Skip to content
Related Articles

Related Articles

Table of Contents

Improve Article
Save Article
Like Article

Total character pairs from two strings, with equal number of set bits in their ascii value

  • Difficulty Level : Easy
  • Last Updated : 14 May, 2021

Given two strings s1 and s2. The task is to take one character from the first string and one character from the second string and check if the ASCII values of both the character have the same number of set bits. Print the total number of such pairs.

Examples:

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 

Approach: 

  • Make two arrays arr1 and arr2 of size 6 with all values initialized to 0 to store the frequency of the number of set bits. Since the maximum number of set bits in lower case alphabets is 6.
  • Traverse the string s1, and find the ASCII value of each character. Store the frequency of the number of set bits of each ASCII value in an array arr1. (For example, if there are 3 characters with 4 set bits, then store 3 at arr[4])
  • Do a similar operation for string s2 and store its value in another array arr2.
  • Initialize a count variable with 0.
  • For total number of pairs, keep on adding (arr1[i] * arr2[i]) in count variable for all valid values of i.

Below is the implementation of the above approach: 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
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



<script>
      // 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));
    </script>
Output: 
17
 

My Personal Notes
arrow_drop_up
Next

Convert the ASCII value sentence to its equivalent string
Recommended Articles
Page :
Replace every character of string by character whose ASCII value is K times more than it
01, Aug 18
Total distinct pairs from two arrays such that second number can be obtained by inverting bits of first
04, Apr 19
Count of pairs {X, Y} from an array such that sum of count of set bits in X ⊕ Y and twice the count of set bits in X & Y is M
12, Mar 21
Count pairs of elements such that number of set bits in their OR is B[i]
17, Jun 19
Count pairs of elements such that number of set bits in their AND is B[i]
31, May 19
Count of pairs in given Array with product of their values equal to sum of their indices (arr[i]*arr[j] = i+j)
18, Nov 21
Count strings having sum of ASCII values of characters equal to a Prime or Armstrong Number
03, Jun 21
Count pairs of characters in a string whose ASCII value difference is K
30, May 19
Sub-strings having exactly k characters that have ASCII value greater than p
03, Nov 18
Minimize swaps of same-indexed characters to make sum of ASCII value of characters of both the strings odd
10, Feb 21
Count of index pairs (i, j) such that string after deleting ith character is equal to string after deleting jth character
30, Nov 21
Convert Hexadecimal value String to ASCII value String
22, May 18
Check whether two strings can be made equal by reversing substring of equal length from both strings
15, Jul 20
Count total set bits in all numbers from 1 to N | Set 3
19, Sep 19
Count total set bits in all numbers from 1 to n | Set 2
10, Jul 19
Count of strings possible by replacing two consecutive same character with new character
30, Mar 20
Flip bits of the sum of count of set bits of two given numbers
03, Jan 21
Repeat last occurrence of each alphanumeric character to their position in character family times
24, Aug 21
Print each word in a sentence with their corresponding average of ASCII values
19, Nov 18
Count new pairs of strings that can be obtained by swapping first characters of pairs of strings from given array
14, Apr 21
Count pairs from 1 to N such that their Sum is divisible by their XOR
17, Mar 20
Minimize ASCII values sum after removing all occurrences of one character
06, Mar 19
Sort the character array based on ASCII % N
17, Jun 19
Program to convert given Binary to its equivalent ASCII character string
09, Dec 21
Article Contributed By :
https://media.geeksforgeeks.org/auth/avatar.png
Shashank_Sharma
@Shashank_Sharma
Vote for difficulty
Current difficulty :
Easy





Improved By :
Article Tags :
Practice Tags :
Report Issue

We use cookies to ensure you have the best browsing experience on our website. By using our site, you
acknowledge that you have read and understood our
Cookie Policy &
        Privacy Policy

Lightbox

Start Your Coding Journey Now!