Total character pairs from two strings, with equal number of set bits in their ascii value
Last Updated :
08 Feb, 2024
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 characters 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 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++
#include <bits/stdc++.h>
using namespace std;
int totalPairs(string s1, string s2)
{
int count = 0;
int arr1[7], arr2[7];
for (int i = 1; i <= 6; i++) {
arr1[i] = 0;
arr2[i] = 0;
}
for (int i = 0; i < s1.length(); i++) {
int set_bits = __builtin_popcount((int)s1[i]);
arr1[set_bits]++;
}
for (int i = 0; i < s2.length(); i++) {
int set_bits = __builtin_popcount((int)s2[i]);
arr2[set_bits]++;
}
for (int i = 1; i <= 6; i++)
count += (arr1[i] * arr2[i]);
return count;
}
int main()
{
string s1 = "geeks";
string s2 = "forgeeks";
cout << totalPairs(s1, s2);
return 0;
}
|
Java
class GFG
{
static int totalPairs(String s1, String s2)
{
int count = 0;
int[] arr1 = new int[7];
int[] arr2 = new int[7];
for (int i = 0; i < s1.length(); i++)
{
int set_bits = Integer.bitCount(s1.charAt(i));
arr1[set_bits]++;
}
for (int i = 0; i < s2.length(); i++)
{
int set_bits = Integer.bitCount(s2.charAt(i));
arr2[set_bits]++;
}
for (int i = 1; i <= 6; i++)
{
count += (arr1[i] * arr2[i]);
}
return count;
}
public static void main(String[] args)
{
String s1 = "geeks";
String s2 = "forgeeks";
System.out.println(totalPairs(s1, s2));
}
}
|
Python3
def countSetBits(n):
count = 0
while (n):
count += n & 1
n >>= 1
return count
def totalPairs(s1, s2) :
count = 0;
arr1 = [0] * 7; arr2 = [0] * 7;
for i in range(len(s1)) :
set_bits = countSetBits(ord(s1[i]))
arr1[set_bits] += 1;
for i in range(len(s2)) :
set_bits = countSetBits(ord(s2[i]));
arr2[set_bits] += 1;
for i in range(1, 7) :
count += (arr1[i] * arr2[i]);
return count;
if __name__ == "__main__" :
s1 = "geeks";
s2 = "forgeeks";
print(totalPairs(s1, s2));
|
C#
using System;
using System.Linq;
class GFG
{
static int totalPairs(string s1, string s2)
{
int count = 0;
int[] arr1 = new int[7];
int[] arr2 = new int[7];
for (int i = 0; i < s1.Length; i++)
{
int set_bits = Convert.ToString((int)s1[i], 2).Count(c => c == '1');
arr1[set_bits]++;
}
for (int i = 0; i < s2.Length; i++)
{
int set_bits = Convert.ToString((int)s2[i], 2).Count(c => c == '1');
arr2[set_bits]++;
}
for (int i = 1; i <= 6; i++)
count += (arr1[i] * arr2[i]);
return count;
}
static void Main()
{
string s1 = "geeks";
string s2 = "forgeeks";
Console.WriteLine(totalPairs(s1, s2));
}
}
|
Javascript
<script>
function countSetBits(n) {
var count = 0;
while (n) {
count += n & 1;
n >>= 1;
}
return count;
}
function totalPairs(s1, s2) {
var count = 0;
var arr1 = new Array(7).fill(0);
var arr2 = new Array(7).fill(0);
for (let i = 0; i < s1.length; i++) {
set_bits = countSetBits(s1[i].charCodeAt(0));
arr1[set_bits] += 1;
}
for (let i = 0; i < s2.length; i++) {
set_bits = countSetBits(s2[i].charCodeAt(0));
arr2[set_bits] += 1;
}
for (let i = 1; i < 7; i++) {
count += arr1[i] * arr2[i];
}
return count;
}
var s1 = "geeks";
var s2 = "forgeeks";
document.write(totalPairs(s1, s2));
</script>
|
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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