来自两个字符串的总字符对，在它们的 ascii 值中具有相同数量的设置位

给定两个字符串s1和s2 。任务是从第一个字符串中取出一个字符，从第二个字符串中取出一个字符，并检查这两个字符的 ASCII 值是否具有相同数量的设置位。打印此类对的总数。

例子：

Input: s1 = “xcd”, s2 = “swa” 
Output: 1 
Only valid pair is (d, a) with ASCII values as 100 and 97 respectively. 
Both of which contains 3 set bits.
Input: s1 = “geeks”, s2 = “forgeeks” 
Output: 17 

方法：

• 制作两个大小为 6 的数组 arr1 和 arr2，所有值都初始化为 0，以存储设置位数的频率。由于小写字母中设置的最大位数为 6。
• 遍历字符串s1，找到每个字符的 ASCII 值。将每个 ASCII 值的设置位数的频率存储在数组 arr1 中。 （例如，如果有 3 个字符具有 4 个设置位，则将 3 存储在 arr[4] 中）
• 对字符串s2 执行类似的操作，并将其值存储在另一个数组 arr2 中。
• 用 0 初始化一个计数变量。
• 对于对的总数，对于 i 的所有有效值，继续在计数变量中添加 (arr1[i] * arr2[i])。

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 valid pairs
int totalPairs(string s1, string s2)
{
    int count = 0;
 
    int arr1[7], arr2[7];
 
    // Initialise both arrays with 0
    for (int i = 1; i <= 6; i++) {
        arr1[i] = 0;
        arr2[i] = 0;
    }
 
    // Store frequency of number of set bits for s1
    for (int i = 0; i < s1.length(); i++) {
        int set_bits = __builtin_popcount((int)s1[i]);
        arr1[set_bits]++;
    }
 
    // Store frequency of number of set bits for s2
    for (int i = 0; i < s2.length(); i++) {
        int set_bits = __builtin_popcount((int)s2[i]);
        arr2[set_bits]++;
    }
 
    // Calculate total pairs
    for (int i = 1; i <= 6; i++)
        count += (arr1[i] * arr2[i]);
 
    // Return the count of valid pairs
    return count;
}
 
// Driver code
int main()
{
    string s1 = "geeks";
    string s2 = "forgeeks";
    cout << totalPairs(s1, s2);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
 
    // Function to return the count of valid pairs
    static int totalPairs(String s1, String s2)
    {
        int count = 0;
 
        int[] arr1 = new int[7];
        int[] arr2 = new int[7];
 
        // Default Initialise both arrays 0
        // Store frequency of number of set bits for s1
        for (int i = 0; i < s1.length(); i++)
        {
            int set_bits = Integer.bitCount(s1.charAt(i));
            arr1[set_bits]++;
        }
 
        // Store frequency of number of set bits for s2
        for (int i = 0; i < s2.length(); i++)
        {
            int set_bits = Integer.bitCount(s2.charAt(i));
            arr2[set_bits]++;
        }
 
        // Calculate total pairs
        for (int i = 1; i <= 6; i++)
        {
            count += (arr1[i] * arr2[i]);
        }
 
        // Return the count of valid pairs
        return count;
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        String s1 = "geeks";
        String s2 = "forgeeks";
        System.out.println(totalPairs(s1, s2));
    }
}
 
// This code has been contributed by 29AjayKumar

Python3

# Python3 implementation of the approach
 
# Function to get no of set bits in binary
# representation of positive integer n
def countSetBits(n):
    count = 0
    while (n):
        count += n & 1
        n >>= 1
    return count
     
# Function to return the count
# of valid pairs
def totalPairs(s1, s2) :
     
    count = 0;
 
    arr1 = [0] * 7; arr2 = [0] * 7;
 
    # Store frequency of number
    # of set bits for s1
    for i in range(len(s1)) :
        set_bits = countSetBits(ord(s1[i]))
        arr1[set_bits] += 1;
     
    # Store frequency of number of
    # set bits for s2
    for i in range(len(s2)) :
        set_bits = countSetBits(ord(s2[i]));
        arr2[set_bits] += 1;
 
    # Calculate total pairs
    for i in range(1, 7) :
        count += (arr1[i] * arr2[i]);
 
    # Return the count of valid pairs
    return count;
 
# Driver code
if __name__ == "__main__" :
 
    s1 = "geeks";
    s2 = "forgeeks";
    print(totalPairs(s1, s2));
 
# This code is contributed by Ryuga

C#

// C# implementation of the approach
using System;
using System.Linq;
 
class GFG
{
     
// Function to return the count of valid pairs
static int totalPairs(string s1, string s2)
{
    int count = 0;
 
    int[] arr1 = new int[7];
    int[] arr2 = new int[7];
 
    // Default Initialise both arrays 0
 
    // Store frequency of number of set bits for s1
    for (int i = 0; i < s1.Length; i++)
    {
        int set_bits = Convert.ToString((int)s1[i], 2).Count(c => c == '1');
        arr1[set_bits]++;
    }
 
    // Store frequency of number of set bits for s2
    for (int i = 0; i < s2.Length; i++)
    {
        int set_bits = Convert.ToString((int)s2[i], 2).Count(c => c == '1');
        arr2[set_bits]++;
    }
 
    // Calculate total pairs
    for (int i = 1; i <= 6; i++)
        count += (arr1[i] * arr2[i]);
 
    // Return the count of valid pairs
    return count;
}
 
// Driver code
static void Main()
{
    string s1 = "geeks";
    string s2 = "forgeeks";
    Console.WriteLine(totalPairs(s1, s2));
}
}
 
// This code is contributed by chandan_jnu

Javascript

      // JavaScript implementation of the approach
      // Function to get no of set bits in binary
      // representation of positive integer n
      function countSetBits(n) {
        var count = 0;
        while (n) {
          count += n & 1;
          n >>= 1;
        }
        return count;
      }
 
      // Function to return the count
      // of valid pairs
      function totalPairs(s1, s2) {
        var count = 0;
 
        var arr1 = new Array(7).fill(0);
        var arr2 = new Array(7).fill(0);
 
        // Store frequency of number
        // of set bits for s1
        for (let i = 0; i < s1.length; i++) {
          set_bits = countSetBits(s1[i].charCodeAt(0));
          arr1[set_bits] += 1;
        }
        // Store frequency of number of
        // set bits for s2
        for (let i = 0; i < s2.length; i++) {
          set_bits = countSetBits(s2[i].charCodeAt(0));
          arr2[set_bits] += 1;
        }
        // Calculate total pairs
        for (let i = 1; i < 7; i++) {
          count += arr1[i] * arr2[i];
        }
 
        // Return the count of valid pairs
        return count;
      }
      // Driver code
      var s1 = "geeks";
      var s2 = "forgeeks";
      document.write(totalPairs(s1, s2));
    

输出：

17