设置位总和等于K的计数对

给定一个数组arr []和一个整数K ，任务是计算置位总和为K的对。
例子：

Input: arr[] = {1, 2, 3, 4, 5}, K = 4 
Output: 1 
(3, 5) is the only valid pair as the count 
of set bits in the integers {1, 2, 3, 4, 5} 
are {1, 1, 2, 1, 2} respectively.
Input: arr[] = {5, 42, 35, 22, 7}, K = 6 
Output: 6 
 

天真的方法：初始化count = 0并运行两个嵌套循环，并检查所有可能的对，并检查count位的总和是否为K。如果是，则增加计数。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the count
// of set bits in n
unsigned int countSetBits(int n)
{
    unsigned int count = 0;
    while (n) {
        n &= (n - 1);
        count++;
    }
    return count;
}
 
// Function to return the count
// of required pairs
int pairs(int arr[], int n, int k)
{
 
    // To store the count
    int count = 0;
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
 
            // Sum of set bits in both the integers
            int sum = countSetBits(arr[i])
                      + countSetBits(arr[j]);
 
            // If current pair satisfies
            // the given condition
            if (sum == k)
                count++;
        }
    }
    return count;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 4;
    cout << pairs(arr, n, k);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
 
// Function to return the count
// of set bits in n
static int countSetBits(int n)
{
    int count = 0;
    while (n > 0)
    {
        n &= (n - 1);
        count++;
    }
    return count;
}
 
// Function to return the count
// of required pairs
static int pairs(int arr[], int n, int k)
{
 
    // To store the count
    int count = 0;
    for (int i = 0; i < n; i++)
    {
        for (int j = i + 1; j < n; j++)
        {
 
            // Sum of set bits in both the integers
            int sum = countSetBits(arr[i])
                    + countSetBits(arr[j]);
 
            // If current pair satisfies
            // the given condition
            if (sum == k)
                count++;
        }
    }
    return count;
}
 
// Driver code
public static void main(String args[])
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int n = arr.length;
    int k = 4;
    System.out.println(pairs(arr, n, k));
}
}
 
// This code has been contributed by 29AjayKumar

Python3

# Python3 implementation of the approach
 
# Function to return the count
# of set bits in n
def countSetBits(n) :
 
    count = 0;
    while (n) :
         
        n &= (n - 1);
        count += 1
         
    return count;
 
 
# Function to return the count
# of required pairs
def pairs(arr, n, k) :
 
    # To store the count
    count = 0;
    for i in range(n) :
        for j in range(i + 1, n) :
 
            # Sum of set bits in both the integers
            sum = countSetBits(arr[i]) + countSetBits(arr[j]);
 
            # If current pair satisfies
            # the given condition
            if (sum == k) :
                count += 1 ;
                 
    return count;
 
 
# Driver code
if __name__ == "__main__" :
 
    arr = [ 1, 2, 3, 4, 5 ];
     
    n = len(arr);
    k = 4;
     
    print(pairs(arr, n, k));
     
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
public class GFG
{
 
// Function to return the count
// of set bits in n
static int countSetBits(int n)
{
    int count = 0;
    while (n > 0)
    {
        n &= (n - 1);
        count++;
    }
    return count;
}
 
// Function to return the count
// of required pairs
static int pairs(int []arr, int n, int k)
{
 
    // To store the count
    int count = 0;
    for (int i = 0; i < n; i++)
    {
        for (int j = i + 1; j < n; j++)
        {
 
            // Sum of set bits in both the integers
            int sum = countSetBits(arr[i])
                    + countSetBits(arr[j]);
 
            // If current pair satisfies
            // the given condition
            if (sum == k)
                count++;
        }
    }
    return count;
}
 
// Driver code
public static void Main(String []args)
{
    int []arr = { 1, 2, 3, 4, 5 };
    int n = arr.Length;
    int k = 4;
    Console.WriteLine(pairs(arr, n, k));
}
}
 
// This code is contributed by Princi Singh

Javascript

 
// javascript implementation of the approach
 
// Function to return the count
// of set bits in n
function countSetBits(n)
{
    var count = 0;
    while (n > 0)
    {
        n &= (n - 1);
        count++;
    }
    return count;
}
 
// Function to return the count
// of required pairs
function pairs(arr , n , k)
{
 
    // To store the count
    var count = 0;
    for (i = 0; i < n; i++)
    {
        for (j = i + 1; j < n; j++)
        {
 
            // Sum of set bits in both the integers
            var sum = countSetBits(arr[i])
                    + countSetBits(arr[j]);
 
            // If current pair satisfies
            // the given condition
            if (sum == k)
                count++;
        }
    }
    return count;
}
 
// Driver code
 
var arr = [ 1, 2, 3, 4, 5 ];
var n = arr.length;
var k = 4;
document.write(pairs(arr, n, k));
 
 
// This code contributed by shikhasingrajput
 

输出：

1

时间复杂度： O(n ² )
高效的方法：假设每个整数都可以使用32位表示，请创建一个大小为32的频率数组freq [] ，其中freq [i]将存储设置位等于i的数量的计数。现在在此频率阵列上运行两个嵌套循环，如果i + j = K，则所有此类i和j的对数将为freq [i] * freq [j] 。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
#define MAX 32
 
// Function to return the count
// of set bits in n
unsigned int countSetBits(int n)
{
    unsigned int count = 0;
    while (n) {
        n &= (n - 1);
        count++;
    }
    return count;
}
 
// Function to return the count
// of required pairs
int pairs(int arr[], int n, int k)
{
 
    // To store the count
    int count = 0;
 
    // Frequency array
    int f[MAX + 1] = { 0 };
    for (int i = 0; i < n; i++)
        f[countSetBits(arr[i])]++;
 
    for (int i = 0; i <= MAX; i++) {
        for (int j = i; j <= MAX; j++) {
 
            // If current pair satisfies
            // the given condition
            if (i + j == k) {
 
                // (arr[i], arr[i]) cannot be a valid pair
                if (i == j)
                    count += ((f[i] * (f[i] - 1)) / 2);
                else
                    count += (f[i] * f[j]);
            }
        }
    }
    return count;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 4;
    cout << pairs(arr, n, k);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG
{
     
static int MAX = 32;
 
// Function to return the count
// of set bits in n
static int countSetBits(int n)
{
    int count = 0;
    while (n > 0)
    {
        n &= (n - 1);
        count++;
    }
    return count;
}
 
// Function to return the count
// of required pairs
static int pairs(int arr[], int n, int k)
{
 
    // To store the count
    int count = 0;
 
    // Frequency array
    int []f = new int[MAX + 1];
    for (int i = 0; i < n; i++)
        f[countSetBits(arr[i])]++;
 
    for (int i = 0; i <= MAX; i++)
    {
        for (int j = i; j <= MAX; j++)
        {
 
            // If current pair satisfies
            // the given condition
            if (i + j == k)
            {
 
                // (arr[i], arr[i]) cannot be a valid pair
                if (i == j)
                    count += ((f[i] * (f[i] - 1)) / 2);
                else
                    count += (f[i] * f[j]);
            }
        }
    }
    return count;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int n = arr.length;
    int k = 4;
    System.out.println(pairs(arr, n, k));
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python implementation of the approach
MAX = 32
 
# Function to return the count
# of set bits in n
def countSetBits(n) :
    count = 0;
    while (n):
        n &= (n - 1);
        count += 1;
 
    return count;
 
# Function to return the count
# of required pairs
def pairs(arr, n, k):
 
    # To store the count
    count = 0;
 
    # Frequency array
    f = [0 for i in range(MAX + 1)]
 
    for i in range(n):
        f[countSetBits(arr[i])] += 1;
 
    for i in range(MAX + 1):
        for j in range(1, MAX + 1):
 
            # If current pair satisfies
            # the given condition
            if (i + j == k):
 
                # (arr[i], arr[i]) cannot be a valid pair
                if (i == j):
                    count += ((f[i] * (f[i] - 1)) / 2);
                else:
                    count += (f[i] * f[j]);
     
    return count;
 
# Driver code
arr = [ 1, 2, 3, 4, 5 ]
n = len(arr)
k = 4
 
print (pairs(arr, n, k))
 
# This code is contributed by CrazyPro

C＃

// C# implementation of the approach
using System;
     
class GFG
{
     
static int MAX = 32;
 
// Function to return the count
// of set bits in n
static int countSetBits(int n)
{
    int count = 0;
    while (n > 0)
    {
        n &= (n - 1);
        count++;
    }
    return count;
}
 
// Function to return the count
// of required pairs
static int pairs(int []arr, int n, int k)
{
 
    // To store the count
    int count = 0;
 
    // Frequency array
    int []f = new int[MAX + 1];
    for (int i = 0; i < n; i++)
        f[countSetBits(arr[i])]++;
 
    for (int i = 0; i <= MAX; i++)
    {
        for (int j = i; j <= MAX; j++)
        {
 
            // If current pair satisfies
            // the given condition
            if (i + j == k)
            {
 
                // (arr[i], arr[i]) cannot be a valid pair
                if (i == j)
                    count += ((f[i] * (f[i] - 1)) / 2);
                else
                    count += (f[i] * f[j]);
            }
        }
    }
    return count;
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 3, 4, 5 };
    int n = arr.Length;
    int k = 4;
    Console.WriteLine(pairs(arr, n, k));
}
}
/* This code is contributed by PrinciRaj1992 */

Java脚本

 
// javascript implementation of the approach
var MAX = 32;
 
// Function to return the count
// of set bits in n
function countSetBits(n)
{
    var count = 0;
    while (n > 0)
    {
        n &= (n - 1);
        count++;
    }
    return count;
}
 
// Function to return the count
// of required pairs
function pairs(arr , n , k)
{
 
    // To store the count
    var count = 0;
 
    // Frequency array
    var f = Array.from({length: MAX + 1}, (_, i) => 0);
    for (i = 0; i < n; i++)
        f[countSetBits(arr[i])]++;
 
    for (i = 0; i <= MAX; i++)
    {
        for (j = i; j <= MAX; j++)
        {
 
            // If current pair satisfies
            // the given condition
            if (i + j == k)
            {
 
                // (arr[i], arr[i]) cannot be a valid pair
                if (i == j)
                    count += ((f[i] * (f[i] - 1)) / 2);
                else
                    count += (f[i] * f[j]);
            }
        }
    }
    return count;
}
 
// Driver code
var arr = [ 1, 2, 3, 4, 5 ];
var n = arr.length;
var k = 4;
document.write(pairs(arr, n, k));
 
// This code contributed by Princi Singh

输出：

1

时间复杂度： O(n)