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# Given a matrix of ‘O’ and ‘X’, find the largest subsquare surrounded by ‘X’

Given a matrix where every element is either ‘O’ or ‘X’, find the largest subsquare surrounded by ‘X’.
In the below article, it is assumed that the given matrix is also a square matrix. The code given below can be easily extended for rectangular matrices.

Examples:

```Input: mat[N][N] = { {'X', 'O', 'X', 'X', 'X'},
{'X', 'X', 'X', 'X', 'X'},
{'X', 'X', 'O', 'X', 'O'},
{'X', 'X', 'X', 'X', 'X'},
{'X', 'X', 'X', 'O', 'O'},
};
Output: 3
The square submatrix starting at (1, 1) is the largest
submatrix surrounded by 'X'

Input: mat[M][N] = { {'X', 'O', 'X', 'X', 'X', 'X'},
{'X', 'O', 'X', 'X', 'O', 'X'},
{'X', 'X', 'X', 'O', 'O', 'X'},
{'X', 'X', 'X', 'X', 'X', 'X'},
{'X', 'X', 'X', 'O', 'X', 'O'},
};
Output: 4
The square submatrix starting at (0, 2) is the largest
submatrix surrounded by 'X'```

A Simple Solution is to consider every square submatrix and check whether is has all corner edges filled with ‘X’. The time complexity of this solution is O(N4).
We can solve this problem in O(N3) time using extra space. The idea is to create two auxiliary arrays hor[N][N] and ver[N][N]. The value stored in hor[i][j] is the number of horizontal continuous ‘X’ characters till mat[i][j] in mat[][]. Similarly, the value stored in ver[i][j] is the number of vertical continuous ‘X’ characters till mat[i][j] in mat[][].

Example:

```mat =  X  O  X  X  X  X
X  O  X  X  O  X
X  X  X  O  O  X
O  X  X  X  X  X
X  X  X  O  X  O
O  O  X  O  O  O

hor = 1  0  1  2  3  4
1  0  1  2  0  1
1  2  3  0  0  1
0  1  2  3  4  5
1  2  3  0  1  0
0  0  1  0  0  0

ver = 1  0  1  1  1  1
2  0  2  2  0  2
3  1  3  0  0  3
0  2  4  1  1  4
1  3  5  0  2  0
0  0  6  0  0  0```

Once we have filled values in hor[][] and ver[][], we start from the bottommost-rightmost corner of matrix and move toward the leftmost-topmost in row by row manner. For every visited entry mat[i][j], we compare the values of hor[i][j] and ver[i][j], and pick the smaller of two as we need a square. Let the smaller of two be ‘small’. After picking smaller of two, we check if both ver[][] and hor[][] for left and up edges respectively. If they have entries for the same, then we found a subsquare. Otherwise we try for small-1.

Below is implementation of the above idea.

## C++

 `// A C++ program to find  the largest subsquare` `// surrounded by 'X' in a given matrix of 'O' and 'X'` `#include ` `using` `namespace` `std;`   `// Size of given matrix is N X N` `#define N 6`   `// A utility function to find minimum of two numbers` `int` `getMin(``int` `x, ``int` `y) { ``return` `(x < y) ? x : y; }`   `// Returns size of maximum size subsquare matrix` `// surrounded by 'X'` `int` `findSubSquare(``int` `mat[][N])` `{` `    ``int` `max = 0; ``// Initialize result`   `    ``// Initialize the left-top value in hor[][] and ver[][]` `    ``int` `hor[N][N], ver[N][N];` `    ``hor = ver = (mat == ``'X'``);`   `    ``// Fill values in hor[][] and ver[][]` `    ``for` `(``int` `i = 0; i < N; i++) ` `    ``{` `        ``for` `(``int` `j = 0; j < N; j++) ` `        ``{` `            ``if` `(mat[i][j] == ``'O'``)` `                ``ver[i][j] = hor[i][j] = 0;` `            ``else` `            ``{` `                ``hor[i][j]` `                    ``= (j == 0) ? 1 : hor[i][j - 1] + 1;` `                ``ver[i][j]` `                    ``= (i == 0) ? 1 : ver[i - 1][j] + 1;` `            ``}` `        ``}` `    ``}`   `    ``// Start from the rightmost-bottommost corner element` `    ``// and find the largest ssubsquare with the help of` `    ``// hor[][] and ver[][]` `    ``for` `(``int` `i = N - 1; i >= 1; i--) ` `    ``{` `        ``for` `(``int` `j = N - 1; j >= 1; j--) ` `        ``{` `            ``// Find smaller of values in hor[][] and ver[][]` `            ``// A Square can only be made by taking smaller` `            ``// value` `            ``int` `small = getMin(hor[i][j], ver[i][j]);`   `            ``// At this point, we are sure that there is a` `            ``// right vertical line and bottom horizontal` `            ``// line of length at least 'small'.`   `            ``// We found a bigger square if following` `            ``// conditions are met: 1)If side of square is` `            ``// greater than max. 2)There is a left vertical` `            ``// line of length >= 'small' 3)There is a top` `            ``// horizontal line of length >= 'small'` `            ``while` `(small > max) ` `            ``{` `                ``if` `(ver[i][j - small + 1] >= small` `                    ``&& hor[i - small + 1][j] >= small) ` `                ``{` `                    ``max = small;` `                ``}` `                ``small--;` `            ``}` `        ``}` `    ``}` `    ``return` `max;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `mat[][N] = {` `        ``{ ``'X'``, ``'O'``, ``'X'``, ``'X'``, ``'X'``, ``'X'` `},` `        ``{ ``'X'``, ``'O'``, ``'X'``, ``'X'``, ``'O'``, ``'X'` `},` `        ``{ ``'X'``, ``'X'``, ``'X'``, ``'O'``, ``'O'``, ``'X'` `},` `        ``{ ``'O'``, ``'X'``, ``'X'``, ``'X'``, ``'X'``, ``'X'` `},` `        ``{ ``'X'``, ``'X'``, ``'X'``, ``'O'``, ``'X'``, ``'O'` `},` `        ``{ ``'O'``, ``'O'``, ``'X'``, ``'O'``, ``'O'``, ``'O'` `},` `    ``};` `  `  `    ``// Function call` `    ``cout << findSubSquare(mat);` `    ``return` `0;` `}`

## Java

 `// A JAVA program to find the` `// largest subsquare surrounded` `// by 'X' in a given matrix of` `// 'O' and 'X'` `import` `java.util.*;`   `class` `GFG ` `{` `    ``// Size of given` `    ``// matrix is N X N` `    ``static` `int` `N = ``6``;`   `    ``// A utility function to` `    ``// find minimum of two numbers` `    ``static` `int` `getMin(``int` `x, ``int` `y)` `    ``{` `        ``return` `(x < y) ? x : y;` `    ``}`   `    ``// Returns size of maximum` `    ``// size subsquare matrix` `    ``// surrounded by 'X'` `    ``static` `int` `findSubSquare(``int` `mat[][])` `    ``{` `        ``int` `max = ``0``; ``// Initialize result`   `        ``// Initialize the left-top` `        ``// value in hor[][] and ver[][]` `        ``int` `hor[][] = ``new` `int``[N][N];` `        ``int` `ver[][] = ``new` `int``[N][N];` `        ``hor[``0``][``0``] = ver[``0``][``0``] = ``'X'``;`   `        ``// Fill values in` `        ``// hor[][] and ver[][]` `        ``for` `(``int` `i = ``0``; i < N; i++) ` `        ``{` `            ``for` `(``int` `j = ``0``; j < N; j++)` `            ``{` `                ``if` `(mat[i][j] == ``'O'``)` `                    ``ver[i][j] = hor[i][j] = ``0``;` `                ``else` `                ``{` `                    ``hor[i][j]` `                        ``= (j == ``0``) ? ``1` `: hor[i][j - ``1``] + ``1``;` `                    ``ver[i][j]` `                        ``= (i == ``0``) ? ``1` `: ver[i - ``1``][j] + ``1``;` `                ``}` `            ``}` `        ``}`   `        ``// Start from the rightmost-` `        ``// bottommost corner element` `        ``// and find the largest` `        ``// subsquare with the help` `        ``// of hor[][] and ver[][]` `        ``for` `(``int` `i = N - ``1``; i >= ``1``; i--) ` `        ``{` `            ``for` `(``int` `j = N - ``1``; j >= ``1``; j--) ` `            ``{` `                ``// Find smaller of values in` `                ``// hor[][] and ver[][] A Square` `                ``// can only be made by taking` `                ``// smaller value` `                ``int` `small = getMin(hor[i][j], ver[i][j]);`   `                ``// At this point, we are sure` `                ``// that there is a right vertical` `                ``// line and bottom horizontal` `                ``// line of length at least 'small'.`   `                ``// We found a bigger square` `                ``// if following conditions` `                ``// are met:` `                ``// 1)If side of square` `                ``//   is greater than max.` `                ``// 2)There is a left vertical` `                ``//   line of length >= 'small'` `                ``// 3)There is a top horizontal` `                ``//   line of length >= 'small'` `                ``while` `(small > max) ` `                ``{` `                    ``if` `(ver[i][j - small + ``1``] >= small` `                        ``&& hor[i - small + ``1``][j] >= small) ` `                    ``{` `                        ``max = small;` `                    ``}` `                    ``small--;` `                ``}` `            ``}` `        ``}` `        ``return` `max;` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``// TODO Auto-generated method stub`   `        ``int` `mat[][] = { { ``'X'``, ``'O'``, ``'X'``, ``'X'``, ``'X'``, ``'X'` `},` `                        ``{ ``'X'``, ``'O'``, ``'X'``, ``'X'``, ``'O'``, ``'X'` `},` `                        ``{ ``'X'``, ``'X'``, ``'X'``, ``'O'``, ``'O'``, ``'X'` `},` `                        ``{ ``'O'``, ``'X'``, ``'X'``, ``'X'``, ``'X'``, ``'X'` `},` `                        ``{ ``'X'``, ``'X'``, ``'X'``, ``'O'``, ``'X'``, ``'O'` `},` `                        ``{ ``'O'``, ``'O'``, ``'X'``, ``'O'``, ``'O'``, ``'O'` `} };` `      `  `        ``// Function call` `        ``System.out.println(findSubSquare(mat));` `    ``}` `}`   `// This code is contributed` `// by ChitraNayal`

## Python3

 `# A Python3 program to find the largest` `# subsquare surrounded by 'X' in a given` `# matrix of 'O' and 'X'` `import` `math as mt`   `# Size of given matrix is N X N` `N ``=` `6`   `# A utility function to find minimum` `# of two numbers`     `def` `getMin(x, y):` `    ``if` `x < y:` `        ``return` `x` `    ``else``:` `        ``return` `y`   `# Returns size of Maximum size` `# subsquare matrix surrounded by 'X'`     `def` `findSubSquare(mat):`   `    ``Max` `=` `0`  `# Initialize result`   `    ``# Initialize the left-top value` `    ``# in hor[][] and ver[][]` `    ``hor ``=` `[[``0` `for` `i ``in` `range``(N)]` `           ``for` `i ``in` `range``(N)]` `    ``ver ``=` `[[``0` `for` `i ``in` `range``(N)]` `           ``for` `i ``in` `range``(N)]`   `    ``if` `mat[``0``][``0``] ``=``=` `'X'``:` `        ``hor[``0``][``0``] ``=` `1` `        ``ver[``0``][``0``] ``=` `1`   `    ``# Fill values in hor[][] and ver[][]` `    ``for` `i ``in` `range``(N):`   `        ``for` `j ``in` `range``(N):`   `            ``if` `(mat[i][j] ``=``=` `'O'``):` `                ``ver[i][j], hor[i][j] ``=` `0``, ``0` `            ``else``:` `                ``if` `j ``=``=` `0``:` `                    ``ver[i][j], hor[i][j] ``=` `1``, ``1` `                ``else``:` `                    ``(ver[i][j],` `                     ``hor[i][j]) ``=` `(ver[i ``-` `1``][j] ``+` `1``,` `                                   ``hor[i][j ``-` `1``] ``+` `1``)`   `    ``# Start from the rightmost-bottommost corner` `    ``# element and find the largest ssubsquare` `    ``# with the help of hor[][] and ver[][]` `    ``for` `i ``in` `range``(N ``-` `1``, ``0``, ``-``1``):`   `        ``for` `j ``in` `range``(N ``-` `1``, ``0``, ``-``1``):`   `            ``# Find smaller of values in hor[][] and` `            ``# ver[][]. A Square can only be made by` `            ``# taking smaller value` `            ``small ``=` `getMin(hor[i][j], ver[i][j])`   `            ``# At this point, we are sure that there` `            ``# is a right vertical line and bottom` `            ``# horizontal line of length at least 'small'.`   `            ``# We found a bigger square if following` `            ``# conditions are met:` `            ``# 1)If side of square is greater than Max.` `            ``# 2)There is a left vertical line` `            ``#   of length >= 'small'` `            ``# 3)There is a top horizontal line` `            ``#   of length >= 'small'` `            ``while` `(small > ``Max``):`   `                ``if` `(ver[i][j ``-` `small ``+` `1``] >``=` `small ``and` `                        ``hor[i ``-` `small ``+` `1``][j] >``=` `small):`   `                    ``Max` `=` `small`   `                ``small ``-``=` `1`   `    ``return` `Max`     `# Driver Code` `mat ``=` `[[``'X'``, ``'O'``, ``'X'``, ``'X'``, ``'X'``, ``'X'``],` `       ``[``'X'``, ``'O'``, ``'X'``, ``'X'``, ``'O'``, ``'X'``],` `       ``[``'X'``, ``'X'``, ``'X'``, ``'O'``, ``'O'``, ``'X'``],` `       ``[``'O'``, ``'X'``, ``'X'``, ``'X'``, ``'X'``, ``'X'``],` `       ``[``'X'``, ``'X'``, ``'X'``, ``'O'``, ``'X'``, ``'O'``],` `       ``[``'O'``, ``'O'``, ``'X'``, ``'O'``, ``'O'``, ``'O'``]]`   `# Function call` `print``(findSubSquare(mat))`   `# This code is contributed by` `# Mohit kumar 29`

## C#

 `// A C# program to find the` `// largest subsquare surrounded` `// by 'X' in a given matrix of` `// 'O' and 'X'` `using` `System;`   `class` `GFG ` `{` `    ``// Size of given` `    ``// matrix is N X N` `    ``static` `int` `N = 6;`   `    ``// A utility function to` `    ``// find minimum of two numbers` `    ``static` `int` `getMin(``int` `x, ``int` `y)` `    ``{` `        ``return` `(x < y) ? x : y;` `    ``}`   `    ``// Returns size of maximum` `    ``// size subsquare matrix` `    ``// surrounded by 'X'` `    ``static` `int` `findSubSquare(``int``[, ] mat)` `    ``{` `        ``int` `max = 0; ``// Initialize result`   `        ``// Initialize the left-top` `        ``// value in hor[][] and ver[][]` `        ``int``[, ] hor = ``new` `int``[N, N];` `        ``int``[, ] ver = ``new` `int``[N, N];` `        ``hor[0, 0] = ver[0, 0] = ``'X'``;`   `        ``// Fill values in` `        ``// hor[][] and ver[][]` `        ``for` `(``int` `i = 0; i < N; i++) ` `        ``{` `            ``for` `(``int` `j = 0; j < N; j++) ` `            ``{` `                ``if` `(mat[i, j] == ``'O'``)` `                    ``ver[i, j] = hor[i, j] = 0;` `                ``else` `                ``{` `                    ``hor[i, j]` `                        ``= (j == 0) ? 1 : hor[i, j - 1] + 1;` `                    ``ver[i, j]` `                        ``= (i == 0) ? 1 : ver[i - 1, j] + 1;` `                ``}` `            ``}` `        ``}`   `        ``// Start from the rightmost-` `        ``// bottommost corner element` `        ``// and find the largest` `        ``// subsquare with the help` `        ``// of hor[][] and ver[][]` `        ``for` `(``int` `i = N - 1; i >= 1; i--) ` `        ``{` `            ``for` `(``int` `j = N - 1; j >= 1; j--) ` `            ``{` `                ``// Find smaller of values in` `                ``// hor[][] and ver[][] A Square` `                ``// can only be made by taking` `                ``// smaller value` `                ``int` `small = getMin(hor[i, j], ver[i, j]);`   `                ``// At this point, we are sure` `                ``// that there is a right vertical` `                ``// line and bottom horizontal` `                ``// line of length at least 'small'.`   `                ``// We found a bigger square` `                ``// if following conditions` `                ``// are met:` `                ``// 1)If side of square` `                ``// is greater than max.` `                ``// 2)There is a left vertical` `                ``// line of length >= 'small'` `                ``// 3)There is a top horizontal` `                ``// line of length >= 'small'` `                ``while` `(small > max)` `                ``{` `                    ``if` `(ver[i, j - small + 1] >= small` `                        ``&& hor[i - small + 1, j] >= small) ` `                    ``{` `                        ``max = small;` `                    ``}` `                    ``small--;` `                ``}` `            ``}` `        ``}` `        ``return` `max;` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `Main()` `    ``{` `        ``// TODO Auto-generated method stub`   `        ``int``[, ] mat = { { ``'X'``, ``'O'``, ``'X'``, ``'X'``, ``'X'``, ``'X'` `},` `                        ``{ ``'X'``, ``'O'``, ``'X'``, ``'X'``, ``'O'``, ``'X'` `},` `                        ``{ ``'X'``, ``'X'``, ``'X'``, ``'O'``, ``'O'``, ``'X'` `},` `                        ``{ ``'O'``, ``'X'``, ``'X'``, ``'X'``, ``'X'``, ``'X'` `},` `                        ``{ ``'X'``, ``'X'``, ``'X'``, ``'O'``, ``'X'``, ``'O'` `},` `                        ``{ ``'O'``, ``'O'``, ``'X'``, ``'O'``, ``'O'``, ``'O'` `} };` `      `  `        ``// Function call` `        ``Console.WriteLine(findSubSquare(mat));` `    ``}` `}`   `// This code is contributed` `// by Akanksha Rai(Abby_akku)`

## PHP

 `= 1; ``\$i``--)` `    ``{` `        ``for` `(``\$j` `= ``\$GLOBALS``[``'N'``] - 1; ``\$j` `>= 1; ``\$j``--)` `        ``{` `            ``// Find smaller of values in ` `            ``// \$hor and \$ver A Square can ` `            ``// only be made by taking ` `            ``// smaller value` `            ``\$small` `= getMin(``\$hor``[``\$i``][``\$j``],` `                            ``\$ver``[``\$i``][``\$j``]);`   `            ``// At this point, we are sure ` `            ``// that there is a right vertical ` `            ``// line and bottom horizontal ` `            ``// line of length at least '\$small'.`   `            ``// We found a bigger square if ` `            ``// following conditions are met:` `            ``// 1)If side of square is ` `            ``//   greater than \$max.` `            ``// 2)There is a left vertical ` `            ``//   line of length >= '\$small'` `            ``// 3)There is a top horizontal` `            ``//   line of length >= '\$small'` `            ``while` `(``\$small` `> ``\$max``)` `            ``{` `                ``if` `(``\$ver``[``\$i``][``\$j` `- ``\$small` `+ 1] >= ``\$small` `&&` `                    ``\$hor``[``\$i` `- ``\$small` `+ 1][``\$j``] >= ``\$small``)` `                ``{` `                    ``\$max` `= ``\$small``;` `                ``}` `                ``\$small``--;` `            ``}` `        ``}` `    ``}` `    ``return` `\$max``;` `}`   `// Driver Code` `\$mat` `= ``array``(``array``(``'X'``, ``'O'``, ``'X'``, ``'X'``, ``'X'``, ``'X'``),` `             ``array``(``'X'``, ``'O'``, ``'X'``, ``'X'``, ``'O'``, ``'X'``),` `             ``array``(``'X'``, ``'X'``, ``'X'``, ``'O'``, ``'O'``, ``'X'``),` `             ``array``(``'O'``, ``'X'``, ``'X'``, ``'X'``, ``'X'``, ``'X'``),` `             ``array``(``'X'``, ``'X'``, ``'X'``, ``'O'``, ``'X'``, ``'O'``),` `             ``array``(``'O'``, ``'O'``, ``'X'``, ``'O'``, ``'O'``, ``'O'``));`   `// Function call` `echo` `findSubSquare(``\$mat``);`   `// This code is contributed` `// by ChitraNayal` `?>`

## Javascript

 ``

Output

`4`

Time complexity: O(N2).
Auxiliary Space: O(N2)

Optimized approach:

A more optimized solution would be to pre-compute the number of contiguous ‘X’ horizontally and vertically, in a matrix of pairs named dp. Now  for every entry of dp we have a pair (int, int) which denotes the maximum contiguous ‘X’ till that point, i.e.

• dp[i][j].first denotes contiguous ‘X’ taken horizontally till that point.
• dp[i][j].second denotes contiguous ‘X’ taken vertically till that point.

Now, a square can be formed with dp[i][j] as the bottom right corner, having sides atmost of length, min(dp[i][j].first, dp[i][j].second)

So, we make another matrix maxside, which will denote the maximum square side formed having the bottom right corner as arr[i][j]. We’ll try to get some intuition from the properties of a square, i.e. all the sides of the square are equal.

Let’s store maximum value that can be obtained, as val = min(dp[i][j].first, dp[i][j].second). From point (i, j), we traverse back horizontally by distance Val, and check if the minimum vertical contiguous ‘X’ till that point is equal to Val.

Similarly, we traverse back vertically by distance Val and check if the minimum horizontal contiguous ‘X’ till that point is equal to Val? Here we are making use of the fact that all sides of square are equal.

Input Matrix:

X  O  X  X  X  X

X  O  X  X  O  X

X  X  X  O  O  X

O  X  X  X  X  X

X  X  X  O  X  O

O  O  X  O  O  O

Value of matrix dp:

(1,1) (0,0) (1,1) (2,1) (3,1) (4,1)

(1,2) (0,0) (1,2) (2,2) (0,0) (1,2)

(1,3) (2,1) (3,3) (0,0) (0,0) (1,3)

(0,0) (1,2) (2,4) (3,1) (4,1) (5,4)

(1,1) (2,3) (3,5) (0,0) (1,2) (0,0)

(0,0) (0,0) (1,6) (0,0) (0,0) (0,0)

Below is the implementation of the above idea:

## C++

 `// A C++ program to find  the largest subsquare` `// surrounded by 'X' in a given matrix of 'O' and 'X'` `#include ` `using` `namespace` `std;`   `// Size of given matrix is N X N` `#define N 6`   `int` `maximumSubSquare(``int` `arr[][N])` `{` `    ``pair<``int``, ``int``> dp;` `    ``int` `maxside;`   `    ``// Initialize maxside with 0` `    ``memset``(maxside, 0, ``sizeof``(maxside));`   `    ``int` `x = 0, y = 0;`   `    ``// Fill the dp matrix horizontally.` `    ``// for contiguous 'X' increment the value of x,` `    ``// otherwise make it 0` `    ``for` `(``int` `i = 0; i < N; i++) ` `    ``{` `        ``x = 0;` `        ``for` `(``int` `j = 0; j < N; j++) ` `        ``{` `            ``if` `(arr[i][j] == ``'X'``)` `                ``x += 1;` `            ``else` `                ``x = 0;` `            ``dp[i][j].first = x;` `        ``}` `    ``}`   `    ``// Fill the dp matrix vertically.` `    ``// For contiguous 'X' increment the value of y,` `    ``// otherwise make it 0` `    ``for` `(``int` `i = 0; i < N; i++) ` `    ``{` `          ``y=0;` `        ``for` `(``int` `j = 0; j < N; j++) ` `        ``{` `            ``if` `(arr[j][i] == ``'X'``)` `                ``y += 1;` `            ``else` `                ``y = 0;` `            ``dp[j][i].second = y;` `        ``}` `    ``}`   `    ``// Now check , for every value of (i, j) if sub-square` `    ``// is possible,` `    ``// traverse back horizontally by value val, and check if` `    ``// vertical contiguous` `    ``// 'X'enfing at (i , j-val+1) is greater than equal to` `    ``// val.` `    ``// Similarly, check if traversing back vertically, the` `    ``// horizontal contiguous` `    ``// 'X'ending at (i-val+1, j) is greater than equal to` `    ``// val.` `    ``int` `maxval = 0, val = 0;` `    ``for` `(``int` `i = 0; i < N; i++) ` `    ``{` `        ``for` `(``int` `j = 0; j < N; j++) ` `        ``{` `            ``val = min(dp[i][j].first, dp[i][j].second);` `            ``if` `(dp[i][j - val + 1].second >= val` `                ``&& dp[i - val + 1][j].first >= val)` `                ``maxside[i][j] = val;` `            ``else` `                ``maxside[i][j] = 0;`   `            ``// store the final answer in maxval` `            ``maxval = max(maxval, maxside[i][j]);` `        ``}` `    ``}`   `    ``// return the final answe.` `    ``return` `maxval;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `mat[][N] = {` `        ``{ ``'X'``, ``'O'``, ``'X'``, ``'X'``, ``'X'``, ``'X'` `},` `        ``{ ``'X'``, ``'O'``, ``'X'``, ``'X'``, ``'O'``, ``'X'` `},` `        ``{ ``'X'``, ``'X'``, ``'X'``, ``'O'``, ``'O'``, ``'X'` `},` `        ``{ ``'O'``, ``'X'``, ``'X'``, ``'X'``, ``'X'``, ``'X'` `},` `        ``{ ``'X'``, ``'X'``, ``'X'``, ``'O'``, ``'X'``, ``'O'` `},` `        ``{ ``'O'``, ``'O'``, ``'X'``, ``'O'``, ``'O'``, ``'O'` `},` `    ``};`   `    ``// Function call` `    ``cout << maximumSubSquare(mat);` `    ``return` `0;` `}`

## Java

 `/*package whatever //do not write package name here */` `import` `java.io.*;` `import` `java.util.*;`   `class` `GFG ` `{` `  ``static` `int` `N = ``6``;`   `  ``static` `int` `maximumSubSquare(``int``[][] arr)` `  ``{` `    ``int``[][][] dp = ``new` `int``[``51``][``51``][``2``];` `    ``int``[][] maxside = ``new` `int``[``51``][``51``];`   `    ``// Initialize maxside with 0` `    ``for` `(``int``[] row : maxside) ` `      ``Arrays.fill(row, ``10``); `   `    ``int` `x = ``0``, y = ``0``;` `    ``// Fill the dp matrix horizontally.` `    ``// for contiguous 'X' increment the value of x,` `    ``// otherwise make it 0` `    ``for` `(``int` `i = ``0``; i < N; i++) ` `    ``{` `      ``x = ``0``;` `      ``for` `(``int` `j = ``0``; j < N; j++) ` `      ``{` `        ``if` `(arr[i][j] == ``'X'``)` `          ``x += ``1``;` `        ``else` `          ``x = ``0``;` `        ``dp[i][j][``0``] = x;` `      ``}` `    ``}`   `    ``// Fill the dp matrix vertically.` `    ``// For contiguous 'X' increment the value of y,` `    ``// otherwise make it 0` `    ``for` `(``int` `i = ``0``; i < N; i++) ` `    ``{` `      ``y = ``0``;` `      ``for` `(``int` `j = ``0``; j < N; j++) ` `      ``{` `        ``if` `(arr[j][i] == ``'X'``)` `          ``y += ``1``;` `        ``else` `          ``y = ``0``;` `        ``dp[j][i][``1``] = y;` `      ``}` `    ``}`   `    ``// Now check , for every value of (i, j) if sub-square` `    ``// is possible,` `    ``// traverse back horizontally by value val, and check if` `    ``// vertical contiguous` `    ``// 'X'enfing at (i , j-val+1) is greater than equal to` `    ``// val.` `    ``// Similarly, check if traversing back vertically, the` `    ``// horizontal contiguous` `    ``// 'X'ending at (i-val+1, j) is greater than equal to` `    ``// val.` `    ``int` `maxval = ``0``, val = ``0``;`   `    ``for` `(``int` `i = ``0``; i < N; i++) ` `    ``{` `      ``for` `(``int` `j = ``0``; j < N; j++) ` `      ``{` `        ``val = Math.min(dp[i][j][``0``], dp[i][j][``1``]);` `        ``if` `(dp[i][j - val + ``1``][``1``] >= val` `            ``&& dp[i - val + ``1``][j][``0``] >= val)` `          ``maxside[i][j] = val;` `        ``else` `          ``maxside[i][j] = ``0``;`   `        ``// store the final answer in maxval` `        ``maxval = Math.max(maxval, maxside[i][j]);` `      ``}` `    ``}`   `    ``// return the final answe.` `    ``return` `maxval;`   `  ``}`   `  ``// Driver code` `  ``public` `static` `void` `main (String[] args) ` `  ``{` `    ``int` `mat[][] = {` `      ``{ ``'X'``, ``'O'``, ``'X'``, ``'X'``, ``'X'``, ``'X'` `},` `      ``{ ``'X'``, ``'O'``, ``'X'``, ``'X'``, ``'O'``, ``'X'` `},` `      ``{ ``'X'``, ``'X'``, ``'X'``, ``'O'``, ``'O'``, ``'X'` `},` `      ``{ ``'O'``, ``'X'``, ``'X'``, ``'X'``, ``'X'``, ``'X'` `},` `      ``{ ``'X'``, ``'X'``, ``'X'``, ``'O'``, ``'X'``, ``'O'` `},` `      ``{ ``'O'``, ``'O'``, ``'X'``, ``'O'``, ``'O'``, ``'O'` `},` `    ``};`   `    ``// Function call` `    ``System.out.println(maximumSubSquare(mat));` `  ``}` `}`   `// This code is contributed by rag2127.`

## Python3

 `# Python3 program to find  the largest ` `# subsquare surrounded by 'X' in a given` `# matrix of 'O' and 'X'`   `# Size of given matrix is N X N` `N ``=` `6`   `def` `maximumSubSquare(arr):` `    `  `    ``dp ``=` `[[[``-``1``, ``-``1``] ``for` `i ``in` `range``(``51``)] ` `                    ``for` `j ``in` `range``(``51``)]`   `    ``# Initialize maxside with 0` `    ``maxside ``=` `[[``0` `for` `i ``in` `range``(``51``)]` `                  ``for` `j ``in` `range``(``51``)]` `                  `  `    ``x ``=` `0` `    ``y ``=` `0` `    `  `    ``# Fill the dp matrix horizontally.` `    ``# for contiguous 'X' increment the ` `    ``# value of x, otherwise make it 0` `    ``for` `i ``in` `range``(N):` `        ``x ``=` `0` `        `  `        ``for` `j ``in` `range``(N):` `            ``if` `(arr[i][j] ``=``=` `'X'``):` `                ``x ``+``=` `1` `            ``else``:` `                ``x ``=` `0` `                `  `            ``dp[i][j][``0``] ``=` `x`   `    ``# Fill the dp matrix vertically.` `    ``# For contiguous 'X' increment` `    ``# the value of y, otherwise ` `    ``# make it 0` `    ``for` `i ``in` `range``(N):` `      ``y ``=` `0` `        ``for` `j ``in` `range``(N):` `            ``if` `(arr[j][i] ``=``=` `'X'``):` `                ``y ``+``=` `1` `            ``else``:` `                ``y ``=` `0` `                `  `            ``dp[j][i][``1``] ``=` `y` `    `  `    ``# Now check , for every value of (i, j) if sub-square` `    ``# is possible,` `    ``# traverse back horizontally by value val, and check if` `    ``# vertical contiguous` `    ``# 'X'enfing at (i , j-val+1) is greater than equal to` `    ``# val.` `    ``# Similarly, check if traversing back vertically, the` `    ``# horizontal contiguous` `    ``# 'X'ending at (i-val+1, j) is greater than equal to` `    ``# val.` `    ``maxval ``=` `0` `    ``val ``=` `0`   `    ``for` `i ``in` `range``(N):` `        ``for` `j ``in` `range``(N):` `            ``val ``=` `min``(dp[i][j][``0``],` `                      ``dp[i][j][``1``])` `                      `  `            ``if` `(dp[i][j ``-` `val ``+` `1``][``1``] >``=` `val ``and` `                ``dp[i ``-` `val ``+` `1``][j][``0``] >``=` `val):` `                ``maxside[i][j] ``=` `val` `            ``else``:` `                ``maxside[i][j] ``=` `0`   `            ``# Store the final answer in maxval` `            ``maxval ``=` `max``(maxval, maxside[i][j])`   `    ``# Return the final answe.` `    ``return` `maxval`   `# Driver code` `mat ``=` `[ [ ``'X'``, ``'O'``, ``'X'``, ``'X'``, ``'X'``, ``'X'` `],` `        ``[ ``'X'``, ``'O'``, ``'X'``, ``'X'``, ``'O'``, ``'X'` `],` `        ``[ ``'X'``, ``'X'``, ``'X'``, ``'O'``, ``'O'``, ``'X'` `],` `        ``[ ``'O'``, ``'X'``, ``'X'``, ``'X'``, ``'X'``, ``'X'` `],` `        ``[ ``'X'``, ``'X'``, ``'X'``, ``'O'``, ``'X'``, ``'O'` `],` `        ``[ ``'O'``, ``'O'``, ``'X'``, ``'O'``, ``'O'``, ``'O'` `] ]` `        `  `# Function call` `print``(maximumSubSquare(mat))`   `# This code is contributed by avanitrachhadiya2155`

## C#

 `// A C# program to find  the largest subsquare` `// surrounded by 'X' in a given matrix of 'O' and 'X'` `using` `System;`   `public` `class` `GFG{` `    `  `    ``static` `int` `N = 6;` ` `  `  ``static` `int` `maximumSubSquare(``int``[,] arr)` `  ``{` `    ``int``[,,] dp = ``new` `int``[51,51,2];` `    ``int``[,] maxside = ``new` `int``[51,51];` ` `  `    ``// Initialize maxside with 0` `    ``for``(``int` `i = 0; i < 51; i++)` `    ``{` `        `  `        ``for``(``int` `j = 0; j < 51; j++)` `        ``{` `            ``maxside[i,j] = 10;` `        ``}` `    ``}` ` `  `    ``int` `x = 0, y = 0;` `    ``// Fill the dp matrix horizontally.` `    ``// for contiguous 'X' increment the value of x,` `    ``// otherwise make it 0` `    ``for` `(``int` `i = 0; i < N; i++)` `    ``{` `      ``x = 0;` `      ``for` `(``int` `j = 0; j < N; j++)` `      ``{` `        ``if` `(arr[i,j] == ``'X'``)` `          ``x += 1;` `        ``else` `          ``x = 0;` `        ``dp[i,j,0] = x;` `      ``}` `    ``}` ` `  `    ``// Fill the dp matrix vertically.` `    ``// For contiguous 'X' increment the value of y,` `    ``// otherwise make it 0` `    ``for` `(``int` `i = 0; i < N; i++)` `    ``{` `      ``y = 0;` `      ``for` `(``int` `j = 0; j < N; j++)` `      ``{` `        ``if` `(arr[j,i] == ``'X'``)` `          ``y += 1;` `        ``else` `          ``y = 0;` `        ``dp[j,i,1] = y;` `      ``}` `    ``}` ` `  `    ``// Now check , for every value of (i, j) if sub-square` `    ``// is possible,` `    ``// traverse back horizontally by value val, and check if` `    ``// vertical contiguous` `    ``// 'X'enfing at (i , j-val+1) is greater than equal to` `    ``// val.` `    ``// Similarly, check if traversing back vertically, the` `    ``// horizontal contiguous` `    ``// 'X'ending at (i-val+1, j) is greater than equal to` `    ``// val.` `    ``int` `maxval = 0, val = 0;` ` `  `    ``for` `(``int` `i = 0; i < N; i++)` `    ``{` `      ``for` `(``int` `j = 0; j < N; j++)` `      ``{` `        ``val = Math.Min(dp[i,j,0], dp[i,j,1]);` `        ``if` `(dp[i,j - val + 1,1] >= val` `            ``&& dp[i - val + 1,j,0] >= val)` `          ``maxside[i,j] = val;` `        ``else` `          ``maxside[i,j] = 0;` ` `  `        ``// store the final answer in maxval` `        ``maxval = Math.Max(maxval, maxside[i,j]);` `      ``}` `    ``}` ` `  `    ``// return the final answe.` `    ``return` `maxval;` ` `  `  ``}` ` `  `  ``// Driver code` `    `  `    ``static` `public` `void` `Main (){` `        `  `        ``int``[,] mat = {` `      ``{ ``'X'``, ``'O'``, ``'X'``, ``'X'``, ``'X'``, ``'X'` `},` `      ``{ ``'X'``, ``'O'``, ``'X'``, ``'X'``, ``'O'``, ``'X'` `},` `      ``{ ``'X'``, ``'X'``, ``'X'``, ``'O'``, ``'O'``, ``'X'` `},` `      ``{ ``'O'``, ``'X'``, ``'X'``, ``'X'``, ``'X'``, ``'X'` `},` `      ``{ ``'X'``, ``'X'``, ``'X'``, ``'O'``, ``'X'``, ``'O'` `},` `      ``{ ``'O'``, ``'O'``, ``'X'``, ``'O'``, ``'O'``, ``'O'` `},` `    ``};` ` `  `    ``// Function call` `    ``Console.WriteLine(maximumSubSquare(mat));` `        `  `    ``}` `}`   `// This code is contributed by patel2127`

## Javascript

 ``

Output

`4`

Time complexity: O(N2
Auxiliary space: O(N2)

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