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: 1
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: 3
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 elementsint 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 elementsint 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 codeint 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 approachclass 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 elementsstatic 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 elementsstatic 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 codepublic 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 approachimport 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 elementsdef 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 elementsdef 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 codeif __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 approachusing 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 elementsstatic 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 elementsstatic 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 codepublic 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
Time Complexity: O(r * c)
Auxiliary Space: O(1)
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