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

24 Jun, 2022
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[6][6] =  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[6][6] = 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[6][6] = 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++
Java
Python3
C#
PHP
Javascript
// A C++ program to find  the largest subsquare// surrounded by 'X' in a given matrix of 'O' and 'X'#include <iostream>using namespace std; // Size of given matrix is N X N#define N 6 // A utility function to find minimum of two numbersint 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[0][0] = ver[0][0] = (mat[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 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 codeint 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;}
// 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
# 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 NN = 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 Codemat = [['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 callprint(findSubSquare(mat)) # This code is contributed by# Mohit kumar 29
// 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// A PHP 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$N = 6; // A utility function to find// minimum of two numbersfunction getMin($x, $y){    return ($x < $y) ? $x : $y;} // Returns size of maximum// size subsquare matrix// surrounded by 'X'function findSubSquare($mat){    $max = 0; // Initialize result    $hor[0][0] = $ver[0][0] = ($mat[0][0] == 'X');     // Fill values in    // $hor and $ver    for ($i = 0; $i < $GLOBALS['N']; $i++)    {        for ($j = 0; $j < $GLOBALS['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 $ver    for ($i = $GLOBALS['N'] - 1; $i >= 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 callecho findSubSquare($mat); // This code is contributed// by ChitraNayal?>
<script> // A JavaScript 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    let N = 6;         // A utility function to    // find minimum of two numbers    function getMin(x,y)    {             return (x < y) ? x : y;    }         // Returns size of maximum    // size subsquare matrix    // surrounded by 'X'    function findSubSquare(mat)    {        let max = 0; // Initialize result          // Initialize the left-top        // value in hor[][] and ver[][]        let hor = new Array(N);        let ver = new Array(N);        for (let i = 0; i < N; i++)        {            hor[i]=new Array(N);            ver[i]=new Array(N);            for (let j = 0; j < N; j++)            {                hor[i][j]="";                ver[i][j]="";            }        }                 hor[0][0] = 'X';        ver[0][0] = 'X';          // Fill values in        // hor[][] and ver[][]        for (let i = 0; i < N; i++)        {            for (let 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 (let i = N - 1; i >= 1; i--)        {            for (let j = N - 1; j >= 1; j--)            {                // Find smaller of values in                // hor[][] and ver[][] A Square                // can only be made by taking                // smaller value                let 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     // TODO Auto-generated method stub    let 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    document.write(findSubSquare(mat))         // This code is contributed by unknown2108     </script>
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) (2,7) (3,1) (4,1)  
(1,2) (0,0) (1,2) (2,8) (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++
Java
Python3
C#
Javascript
// A C++ program to find  the largest subsquare// surrounded by 'X' in a given matrix of 'O' and 'X'#include <bits/stdc++.h>using namespace std; // Size of given matrix is N X N#define N 6 int maximumSubSquare(int arr[][N]){    pair<int, int> dp[51][51];    int maxside[51][51];     // 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++)    {        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 codeint 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;}
/*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++)    {      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 program to find  the largest# subsquare surrounded by 'X' in a given# matrix of 'O' and 'X' # Size of given matrix is N X NN = 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):        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 codemat = [ [ '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 callprint(maximumSubSquare(mat)) # This code is contributed by avanitrachhadiya2155
// 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++)    {      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
<script>/*package whatever //do not write package name here */let N = 6; function maximumSubSquare(arr){    let dp = new Array(51);    for(let i = 0; i < 51; i++)    {        dp[i] = new Array(51);        for(let j = 0; j < 51; j++)        {            dp[i][j] = new Array(2);            for(let k = 0; k < 2; k++)            {                dp[i][j][k] = 0;            }        }    }         let maxside = new Array(51);    for(let i = 0; i < 51; i++)    {        maxside[i] = new Array(51);        for(let j = 0; j < 51; j++)        {            maxside[i][j] = 10;        }    }         let x = 0, y = 0;         // Fill the dp matrix horizontally.    // for contiguous 'X' increment the value of x,    // otherwise make it 0    for (let i = 0; i < N; i++)    {      x = 0;      for (let 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 (let i = 0; i < N; i++)    {      for (let 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.    let maxval = 0, val = 0;      for (let i = 0; i < N; i++)    {      for (let 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 codelet 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 calldocument.write(maximumSubSquare(mat)); // This code is contributed by ab2127</script>
4
Time complexity: O(N2) Auxiliary space: O(N2)
ukasp
Suryaveer Singh
Akanksha_Rai
mohit kumar 29
Ar26
avanitrachhadiya2155
rag2127
unknown2108
ab2127
patel2127
codewithshinchan
hardikkoriintern
D-E-Shaw
Matrix
D-E-Shaw
Matrix
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