需要翻转数组元素的最小位数以使所有数组元素相等

给定一个由N 个正整数组成的数组arr[] ，任务是找到使所有数组元素相等所需翻转的数组元素的最小位数。

例子：

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

方法：可以通过修改数组元素来解决给定的问题，即在所有数组元素之间的每个位置处设置位和未设置位的数量。请按照以下步骤解决问题：

• 初始化两个频率数组，比如大小为32 的fre0[]和fre1[] ，用于对数组元素的每一位计算 0 和 1 的频率。
• 遍历给定的阵列和用于每个数组元素，ARR [I]如果^第i比特ARR的第j [i]是一组位，然后通过递增1 fre1 [j]的频率。否则，将fre0[j]的频率增加1 。
• 完成上述步骤后，为[0, 32]范围内的每个位i打印fre0[i]和fre1[i]的最小值之和。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
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#

// 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

// 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

输出：

2

时间复杂度： O(N * log N)
辅助空间： O(1)

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