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

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 character have the same number of set bits. Print the total number of such pairs.

Examples:

Input: s1 = “xcd”, s2 = “swa”
Output:
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 arr1[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 ``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

 ``

Output:

`17`

Time Complexity: O(32 * (n1 + n2)), where n1 and n2 are the length of the given two strings.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

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