Minimum number of bits of array elements required to be flipped to make all array elements equal

Last Updated : 06 May, 2021

Given an array arr[] consisting of N positive integers, the task is to find the minimum number of bits of array elements required to be flipped to make all array elements equal.

Examples:

Input: arr[] = {3, 5}
Output: 2
Explanation:
Following are the flipping of bits of the array elements required:

For element arr[0](= 3): Flipping the 3rd bit from the right modifies arr[0] to (= 7(111)₂). Now, the array becomes {7, 5}.
For element arr[0](= 7): Flipping the 2nd bit from the right modifies arr[0] to 5 (= (101)₂). Now, the array becomes {5, 5}.

After performing the above operations, all the array elements are equal. Therefore, the total number of flips of bits required is 2.

Input: arr[] = {4, 6, 3, 4, 5}
Output: 5

Approach: The given problem can be solved by modifying the array element in such a way that the number of set bits and unset bits at every position between all array elements. Follow the below steps to solve the problem:

Initialize two frequency arrays say fre0[] and fre1[] of size 32 for counting the frequency of 0 and 1 for every bit of array elements.
Traverse the given array and for each array element, arr[i] if the j^th bit of arr[i] is a set bit, then increment the frequency of fre1[j] by 1. Otherwise, increment the frequency of fre0[j] by 1.
After completing the above steps, print the sum of the minimum of fre0[i] and fre1[i] for each bit i over the range [0, 32].

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to count minimum number
// of bits required to be flipped
// to make all array elements equal
int makeEqual(int* arr, int n)
{
    // Stores the count of unset bits
    int fre0[33] = { 0 };

    // Stores the count of set bits
    int fre1[33] = { 0 };

    // Traverse the array
    for (int i = 0; i < n; i++) {

        int x = arr[i];

        // Traverse the bit of arr[i]
        for (int j = 0; j < 33; j++) {

            // If current bit is set
            if (x & 1) {

                // Increment fre1[j]
                fre1[j] += 1;
            }

            // Otherwise
            else {

                // Increment fre0[j]
                fre0[j] += 1;
            }

            // Right shift x by 1
            x = x >> 1;
        }
    }

    // Stores the count of total moves
    int ans = 0;

    // Traverse the range [0, 32]
    for (int i = 0; i < 33; i++) {

        // Update the value of ans
        ans += min(fre0[i], fre1[i]);
    }

    // Return the minimum number of
    // flips required
    return ans;
}

// Driver Code
int main()
{
    int arr[] = { 3, 5 };
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << makeEqual(arr, N);

    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;

class GFG{

// Function to count minimum number
// of bits required to be flipped
// to make all array elements equal
static int makeEqual(int arr[], int n)
{
    
    // Stores the count of unset bits
    int fre0[] = new int[33];

    // Stores the count of set bits
    int fre1[] = new int[33];

    // Traverse the array
    for(int i = 0; i < n; i++)
    {
        int x = arr[i];

        // Traverse the bit of arr[i]
        for(int j = 0; j < 33; j++) 
        {

            // If current bit is set
            if ((x & 1) != 0) 
            {
                
                // Increment fre1[j]
                fre1[j] += 1;
            }

            // Otherwise
            else
            {
                
                // Increment fre0[j]
                fre0[j] += 1;
            }

            // Right shift x by 1
            x = x >> 1;
        }
    }

    // Stores the count of total moves
    int ans = 0;

    // Traverse the range [0, 32]
    for(int i = 0; i < 33; i++)
    {
        
        // Update the value of ans
        ans += Math.min(fre0[i], fre1[i]);
    }

    // Return the minimum number of
    // flips required
    return ans;
}

// Driver Code
public static void main(String[] args)
{
    int arr[] = { 3, 5 };
    int N = arr.length;
    
    System.out.print(makeEqual(arr, N));
}
}

// This code is contributed by Kingash

Python3

# Python3 program for the above approach

# Function to count minimum number
# of bits required to be flipped
# to make all array elements equal
def makeEqual(arr, n):
  
    # Stores the count of unset bits
    fre0 = [0]*33

    # Stores the count of set bits
    fre1 = [0]*33
    
    # Traverse the array
    for i in range(n):

        x = arr[i]

        # Traverse the bit of arr[i]
        for j in range(33):

            # If current bit is set
            if (x & 1):
                # Increment fre1[j]
                fre1[j] += 1
            # Otherwise
            else:
                # Increment fre0[j]
                fre0[j] += 1
            # Right shift x by 1
            x = x >> 1

    # Stores the count of total moves
    ans = 0

    # Traverse the range [0, 32]
    for i in range(33):
      
        # Update the value of ans
        ans += min(fre0[i], fre1[i])

    # Return the minimum number of
    # flips required
    return ans

# Driver Code
if __name__ == '__main__':
    arr= [3, 5]
    N = len(arr)
    print(makeEqual(arr, N))

# This code is contributed by mohit kumar 29.

// C# program for the above approach
using System;
class GFG {

    // Function to count minimum number
    // of bits required to be flipped
    // to make all array elements equal
    static int makeEqual(int[] arr, int n)
    {

        // Stores the count of unset bits
        int[] fre0 = new int[33];

        // Stores the count of set bits
        int[] fre1 = new int[33];

        // Traverse the array
        for (int i = 0; i < n; i++) {
            int x = arr[i];

            // Traverse the bit of arr[i]
            for (int j = 0; j < 33; j++) {

                // If current bit is set
                if ((x & 1) != 0) {

                    // Increment fre1[j]
                    fre1[j] += 1;
                }

                // Otherwise
                else {

                    // Increment fre0[j]
                    fre0[j] += 1;
                }

                // Right shift x by 1
                x = x >> 1;
            }
        }

        // Stores the count of total moves
        int ans = 0;

        // Traverse the range [0, 32]
        for (int i = 0; i < 33; i++) {

            // Update the value of ans
            ans += Math.Min(fre0[i], fre1[i]);
        }

        // Return the minimum number of
        // flips required
        return ans;
    }

    // Driver Code
    public static void Main()
    {
        int[] arr = { 3, 5 };
        int N = arr.Length;

        Console.WriteLine(makeEqual(arr, N));
    }
}

// This code is contributed by ukasp.

JavaScript

<script>
// javascript program for the above approach
    // Function to count minimum number
    // of bits required to be flipped
    // to make all array elements equal
    function makeEqual(arr , n)
    {

        // Stores the count of unset bits
        var fre0 = Array(33).fill(0);

        // Stores the count of set bits
        var fre1 = Array(33).fill(0);

        // Traverse the array
        for (i = 0; i < n; i++) {
            var x = arr[i];

            // Traverse the bit of arr[i]
            for (j = 0; j < 33; j++) {

                // If current bit is set
                if ((x & 1) != 0) {

                    // Increment fre1[j]
                    fre1[j] += 1;
                }

                // Otherwise
                else {

                    // Increment fre0[j]
                    fre0[j] += 1;
                }

                // Right shift x by 1
                x = x >> 1;
            }
        }

        // Stores the count of total moves
        var ans = 0;

        // Traverse the range [0, 32]
        for (i = 0; i < 33; i++) {

            // Update the value of ans
            ans += Math.min(fre0[i], fre1[i]);
        }

        // Return the minimum number of
        // flips required
        return ans;
    }

    // Driver Code
        var arr = [ 3, 5 ];
        var N = arr.length;

        document.write(makeEqual(arr, N));

// This code is contributed by aashish1995
</script>

Output:

Time Complexity: O(N * log N)
Auxiliary Space: O(1)

Find the minimum number of operations required to make all array elements equal

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