Given a binary matrix of R rows and C columns. We are allowed flip to any size of sub matrix. Flipping means changing 1 to 0 and 0 to 1. The task is maximize the number of 1s in the matrix. Output the maximum number of 1s.
Examples:
Input : R = 3, C =3
mat[][] = { 0, 0, 1,
0, 0, 1,
1, 0, 1 }
Output : 8
Flip
0 0 1
0 0 1
1 0 1
to get
1 1 1
1 1 1
0 1 1
Input : R = 2, C = 3
mat[][] = { 0, 0, 0,
0, 0, 0 }
Output : 6
Create a matrix ones[][] of R rows and C columns, which precomputes the number of ones in the submatrix from (0, 0) to (i, j) by
// Common elements in ones[i-1][j] and
// ones[i][j-1] are ones[i-1][j-1]
ones[i][j] = ones[i-1][j] + ones[i][j-1] -
ones[i-1][j-1] + (mat[i][j] == 1)
Since, we are allowed to flip sub matrix only once. We iterate over all possible submatrices of all possible sizes for each cell (i, j) to (i + k – 1, i + k – 1). We calculate the total number of ones after the digits are filliped in the chosen submatrix.
Total number of ones in the final matrix after flipping submatrix (i, j) to (i + k – 1) will be Ones in the whole matrix – Ones in the chosen submatrix + Zeroes in the chosen sub matrix. That comes out to be :-
ones[R][C] – cal(i, j, i + k, j + k – 1) + k*k – cal(i, j, i + k – 1, j + k – 1)
where cal(a, b, c, d) denotes the number of ones in square submatrix of length c – a.
Now cal(x1, y1, x2, y2) can be define by:
ones[x2][y2] – ones[x2][y1 – 1] – ones[x1 – 1][y2] + ones[x1 – 1][y1 – 1].
Below is the implementation of this approach:
C++
// C++ program to find maximum number of ones after // one flipping in Binary Matrix #include <bits/stdc++.h> #define R 3 #define C 3 using namespace std; // Return number of ones in square submatrix of size // k x k starting from (x, y) int cal(int ones[R + 1][C + 1], int x, int y, int k) { return ones[x + k - 1][y + k - 1] - ones[x - 1][y + k - 1] - ones[x + k - 1][y - 1] + ones[x - 1][y - 1]; } // Return maximum number of 1s after flipping a submatrix int sol(int mat[R][C]) { int ans = 0; // Precomputing the number of 1s int ones[R + 1][C + 1] = {0}; for (int i = 1; i <= R; i++) for (int j = 1; j <= C; j++) ones[i][j] = ones[i - 1][j] + ones[i][j - 1] - ones[i - 1][j - 1] + (mat[i - 1][j - 1] == 1); // Finding the maximum number of 1s after flipping for (int k = 1; k <= min(R, C); k++) for (int i = 1; i + k - 1 <= R; i++) for (int j = 1; j + k - 1 <= C; j++) ans = max(ans, (ones[R][C] + k * k - 2 * cal(ones, i, j, k))); return ans; } // Driver code int main() { int mat[R][C] = {{0, 0, 1}, { 0, 0, 1}, { 1, 0, 1 } }; cout << sol(mat) << endl; return 0; } |
Java
// Java program to find maximum number of ones after // one flipping in Binary Matrix class GFG { static final int R = 3; static final int C = 3 ; // Return number of ones in square submatrix of size // k x k starting from (x, y) static int cal(int ones[][], int x, int y, int k) { return ones[x + k - 1][y + k - 1] - ones[x - 1][y + k - 1] - ones[x + k - 1][y - 1] + ones[x - 1][y - 1]; } // Return maximum number of 1s after flipping a submatrix static int sol(int mat[][]) { int ans = 0; int val =0; // Precomputing the number of 1s int ones[][] = new int[R + 1][C + 1]; for (int i = 1; i <= R; i++) for (int j = 1; j <= C; j++) { if(mat[i - 1][j - 1] == 1) val=1; ones[i][j] = ones[i - 1][j] + ones[i][j - 1] - ones[i - 1][j - 1] + (val); } // Finding the maximum number of 1s after flipping for (int k = 1; k <= Math.min(R, C); k++) for (int i = 1; i + k - 1 <= R; i++) for (int j = 1; j + k - 1 <= C; j++) ans = Math.max(ans, (ones[R][C] + k * k - 2 * cal(ones, i, j, k))); return ans; } // Driver code static public void main(String[] args) { int mat[][] = {{0, 0, 1}, { 0, 0, 1}, { 1, 0, 1 } }; System.out.println(sol(mat)); } } // This code is contributed by Rajput-Ji |
Python3
# Python 3 program to find maximum number of # ones after one flipping in Binary Matrix R = 3C = 3 # Return number of ones in square submatrix # of size k x k starting from (x, y) def cal(ones, x, y, k): return (ones[x + k - 1][y + k - 1] - ones[x - 1][y + k - 1] - ones[x + k - 1][y - 1] + ones[x - 1][y - 1]) # Return maximum number of 1s after # flipping a submatrix def sol(mat): ans = 0 # Precomputing the number of 1s ones = [[0 for i in range(C + 1)] for i in range(R + 1)] for i in range(1, R + 1, 1): for j in range(1, C + 1, 1): ones[i][j] = (ones[i - 1][j] + ones[i][j - 1] - ones[i - 1][j - 1] + (mat[i - 1][j - 1] == 1)) # Finding the maximum number of 1s # after flipping for k in range(1, min(R, C) + 1, 1): for i in range(1, R - k + 2, 1): for j in range(1, C - k + 2, 1): ans = max(ans, (ones[R][C] + k * k - 2 * cal(ones, i, j, k))) return ans # Driver code if __name__ == '__main__': mat = [[0, 0, 1], [0, 0, 1], [1, 0, 1]] print(sol(mat)) # This code is contributed by # Sahil_Shelangia |
C#
// C# program to find maximum number of ones after // one flipping in Binary Matrix using System; public class GFG { static readonly int R = 3; static readonly int C = 3 ; // Return number of ones in square submatrix of size // k x k starting from (x, y) static int cal(int [,]ones, int x, int y, int k) { return ones[x + k - 1,y + k - 1] - ones[x - 1,y + k - 1] - ones[x + k - 1,y - 1] + ones[x - 1,y - 1]; } // Return maximum number of 1s after flipping a submatrix static int sol(int [,]mat) { int ans = 0; int val =0; // Precomputing the number of 1s int [,]ones = new int[R + 1,C + 1]; for (int i = 1; i <= R; i++) for (int j = 1; j <= C; j++) { if(mat[i - 1,j - 1] == 1) val=1; ones[i,j] = ones[i - 1,j] + ones[i,j - 1] - ones[i - 1,j - 1] + (val); } // Finding the maximum number of 1s after flipping for (int k = 1; k <= Math.Min(R, C); k++) for (int i = 1; i + k - 1 <= R; i++) for (int j = 1; j + k - 1 <= C; j++) ans = Math.Max(ans, (ones[R,C] + k * k - 2 * cal(ones, i, j, k))); return ans; } // Driver code static public void Main() { int [,]mat = {{0, 0, 1}, { 0, 0, 1}, { 1, 0, 1 } }; Console.WriteLine(sol(mat)); } } // This code is contributed by 29AjayKumar |
PHP
<?php // PHP program to find maximum number of ones after // one flipping in Binary Matrix $R = 3; $C = 3; // Return number of ones in square submatrix of size // k x k starting from (x, y) function cal($ones, $x, $y, $k) { return $ones[$x + $k - 1][$y + $k - 1] - $ones[$x - 1][$y + $k - 1] - $ones[$x + $k - 1][$y - 1] + $ones[$x - 1][$y - 1]; } // Return maximum number of 1s after flipping a submatrix function sol($mat) { global $C,$R; $ans = 0; // Precomputing the number of 1s $ones=array_fill(0,$R + 1,array_fill(0,$C + 1,0)); for ($i = 1; $i <= $R; $i++) for ($j = 1; $j <= $C; $j++) $ones[$i][$j] = $ones[$i - 1][$j] + $ones[$i][$j - 1] - $ones[$i - 1][$j - 1] + (int)($mat[$i - 1][$j - 1] == 1); // Finding the maximum number of 1s after flipping for ($k = 1; $k <= min($R, $C); $k++) for ($i = 1; $i + $k - 1 <= $R; $i++) for ($j = 1; $j + $k - 1 <= $C; $j++) $ans = max($ans, ($ones[$R][$C] + $k * $k - 2 * cal($ones, $i, $j, $k))); return $ans; } // Driver code $mat = array(array(0, 0, 1), array( 0, 0, 1), array( 1, 0, 1 )); echo sol($mat); // This code is contributed by mits ?> |
Output:
8
Time Complexity: O(R*C*min(R, C))
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