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[][]. 

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. 

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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 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[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 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;
}
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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)
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

<?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 numbers
function 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 call
echo findSubSquare($mat);
 
// This code is contributed
// by ChitraNayal
?>
chevron_right

Output
4

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.

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:

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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 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;
}
chevron_right

Output
4

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

This article is contributed by Abhinav Rastogi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.





Article Tags :
Practice Tags :