# Find sub-matrix with the given sum

Given an N x N matrix and two integers S and K, the task is to find whether there exists a K x K sub-matrix with sum equal to S.

Examples:

Input: K = 2, S = 14, mat[][] = {
{ 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 },
{ 13, 14, 15, 16 }}
Output: Yes
1 2
5 6
is the required 2 x 2 sub-matrix with element sum = 14

Input: K = 1, S = 5, mat[][] = {{1, 2}, {7, 8}}
Output: No

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

Approach: Dynamic programming can be used to solve this problem,

• Create an array dp[N + 1][N + 1] where dp[i][j] stores the sum of all the elements with row between 1 to i and column between 1 to j.
• Once the 2-D matrix is generated, now suppose we wish to find sum of square starting with (i, j) to (i + x, j + x). The required sum will be dp[i + x][j + x] – dp[i][j + x] – dp[i + x][j] + dp[i][j] where,
1. First term denotes the sum of all the elements present in rows between 1 to i + x and columns between 1 to j + x. This area has our required square.
2. Second two terms is to remove the area which is outside our required region but inside the region calculated in the first step.
3. Sum of elements of rows between 1 to i and columns between 1 to j is subtracted twice in the second step, so it is added once.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` `#define ll long long int ` `#define N 4 ` ` `  `// Function to return the sum of the sub-matrix ` `int` `getSum(``int` `r1, ``int` `r2, ``int` `c1, ``int` `c2, ` `           ``int` `dp[N + 1][N + 1]) ` `{ ` `    ``return` `dp[r2][c2] - dp[r2][c1] ` `           ``- dp[r1][c2] + dp[r1][c1]; ` `} ` ` `  `// Function that returns true if it is possible ` `// to find the sub-matrix with required sum ` `bool` `sumFound(``int` `K, ``int` `S, ``int` `grid[N][N]) ` `{ ` ` `  `    ``// 2-D array to store the sum of ` `    ``// all the sub-matrices ` `    ``int` `dp[N + 1][N + 1]; ` ` `  `    ``// Filling of dp[][] array ` `    ``for` `(``int` `i = 0; i < N; i++) ` `        ``for` `(``int` `j = 0; j < N; j++) ` `            ``dp[i + 1][j + 1] = dp[i + 1][j] + dp[i][j + 1] ` `                               ``- dp[i][j] + grid[i][j]; ` ` `  `    ``// Checking for each possible sub-matrix of size k X k ` `    ``for` `(``int` `i = 0; i < N; i += K) ` `        ``for` `(``int` `j = 0; j < N; j += K) { ` `            ``int` `sum = getSum(i, i + K, j, j + K, dp); ` ` `  `            ``if` `(sum == S) ` `                ``return` `true``; ` `        ``} ` ` `  `    ``// Sub-matrix with the given sum not found ` `    ``return` `false``; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `grid[N][N] = { { 1, 2, 3, 4 }, ` `                       ``{ 5, 6, 7, 8 }, ` `                       ``{ 9, 10, 11, 12 }, ` `                       ``{ 13, 14, 15, 16 } }; ` `    ``int` `K = 2; ` `    ``int` `S = 14; ` ` `  `    ``if` `(sumFound(K, S, grid)) ` `        ``cout << ``"Yes"` `<< endl; ` `    ``else` `        ``cout << ``"No"` `<< endl; ` `} `

## PHP

 ` `

Output:

```Yes
```

My Personal Notes arrow_drop_up