需要翻转以使所有数组元素相等的数组元素的最小位数

给定一个由N个正整数组成的数组arr [] ，任务是找到需要翻转以使所有数组元素相等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以使所有数组元素之间的每个位置的设置位和未设置位的数量变大来解决给定的问题。请按照以下步骤解决问题：

• 初始化两个频率数组，例如大小为32的fre0 []和fre1 [] ，以对数组元素的每一位计数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，在范围[ 0，32]内为每个位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

输出：

2

时间复杂度： O(N * log N)
辅助空间： O(1)