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 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[][]. Following is an 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.

div class="responsive-tabs">

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=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 program to test above function 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'},                     };     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 = ver = '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'}};             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 == 'X':         hor = 1         ver = 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']] 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'}};     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')); echo findSubSquare(\$mat);    // This code is contributed // by ChitraNayal ?>

Output:

4