# Finding the maximum square sub-matrix with all equal elements

Given a N x N matrix, determine the maximum K such that K x K is a submatrix with all equal elements i.e., all the elements in this submatrix must be same.

Constraints:
1 <= N <= 1000
0 <= Ai , j <= 109

Examples:

```
Input : a[][] = {{2, 3, 3},
{2, 3, 3},
{2, 2, 2}}
Output : 2
Explanation: A 2x2 matrix is formed from index
A0,1 to A1,2

Input : a[][]  = {{9, 9, 9, 8},
{9, 9, 9, 6},
{9, 9, 9, 3},
{2, 2, 2, 2}
Output : 3
Explanation : A 3x3 matrix is formed from index
A0,0 to A2,2
```

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

Method I ( Naive approach )
We can easily find all the square submatrices in O(n3) time and check whether each submatrix contains equal elements or not in O(n2) time Which makes the total running time of the algorithm as O(n5).

Method II ( Dynamic Programming )
For each cell (i, j), we store the largest value of K such that K x K is a submatrix with all equal elements and position of (i, j) being the bottom-right most element.

And DPi,j depends upon {DPi-1, j, DPi, j-1, DPi-1, j-1}

```If Ai, j is equal to {Ai-1, j, Ai, j-1, Ai-1, j-1},
all the three values:
DPi, j = min(DPi-1, j, DPi, j-1, DPi-1, j-1) + 1
Else
DPi, j = 1  // Matrix Size 1

The answer would be the maximum of all DPi, j's
```

Below is C++ implementation of above steps.

```// C++ program to find maximum K such that K x K
// is a submatrix with equal elements.
#include<bits/stdc++.h>
#define Row 6
#define Col 6
using namespace std;

// Returns size of the largest square sub-matrix
// with all same elements.
int largestKSubmatrix(int a[][Col])
{
int dp[Row][Col];
memset(dp, sizeof(dp), 0);

int result = 0;
for (int i = 0 ; i < Row ; i++)
{
for (int j = 0 ; j < Col ; j++)
{
// If elements is at top row or first
// column, it wont form a  square
// matrix's bottom-right
if (i == 0 || j == 0)
dp[i][j] = 1;

else
{
// Check if adjacent elements are equal
if (a[i][j] == a[i-1][j] &&
a[i][j] == a[i][j-1] &&
a[i][j] == a[i-1][j-1] )
dp[i][j] = min(min(dp[i-1][j], dp[i][j-1]),
dp[i-1][j-1] ) + 1;

// If not equal, then it will form a 1x1
// submatrix
else dp[i][j] = 1;
}

// Update result at each (i,j)
result = max(result, dp[i][j]);
}
}

return result;
}

// Driven Program
int main()
{
int a[Row][Col] = { 2, 2, 3, 3, 4, 4,
5, 5, 7, 7, 7, 4,
1, 2, 7, 7, 7, 4,
4, 4, 7, 7, 7, 4,
5, 5, 5, 1, 2, 7,
8, 7, 9, 4, 4, 4
};

cout << largestKSubmatrix(a) << endl;

return 0;
}
```

Output:

```3
```

Time Complexity : O(Row * Col)
Auxiliary Space : O(Row * Col)

This article is contributed by Shubham Gupta. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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