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Minimum cells to be flipped to get a 2*2 submatrix with equal elements

  • Last Updated : 07 Apr, 2021

Given a matrix of size M * N, the task is to find the count of the minimum number of cells that must be flipped such that there is at least a submatrix of size 2*2 with all equal elements.
Examples: 
 

Input: mat[] = {“00000”, “10111”, “00000”, “11111”} 
Output:
One of the possible submatrix could be {{0, 0}, {1, 0}} 
where only a single element has to be flipped.
Input: mat[] = {“0101”, “0101”, “0101”} 
Output:
 

 

Approach: For every submatrix of size 2*2, count the number of 0s and the number of 1s in it and the minimum of these two will be the count of flips required to get the matrix with all equal elements. The minimum of this value for all the submatrices is the required answer.
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the minimum flips
// required such that the submatrix from
// mat[i][j] to mat[i + 1][j + 1]
// contains all equal elements
int minFlipsSub(string mat[], int i, int j)
{
    int cnt0 = 0, cnt1 = 0;
 
    if (mat[i][j] == '1')
        cnt1++;
    else
        cnt0++;
 
    if (mat[i][j + 1] == '1')
        cnt1++;
    else
        cnt0++;
 
    if (mat[i + 1][j] == '1')
        cnt1++;
    else
        cnt0++;
 
    if (mat[i + 1][j + 1] == '1')
        cnt1++;
    else
        cnt0++;
 
    return min(cnt0, cnt1);
}
 
// Function to return the minimum number
// of slips required such that the matrix
// contains at least a single submatrix
// of size 2*2 with all equal elements
int minFlips(string mat[], int r, int c)
{
 
    // To store the result
    int res = INT_MAX;
 
    // For every submatrix of size 2*2
    for (int i = 0; i < r - 1; i++) {
        for (int j = 0; j < c - 1; j++) {
 
            // Update the count of flips required
            // for the current submatrix
            res = min(res, minFlipsSub(mat, i, j));
        }
    }
 
    return res;
}
 
// Driver code
int main()
{
    string mat[] = { "0101", "0101", "0101" };
    int r = sizeof(mat) / sizeof(string);
    int c = mat[0].length();
 
    cout << minFlips(mat, r, c);
 
    return 0;
}

Java




// Java implementation of the approach
class GFG
{
 
// Function to return the minimum flips
// required such that the submatrix from
// mat[i][j] to mat[i + 1][j + 1]
// contains all equal elements
static int minFlipsSub(String mat[], int i, int j)
{
    int cnt0 = 0, cnt1 = 0;
 
    if (mat[i].charAt(j) == '1')
        cnt1++;
    else
        cnt0++;
 
    if (mat[i].charAt(j+1) == '1')
        cnt1++;
    else
        cnt0++;
 
    if (mat[i + 1].charAt(j) == '1')
        cnt1++;
    else
        cnt0++;
 
    if (mat[i + 1].charAt(j+1) == '1')
        cnt1++;
    else
        cnt0++;
 
    return Math.min(cnt0, cnt1);
}
 
// Function to return the minimum number
// of slips required such that the matrix
// contains at least a single submatrix
// of size 2*2 with all equal elements
static int minFlips(String mat[], int r, int c)
{
    // To store the result
    int res = Integer.MAX_VALUE;
 
    // For every submatrix of size 2*2
    for (int i = 0; i < r - 1; i++)
    {
        for (int j = 0; j < c - 1; j++)
        {
            // Update the count of flips required
            // for the current submatrix
            res = Math.min(res, minFlipsSub(mat, i, j));
        }
    }
    return res;
}
 
// Driver code
public static void main(String[] args)
{
    String mat[] = { "0101", "0101", "0101" };
    int r = mat.length;
    int c = mat[0].length();
 
    System.out.print(minFlips(mat, r, c));
}
}
 
// This code is contributed by 29AjayKumar

Python 3




# Python 3 implementation of the approach
import sys
 
# Function to return the minimum flips
# required such that the submatrix from
# mat[i][j] to mat[i + 1][j + 1]
# contains all equal elements
def minFlipsSub(mat, i, j):
    cnt0 = 0
    cnt1 = 0
 
    if (mat[i][j] == '1'):
        cnt1 += 1
    else:
        cnt0 += 1
 
    if (mat[i][j + 1] == '1'):
        cnt1 += 1
    else:
        cnt0 += 1
 
    if (mat[i + 1][j] == '1'):
        cnt1 += 1
    else:
        cnt0 += 1
 
    if (mat[i + 1][j + 1] == '1'):
        cnt1 += 1
    else:
        cnt0 += 1
 
    return min(cnt0, cnt1)
 
# Function to return the minimum number
# of slips required such that the matrix
# contains at least a single submatrix
# of size 2*2 with all equal elements
def minFlips(mat, r, c):
     
    # To store the result
    res = sys.maxsize
 
    # For every submatrix of size 2*2
    for i in range(r - 1):
        for j in range(c - 1):
             
            # Update the count of flips required
            # for the current submatrix
            res = min(res, minFlipsSub(mat, i, j))
 
    return res
 
# Driver code
if __name__ == '__main__':
    mat = ["0101", "0101", "0101"]
    r = len(mat)
    c = len(mat[0])
 
    print(minFlips(mat, r, c))
     
# This code is contributed by Surendra_Gangwar

C#




// C# implementation of the approach
using System;
 
class GFG
{
 
// Function to return the minimum flips
// required such that the submatrix from
// mat[i,j] to mat[i + 1,j + 1]
// contains all equal elements
static int minFlipsSub(String []mat,
                       int i, int j)
{
    int cnt0 = 0, cnt1 = 0;
 
    if (mat[i][j] == '1')
        cnt1++;
    else
        cnt0++;
 
    if (mat[i][j + 1] == '1')
        cnt1++;
    else
        cnt0++;
 
    if (mat[i + 1][j] == '1')
        cnt1++;
    else
        cnt0++;
 
    if (mat[i + 1][j + 1] == '1')
        cnt1++;
    else
        cnt0++;
 
    return Math.Min(cnt0, cnt1);
}
 
// Function to return the minimum number
// of slips required such that the matrix
// contains at least a single submatrix
// of size 2*2 with all equal elements
static int minFlips(String []mat,
                    int r, int c)
{
    // To store the result
    int res = int.MaxValue;
 
    // For every submatrix of size 2*2
    for (int i = 0; i < r - 1; i++)
    {
        for (int j = 0; j < c - 1; j++)
        {
            // Update the count of flips required
            // for the current submatrix
            res = Math.Min(res, minFlipsSub(mat, i, j));
        }
    }
    return res;
}
 
// Driver code
public static void Main(String[] args)
{
    String []mat = { "0101", "0101", "0101" };
    int r = mat.Length;
    int c = mat.GetLength(0);
 
    Console.Write(minFlips(mat, r, c));
}
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
// javascript implementation of the approach   
// Function to return the minimum flips
    // required such that the submatrix from
    // mat[i][j] to mat[i + 1][j + 1]
    // contains all equal elements
    function minFlipsSub( mat , i , j)
    {
        var cnt0 = 0, cnt1 = 0;
 
        if (mat[i].charAt(j) == '1')
            cnt1++;
        else
            cnt0++;
 
        if (mat[i].charAt(j + 1) == '1')
            cnt1++;
        else
            cnt0++;
 
        if (mat[i + 1].charAt(j) == '1')
            cnt1++;
        else
            cnt0++;
 
        if (mat[i + 1].charAt(j + 1) == '1')
            cnt1++;
        else
            cnt0++;
 
        return Math.min(cnt0, cnt1);
    }
 
    // Function to return the minimum number
    // of slips required such that the matrix
    // contains at least a single submatrix
    // of size 2*2 with all equal elements
    function minFlips(mat , r , c)
    {
     
        // To store the result
        var res = Number.MAX_VALUE;
 
        // For every submatrix of size 2*2
        for (i = 0; i < r - 1; i++)
        {
            for (j = 0; j < c - 1; j++)
            {
             
                // Update the count of flips required
                // for the current submatrix
                res = Math.min(res, minFlipsSub(mat, i, j));
            }
        }
        return res;
    }
 
    // Driver code
        var mat = [ "0101", "0101", "0101" ];
        var r = mat.length;
        var c = mat[0].length;
 
        document.write(minFlips(mat, r, c));
         
// This code is contributed by Rajput-Ji
</script>
Output: 
2

 

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