Minimum operations required to set all elements of binary matrix
Given a binary matrix of N rows and M columns. The operation allowed on the matrix is to choose any index (x, y) and toggle all the elements between the rectangle having top-left as (0, 0) and bottom-right as (x-1, y-1). Toggling the element means changing 1 to 0 and 0 to 1. The task is to find minimum operations required to make set all the elements of the matrix i.e make all elements as 1.
Examples:
Input : mat[][] = 0 0 0 1 1 0 0 0 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 Output : 1 In one move, choose (3, 3) to make the whole matrix consisting of only 1s. Input : mat[][] = 0 0 1 1 1 0 0 0 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 Output : 3
The idea is to start from the end point (N – 1, M – 1) and traverse the matrix in reverse order. Whenever we encounter a cell which has a value of 0, flip it.
Why traversing from end point ?
Suppose there are 0 at (x, y) and (x + 1, y + 1) cell. You shouldn’t flip a cell (x + 1, y + 1) after cell (x, y) because after you flipped (x, y) to 1, in next move to flip (x + 1, y + 1) cell, you will flip again (x, y) to 0. So there is no benefit from the first move for flipping (x, y) cell.
Below is implementation of this approach:
C++
// C++ program to find minimum operations required // to set all the element of binary matrix #include <bits/stdc++.h> #define N 5 #define M 5 using namespace std; // Return minimum operation required to make all 1s. int minOperation( bool arr[N][M]) { int ans = 0; for ( int i = N - 1; i >= 0; i--) { for ( int j = M - 1; j >= 0; j--) { // check if this cell equals 0 if (arr[i][j] == 0) { // increase the number of moves ans++; // flip from this cell to the start point for ( int k = 0; k <= i; k++) { for ( int h = 0; h <= j; h++) { // flip the cell if (arr[k][h] == 1) arr[k][h] = 0; else arr[k][h] = 1; } } } } } return ans; } // Driven Program int main() { bool mat[N][M] = { 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 }; cout << minOperation(mat) << endl; return 0; } |
Java
// Java program to find minimum operations required // to set all the element of binary matrix class GFG { static final int N = 5 ; static final int M = 5 ; // Return minimum operation required to make all 1s. static int minOperation( boolean arr[][]) { int ans = 0 ; for ( int i = N - 1 ; i >= 0 ; i--) { for ( int j = M - 1 ; j >= 0 ; j--) { // check if this cell equals 0 if (arr[i][j] == false ) { // increase the number of moves ans++; // flip from this cell to the start point for ( int k = 0 ; k <= i; k++) { for ( int h = 0 ; h <= j; h++) { // flip the cell if (arr[k][h] == true ) { arr[k][h] = false ; } else { arr[k][h] = true ; } } } } } } return ans; } // Driven Program public static void main(String[] args) { boolean mat[][] = { { false , false , true , true , true }, { false , false , false , true , true }, { false , false , false , true , true }, { true , true , true , true , true }, { true , true , true , true , true } }; System.out.println(minOperation(mat)); } } // This code is contributed // by PrinciRaj1992 |
Python 3
# Python 3 program to find # minimum operations required # to set all the element of # binary matrix # Return minimum operation # required to make all 1s. def minOperation(arr): ans = 0 for i in range (N - 1 , - 1 , - 1 ): for j in range (M - 1 , - 1 , - 1 ): # check if this # cell equals 0 if (arr[i][j] = = 0 ): # increase the # number of moves ans + = 1 # flip from this cell # to the start point for k in range (i + 1 ): for h in range (j + 1 ): # flip the cell if (arr[k][h] = = 1 ): arr[k][h] = 0 else : arr[k][h] = 1 return ans # Driver Code mat = [[ 0 , 0 , 1 , 1 , 1 ], [ 0 , 0 , 0 , 1 , 1 ], [ 0 , 0 , 0 , 1 , 1 ], [ 1 , 1 , 1 , 1 , 1 ], [ 1 , 1 , 1 , 1 , 1 ]] M = 5 N = 5 print (minOperation(mat)) # This code is contributed # by ChitraNayal |
C#
using System; // C# program to find minimum operations required // to set all the element of binary matrix public class GFG { public const int N = 5; public const int M = 5; // Return minimum operation required to make all 1s. public static int minOperation( bool [][] arr) { int ans = 0; for ( int i = N - 1; i >= 0; i--) { for ( int j = M - 1; j >= 0; j--) { // check if this cell equals 0 if (arr[i][j] == false ) { // increase the number of moves ans++; // flip from this cell to the start point for ( int k = 0; k <= i; k++) { for ( int h = 0; h <= j; h++) { // flip the cell if (arr[k][h] == true ) { arr[k][h] = false ; } else { arr[k][h] = true ; } } } } } } return ans; } // Driven Program public static void Main( string [] args) { bool [][] mat = new bool [][] { new bool [] { false , false , true , true , true }, new bool [] { false , false , false , true , true }, new bool [] { false , false , false , true , true }, new bool [] { true , true , true , true , true }, new bool [] { true , true , true , true , true } }; Console.WriteLine(minOperation(mat)); } } // This code is contributed by Shrikant13 |
PHP
<?php // PHP program to find minimum // operations required to set // all the element of binary matrix $N = 5; $M = 5; // Return minimum operation // required to make all 1s. function minOperation(& $arr ) { global $N , $M ; $ans = 0; for ( $i = $N - 1; $i >= 0; $i --) { for ( $j = $M - 1; $j >= 0; $j --) { // check if this // cell equals 0 if ( $arr [ $i ][ $j ] == 0) { // increase the // number of moves $ans ++; // flip from this cell // to the start point for ( $k = 0; $k <= $i ; $k ++) { for ( $h = 0; $h <= $j ; $h ++) { // flip the cell if ( $arr [ $k ][ $h ] == 1) $arr [ $k ][ $h ] = 0; else $arr [ $k ][ $h ] = 1; } } } } } return $ans ; } // Driver Code $mat = array ( array (0, 0, 1, 1, 1), array (0, 0, 0, 1, 1), array (0, 0, 0, 1, 1), array (1, 1, 1, 1, 1), array (1, 1, 1, 1, 1)); echo minOperation( $mat ); // This code is contributed // by ChitraNayal ?> |
Output:
3
Time Complexity: O(N2 * M2).
Space Complexity: O(N*M).
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