来自数组的对 {X, Y} 的计数，使得 X ⊕ Y 中设置位的计数和 X & Y 中设置位计数的两倍之和为 M

给定一个由N 个非负整数和一个整数M组成的数组arr[] ，任务是找到满足条件setBits(X ⊕ Y) + 2 * setBits的数组元素的无序对{X, Y}的计数(X & Y) = M ，其中⊕表示按位异或， &表示按位与。

例子：

Input: arr[] = {30, 0, 4, 5 }, M = 2
Output: 2
Explanation: The pairs satisfying the necessary conditions are {{3, 0}, {0, 5}}.

Input: arr[] = {1, 2, 3, 4}, M = 3
Output: 3
Explanation: The pairs satisfying the necessary conditions are {{1, 3}, {2, 3}, {3, 4}}.

朴素的方法：最简单的方法是从给定的数组中生成所有可能的对，对于每一对，检查是否满足必要条件。增加满足给定条件的对的计数，最后打印对的计数。

时间复杂度： O(N ² )
辅助空间： O(1)

高效的方法：上述方法可以基于以下观察进行优化：

• From the property of Bitwise XOR:
  • setBits( a⊕ b ) = setBits( a|b ) – setBits( a&b )
  • setBits( a|b ) = setBits(a) + setBits(b) – setBits( a&b )
• Therefore, the given equation becomes:
  • ( setBits( X|Y ) – setBits( X&Y ) )+( 2 × setBits( X&Y ) ) = M
  • setBits( X ) + setBits ( Y ) – setBits( X&Y ) – setBits( X&Y ) + ( 2 × setBits ( X&Y ) ) = M
  • setBits( X ) + setBits( Y ) = M
• Therefore, the task reduces to counting the pairs of elements whose sum of set bits is equal to M.

请按照以下步骤解决问题：

• 首先，遍历数组arr[]。
• 对于每个数组元素arr[i]，使用其中存在的设置位计数更新arr[i] 。
• 现在打印数组中总和等于 M 的对的计数。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to count number
// of setbits in the number n
int countsetbits(int n)
{
    // Stores the count of setbits
    int count = 0;
 
    // Iterate while N is
    // greater than 0
    while (n) {
 
        // Increment count by 1
        // if N is odd
        count += n & 1;
 
        // Right shift N
        n >>= 1;
    }
 
    // Return the count of set bits
    return (count);
}
 
// Function to find total count of
// given pairs satisfying the equation
int countPairs(int a[], int N, int M)
{
 
    for (int i = 0; i < N; i++) {
 
        // Update arr[i] with the count
        // of set bits of arr[i]
        a[i] = countsetbits(a[i]);
    }
 
    // Stores the frequency
    // of each array element
    unordered_map mp;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // Increment the count
        // of arr[i] in mp
        mp[a[i]]++;
    }
 
    // Stores the total count of pairs
    int count = 0;
 
    // Traverse the array arr[]
    for (int i = 0; i < N; i++) {
 
        // Increment count by mp[M - a[i]]
        count += mp[M - a[i]];
 
        // If a[i] is equal to M-a[i]
        if (a[i] == M - a[i]) {
 
            // Decrment count by 1
            count--;
        }
    }
 
    // Return count/2
    return (count / 2);
}
 
// Driver Code
int main()
{
    // Input
    int arr[] = { 3, 0, 4, 5 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int M = 2;
 
    cout << countPairs(arr, N, M);
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG{
 
// Function to count number
// of setbits in the number n
static int countsetbits(int n)
{
     
    // Stores the count of setbits
    int count = 0;
 
    // Iterate while N is
    // greater than 0
    while (n != 0)
    {
         
        // Increment count by 1
        // if N is odd
        count += n & 1;
 
        // Right shift N
        n >>= 1;
    }
 
    // Return the count of set bits
    return (count);
}
 
// Function to find total count of
// given pairs satisfying the equation
static int countPairs(int[] a, int N, int M)
{
    for(int i = 0; i < N; i++)
    {
         
        // Update arr[i] with the count
        // of set bits of arr[i]
        a[i] = countsetbits(a[i]);
    }
 
    // Stores the frequency
    // of each array element
     HashMap mp = new HashMap();
 
    // Traverse the array
    for (int i = 0; i < N; i++)
    {
        if (mp.containsKey(a[i]))
        {
            mp.put(a[i], mp.get(a[i]) + 1);
        }
        else
        {
            mp.put(a[i], 1);
        }
    }
     
    // Stores the total count of pairs
    int count = 0;
 
    // Traverse the array arr[]
    for(int i = 0; i < N; i++)
    {
         
        // Increment count by mp[M - a[i]]
        count += mp.get(M - a[i]);
 
        // If a[i] is equal to M-a[i]
        if (a[i] == M - a[i])
        {
             
            // Decrment count by 1
            count--;
        }
    }
 
    // Return count/2
    return (count / 2);
}   
 
// Driver Code
public static void main(String[] args)
{
     
    // Input
    int[] arr = { 3, 0, 4, 5 };
    int N = arr.length;
    int M = 2;
 
    System.out.println(countPairs(arr, N, M));
}
}
 
// This code is contributed by avijitmondal1998

Python3

# Python3 Program for the above approach
 
# Function to count number
# of setbits in the number n
def  countSetBits(n):
   
    # Stores the count of setbits
    count = 0
     
    # Iterate while N is
    # greater than 0
    while (n):
       
        # Increment count by 1
        # if N is odd
        count += n & 1
         
        # Right shift N
        n >>= 1
         
    # Return the count of set bits
    return count
 
def countPairs(arr, N, M):
    for i in range(0, N):
       
        # Update arr[i] with the count
        # of set bits of arr[i]
        arr[i] = countSetBits(arr[i])
         
    # Store counts of all elements in a dictionary
    mp = {}
    for i in range(0, N):
        if arr[i] in mp:
            mp[arr[i]] += 1
        else:
            mp[arr[i]] = 1
    twice_count = 0
     
    # Iterate through each element and increment
    # the count (Notice that every pair is
    # counted twice)
    for i in range(0, N):
        if M - arr[i] in mp.keys():
            twice_count += mp[M - arr[i]]
             
        # if (arr[i], arr[i]) pair satisfies the
        # condition, then we need to ensure that
        # the count is  decreased by one such
        # that the (arr[i], arr[i]) pair is not
        # considered
        if (M - arr[i] == arr[i]):
            twice_count -= 1
  
    # return the half of twice_count
    return int(twice_count / 2)
         
# Driver code
# Input
arr = [ 3, 0, 4, 5]
N = len(arr)
M = 2
print(countPairs(arr, N, M))
 
# This cose is contributed by santhoshcharan.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to count number
// of setbits in the number n
static int countsetbits(int n)
{
     
    // Stores the count of setbits
    int count = 0;
 
    // Iterate while N is
    // greater than 0
    while (n != 0)
    {
         
        // Increment count by 1
        // if N is odd
        count += n & 1;
 
        // Right shift N
        n >>= 1;
    }
 
    // Return the count of set bits
    return (count);
}
 
// Function to find total count of
// given pairs satisfying the equation
static int countPairs(int[] a, int N, int M)
{
    for(int i = 0; i < N; i++)
    {
         
        // Update arr[i] with the count
        // of set bits of arr[i]
        a[i] = countsetbits(a[i]);
    }
 
    // Stores the frequency
    // of each array element
    Dictionary mp = new Dictionary();
 
    // Traverse the array
    for(int i = 0; i < N; ++i)
    {
            
        // Update frequency of
        // each array element
        if (mp.ContainsKey(a[i]) == true)
            mp[a[i]] += 1;
        else
            mp[a[i]] = 1;
    }
     
    // Stores the total count of pairs
    int count = 0;
 
    // Traverse the array arr[]
    for(int i = 0; i < N; i++)
    {
         
        // Increment count by mp[M - a[i]]
        count += mp[M - a[i]];
 
        // If a[i] is equal to M-a[i]
        if (a[i] == M - a[i])
        {
             
            // Decrment count by 1
            count--;
        }
    }
 
    // Return count/2
    return (count / 2);
}
 
// Driver Code
static public void Main()
{
     
    // Input
    int[] arr = { 3, 0, 4, 5 };
    int N = arr.Length;
    int M = 2;
 
    Console.WriteLine(countPairs(arr, N, M));
}
}
 
// This code is contributed by sanjoy_62

Javascript

// Javascript program for the above approach
 
// Function to count number
// of setbits in the number n
function countsetbits(n)
{
    // Stores the count of setbits
    var count = 0;
 
    // Iterate while N is
    // greater than 0
    while (n) {
 
        // Increment count by 1
        // if N is odd
        count += n & 1;
 
        // Right shift N
        n >>= 1;
    }
 
    // Return the count of set bits
    return (count);
}
 
// Function to find total count of
// given pairs satisfying the equation
function countPairs(a, N, M)
{
 
    for (var i = 0; i < N; i++) {
 
        // Update arr[i] with the count
        // of set bits of arr[i]
        a[i] = countsetbits(a[i]);
    }
 
    // Stores the frequency
    // of each array element
    var mp = new Map();
 
    // Traverse the array
    for (var i = 0; i < N; i++) {
 
        // Increment the count
        // of arr[i] in mp
        if(mp.has(a[i]))
            mp.set(a[i], mp.get(a[i])+1)
        else
            mp.set(a[i], 1)
    }
 
    // Stores the total count of pairs
    var count = 0;
 
    // Traverse the array arr[]
    for (var i = 0; i < N; i++) {
 
        // Increment count by mp[M - a[i]]
        count += mp.get(M - a[i]);
 
        // If a[i] is equal to M-a[i]
        if (a[i] == M - a[i]) {
 
            // Decrment count by 1
            count--;
        }
    }
 
    // Return count/2
    return (count / 2);
}
 
// Driver Code
// Input
var arr = [3, 0, 4, 5];
var N = arr.length;
var M = 2;
document.write( countPairs(arr, N, M));
 

输出：

2

时间复杂度： O(N)
辅助空间： O(1)

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