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Minimum operations required to set all elements of binary matrix

  • Difficulty Level : Medium
  • Last Updated : 22 Apr, 2021

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

 

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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
?>

Javascript




<script>
// Javascript program to find minimum operations required
// to set all the element of binary matrix
     
    let N = 5;
    let M = 5;
     
    // Return minimum operation required to make all 1s.
    function minOperation(arr)
    {
        let ans = 0;
        for (let i = N - 1; i >= 0; i--)
        {
            for (let 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 (let k = 0; k <= i; k++)
                    {
                        for (let 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
    let 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]]
    document.write(minOperation(mat))
     
    // This code is contributed by avanitrachhadiya2155
</script>

Output: 
 

3

Time Complexity: O(N2 * M2). 
Space Complexity: O(N*M).
This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 




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