namespace` `std; ` ` `  `    ``// Function to return maximum  ` `    ``// length of square submatrix ` `    ``// having sum of elements at-most K ` `    ``int` `maxLengthSquare(``int` `row, ``int` `column,  ` `                        ``int` `arr[], ``int` `k) ` `    ``{ ` `        ``// Matrix to store prefix sum ` `        ``int` `sum[row + 1][column + 1] ; ` `     `  `        ``for``(``int` `i = 1; i <= row; i++) ` `            ``for``(``int` `j = 0; j <= column; j++) ` `                ``sum[i][j] = 0; ` `                 `  `        ``// Current maximum length ` `        ``int` `cur_max = 1; ` `     `  `        ``// Variable for storing  ` `        ``// maximum length of square ` `        ``int` `max = 0; ` `             `  `        ``for` `(``int` `i = 1; i <= row; i++)  ` `        ``{ ` `            ``for` `(``int` `j = 1; j <= column; j++)  ` `            ``{ ` `                ``// Calculating prefix sum ` `                ``sum[i][j] = sum[i - 1][j] + sum[i][j - 1] +  ` `                            ``arr[i - 1][j - 1] - sum[i - 1][j - 1]; ` `         `  `                ``// Checking whether there  ` `                ``// exits square with length  ` `                ``// cur_max+1 or not ` `                ``if``(i >= cur_max && j >= cur_max &&  ` `                    ``sum[i][j] - sum[i - cur_max][j] ` `                    ``- sum[i][j - cur_max] +  ` `                    ``sum[i - cur_max][j - cur_max] <= k) ` `                ``{ ` `                    ``max = cur_max++; ` `                ``} ` `            ``} ` `        ``} ` `     `  `        ``// Returning the  ` `        ``// maximum length ` `        ``return` `max; ` `    ``} ` ` `  `    ``// Driver code ` `    ``int` `main()  ` `    ``{ ` `         `  `        ``int` `row = 4, column = 4; ` `        ``int` `matrix = { {1, 1, 1, 1}, ` `                        ``{1, 0, 0, 0}, ` `                        ``{1, 0, 0, 0}, ` `                        ``{1, 0, 0, 0} ` `                        ``}; ` `     `  `        ``int` `k = 6;  ` `        ``int` `ans = maxLengthSquare(row, column, matrix, k); ` `        ``cout << ans; ` `         `  `        ``return` `0; ` `    ``} ` ` `  `// This code is contributed by AnkitRai01 `

## Java

 `// Java implementation of  ` `// the above approach ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` `    ``// Function to return maximum  ` `    ``// length of square submatrix ` `    ``// having sum of elements at-most K ` `    ``public` `static` `int` `maxLengthSquare(``int` `row,``int` `column, ` `                                        ``int``[][] arr, ``int` `k) ` `    ``{ ` `        ``// Matrix to store prefix sum ` `        ``int` `sum[][] = ``new` `int``[row + ``1``][column + ``1``]; ` `     `  `        ``// Current maximum length ` `        ``int` `cur_max = ``1``; ` `     `  `        ``// Variable for storing  ` `        ``// maximum length of square ` `        ``int` `max = ``0``; ` `             `  `        ``for` `(``int` `i = ``1``; i <= row; i++)  ` `        ``{ ` `            ``for` `(``int` `j = ``1``; j <= column; j++)  ` `            ``{ ` `                ``// Calculating prefix sum ` `                ``sum[i][j] = sum[i - ``1``][j] + sum[i][j - ``1``] +  ` `                            ``arr[i - ``1``][j - ``1``] - sum[i - ``1``][j - ``1``]; ` `         `  `                ``// Checking whether there  ` `                ``// exits square with length  ` `                ``// cur_max+1 or not ` `                ``if``(i >=cur_max&&j>=cur_max&&sum[i][j]-sum[i - cur_max][j] ` `                            ``- sum[i][j - cur_max]  ` `                            ``+ sum[i - cur_max][j - cur_max] <= k){ ` `                    ``max = cur_max++; ` `                ``} ` `            ``} ` `        ``} ` `     `  `        ``// Returning the  ` `        ``// maximum length ` `        ``return` `max; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String args[])  ` `    ``{ ` `         `  `        ``int` `row = ``4` `, column = ``4``; ` `        ``int` `matrix[][] = { {``1``, ``1``, ``1``, ``1``}, ` `                        ``{``1``, ``0``, ``0``, ``0``}, ` `                        ``{``1``, ``0``, ``0``, ``0``}, ` `                        ``{``1``, ``0``, ``0``, ``0``} ` `                        ``}; ` `     `  `        ``int` `k = ``6``;  ` `        ``int` `ans = maxLengthSquare(row,column,matrix, k); ` `        ``System.out.println(ans); ` `    ``} ` `} `

## Python3

 `# Python3 implementation of the above approach  ` `import` `numpy as np ` ` `  `# Function to return maximum  ` `# length of square submatrix  ` `# having sum of elements at-most K  ` `def` `maxLengthSquare(row, column, arr, k) : ` `     `  `    ``# Matrix to store prefix sum  ` `    ``sum` `=` `np.zeros((row ``+` `1``, column ``+` `1``));  ` `     `  `    ``# Current maximum length  ` `    ``cur_max ``=` `1``;  ` `     `  `    ``# Variable for storing  ` `    ``# maximum length of square  ` `    ``max` `=` `0``;  ` `             `  `    ``for` `i ``in` `range``(``1``, row ``+` `1``) : ` `        ``for` `j ``in` `range``(``1``, column ``+` `1``) : ` `             `  `            ``# Calculating prefix sum ` `            ``sum``[i][j] ``=` `sum``[i ``-` `1``][j] ``+` `sum``[i][j ``-` `1``] ``+` `\ ` `                        ``arr[i ``-` `1``][j ``-` `1``] ``-` `\ ` `                        ``sum``[i ``-` `1``][j ``-` `1``]; ` `             `  `            ``# Checking whether there ` `            ``# exits square with length ` `            ``# cur_max+1 or not ` `            ``if``(i >``=` `cur_max ``and` `j >``=` `cur_max ``and`  `                 ``sum``[i][j] ``-` `sum``[i ``-` `cur_max][j] ``-` `sum``[i][j ``-`  `                                     ``cur_max] ``+` `sum``[i ``-`  `                                     ``cur_max][j ``-` `cur_max] <``=` `k) : ` `                ``max` `=` `cur_max;  ` `                ``cur_max ``+``=` `1``; ` `     `  `    ``# Returning the maximum length  ` `    ``return` `max``;  ` `     `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `: ` `     `  `    ``row ``=` `4` `; ` `    ``column ``=` `4``; ` `    ``matrix ``=` `[[``1``, ``1``, ``1``, ``1``], ` `              ``[``1``, ``0``, ``0``, ``0``], ` `              ``[``1``, ``0``, ``0``, ``0``], ` `              ``[``1``, ``0``, ``0``, ``0``]]; ` `    ``k ``=` `6``; ` `    ``ans ``=` `maxLengthSquare(row, column, matrix, k); ` `    ``print``(ans);  ` `     `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# implementation of the above approach  ` `using` `System; ` ` `  `class` `GFG  ` `{  ` `    ``// Function to return maximum  ` `    ``// length of square submatrix  ` `    ``// having sum of elements at-most K  ` `    ``public` `static` `int` `maxLengthSquare(``int` `row,``int` `column,  ` `                                        ``int``[,] arr, ``int` `k)  ` `    ``{  ` `        ``// Matrix to store prefix sum  ` `        ``int` `[,]sum = ``new` `int``[row + 1,column + 1];  ` `     `  `        ``// Current maximum length  ` `        ``int` `cur_max = 1;  ` `     `  `        ``// Variable for storing  ` `        ``// maximum length of square  ` `        ``int` `max = 0;  ` `             `  `        ``for` `(``int` `i = 1; i <= row; i++)  ` `        ``{  ` `            ``for` `(``int` `j = 1; j <= column; j++)  ` `            ``{  ` `                ``// Calculating prefix sum  ` `                ``sum[i, j] = sum[i - 1, j] + sum[i, j - 1] +  ` `                            ``arr[i - 1, j - 1] - sum[i - 1, j - 1];  ` `         `  `                ``// Checking whether there  ` `                ``// exits square with length  ` `                ``// cur_max+1 or not  ` `                ``if``(i >=cur_max && j>=cur_max && sum[i, j]-sum[i - cur_max, j]  ` `                            ``- sum[i, j - cur_max]  ` `                            ``+ sum[i - cur_max, j - cur_max] <= k) ` `                ``{  ` `                    ``max = cur_max++;  ` `                ``}  ` `            ``}  ` `        ``}  ` `     `  `        ``// Returning the  ` `        ``// maximum length  ` `        ``return` `max;  ` `    ``}  ` ` `  `    ``// Driver code  ` `    ``public` `static` `void` `Main()  ` `    ``{  ` `         `  `        ``int` `row = 4 , column = 4;  ` `        ``int` `[,]matrix = { {1, 1, 1, 1},  ` `                        ``{1, 0, 0, 0},  ` `                        ``{1, 0, 0, 0},  ` `                        ``{1, 0, 0, 0}  ` `                        ``};  ` `     `  `        ``int` `k = 6;  ` `        ``int` `ans = maxLengthSquare(row, column, matrix, k);  ` `        ``Console.WriteLine(ans);  ` `    ``}  ` `}  ` ` `  `// This code is contributed by AnkitRai01 `

Output:

```3
```

Time Complexity: O(N x M)

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Improved By : AnkitRai01