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

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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.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 `

Output:

```2
```

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