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Count number of equal pairs in a string

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Given a string s, find the number of pairs of characters that are same. Pairs (s[i], s[j]), (s[j], s[i]), (s[i], s[i]), (s[j], s[j]) should be considered different. 

Examples :

Input: air
Output: 3
Explanation :
3 pairs that are equal are (a, a), (i, i) and (r, r)
Input : geeksforgeeks
Output : 31

The naive approach will be you to run two nested for loops and find out all pairs and keep a count of all pairs. 

Steps to implement-

  • Find the size of the given string
  • Initialize a variable let us say “ans” with value 0 to store the answer
  • Run two nested “for loop” from 0 to the size of string-1
  • In that nested loop if two pair of characters are the same then increment that “ans” by 1
  • In last return or print the value stored in “ans”

Code-

C++

// CPP program to count the number of pairs
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the number of equal pairs
int countPairs(string s)
{
    // Length of string
    int n = s.size();
 
    // Stores the answer
    int ans = 0;
 
    // Running nested loops to check equal pairs
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            // When such pair got found
            if (s[i] == s[j]) {
                ans++;
            }
        }
    }
 
    return ans;
}
 
// Driver Code
int main()
{
    string s = "geeksforgeeks";
    cout << countPairs(s);
    return 0;
}

                    

Java

public class Main {
    // Function to count the number of equal pairs
    public static int countPairs(String s) {
        // Length of string
        int n = s.length();
 
        // Stores the answer
        int ans = 0;
 
        // Running nested loops to check equal pairs
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                // When such pair got found
                if (s.charAt(i) == s.charAt(j)) {
                    ans++;
                }
            }
        }
 
        return ans;
    }
 
    // Driver Code
    public static void main(String[] args) {
        String s = "geeksforgeeks";
        System.out.println(countPairs(s));
    }
}
// This code is contributed by rambabuguphka

                    

Python

# Function to count the number of equal pairs
def countPairs(s):
    # Length of the string
    n = len(s)
 
    # Initialize the answer
    ans = 0
 
    # Running nested loops to check equal pairs
    for i in range(n):
        for j in range(n):
            # When such pair is found
            if s[i] == s[j]:
                ans += 1
 
    return ans
 
# Driver Code
if __name__ == "__main__":
    s = "geeksforgeeks"
    print(countPairs(s))

                    

C#

using System;
 
public class GFG
{
    // Function to count the number of equal pairs
    public static int CountPairs(string s)
    {
        // Length of the string
        int n = s.Length;
 
        // Stores the answer
        int ans = 0;
 
        // Running nested loops to check equal pairs
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < n; j++)
            {
                // When such a pair is found
                if (s[i] == s[j])
                {
                    ans++;
                }
            }
        }
 
        return ans;
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        string s = "geeksforgeeks";
        Console.WriteLine(CountPairs(s));
    }
}

                    

Javascript

// JavaScript program to count the number of pairs
 
// Function to count the number of equal pairs
function countPairs(s)
{
    // Length of string
    let n = s.length;
  
    // Stores the answer
    let ans = 0;
  
    // Running nested loops to check equal pairs
    for (let i = 0; i < n; i++) {
        for (let j = 0; j < n; j++) {
            // When such pair got found
            if (s[i] == s[j]) {
                ans++;
            }
        }
    }
  
    return ans;
}
 
// Driver Code
let s = "geeksforgeeks";
console.log(countPairs(s));

                    

PHP

<?php
 
// Function to count the number of equal pairs
function countPairs($s) {
    // Length of the string
    $n = strlen($s);
 
    // Initialize the answer
    $ans = 0;
 
    // Running nested loops to check equal pairs
    for ($i = 0; $i < $n; $i++) {
        for ($j = 0; $j < $n; $j++) {
            // When such pair is found
            if ($s[$i] == $s[$j]) {
                $ans += 1;
            }
        }
    }
 
    return $ans;
}
 
// Driver Code
$s = "geeksforgeeks";
echo countPairs($s);
 
?>

                    

Output-

31

Time Complexity: O(N2), because of two nested loops
Auxiliary Space: O(1), because no extra space has been used

For an efficient approach, we need to count the number of equal pairs in linear time. Since pairs (x, y) and pairs (y, x) are considered different. We need to use a hash table to store the count of all occurrences of a character.So we know if a character occurs twice, then it will have 4 pairs – (i, i), (j, j), (i, j), (j, i). So using a hash function, store the occurrence of each character, then for each character the number of pairs will be occurrence^2. Hash table will be 256 in length as we have 256 characters. 

Below is the implementation of the above approach : 

C++

// CPP program to count the number of pairs
#include <bits/stdc++.h>
using namespace std;
#define MAX 256
 
// Function to count the number of equal pairs
int countPairs(string s)
{
    // Hash table
    int cnt[MAX] = { 0 };
 
    // Traverse the string and count occurrence
    for (int i = 0; i < s.length(); i++)
        cnt[s[i]]++;
 
    // Stores the answer
    int ans = 0;
 
    // Traverse and check the occurrence of every character
    for (int i = 0; i < MAX; i++)
        ans += cnt[i] * cnt[i];
 
    return ans;
}
 
// Driver Code
int main()
{
    string s = "geeksforgeeks";
    cout << countPairs(s);
    return 0;
}

                    

C

// C program to count the number of equal pairs
#include <stdio.h>
#define MAX 256
 
// Function to count the number of equal pairs
int countPairs(char s[])
{
    // Hash table
    int cnt[MAX] = { 0 };
 
    // Traverse the string and count occurrence
    for (int i = 0; s[i] != '\0'; i++)
        cnt[s[i]]++;
 
    // Stores the answer
    int ans = 0;
 
    // Traverse and check the occurrence of every character
    for (int i = 0; i < MAX; i++)
        ans += cnt[i] * cnt[i];
 
    return ans;
}
 
// Driver Code
int main()
{
    char s[] = "geeksforgeeks";
    printf("%d",countPairs(s));
    return 0;
}
 
// This code is contributed by sandeepkrsuman

                    

Java

// Java program to count the number of pairs
import java.io.*;
 
class GFG {
 
    static int MAX = 256;
     
    // Function to count the number of equal pairs
    static int countPairs(String s)
    {
        // Hash table
        int cnt[] = new int[MAX];
     
        // Traverse the string and count occurrence
        for (int i = 0; i < s.length(); i++)
            cnt[s.charAt(i)]++;
     
        // Stores the answer
        int ans = 0;
     
        // Traverse and check the occurrence
        // of every character
        for (int i = 0; i < MAX; i++)
            ans += cnt[i] * cnt[i];
     
        return ans;
    }
     
    // Driver Code
    public static void main (String[] args)
    {
        String s = "geeksforgeeks";
        System.out.println(countPairs(s));
    }
}
 
// This code is contributed by vt_m

                    

Python 3

# Python3 program to count the
# number of pairs
MAX = 256
 
# Function to count the number
# of equal pairs
def countPairs(s):
     
    # Hash table
    cnt = [0 for i in range(0, MAX)]
 
    # Traverse the string and count
    # occurrence
    for i in range(len(s)):
        cnt[ord(s[i]) - 97] += 1
 
    # Stores the answer
    ans = 0
 
    # Traverse and check the occurrence
    # of every character
    for i in range(0, MAX):
        ans += cnt[i] * cnt[i]
 
    return ans
 
# Driver code
if __name__=="__main__":
    s = "geeksforgeeks"
    print(countPairs(s))
 
# This code is contributed
# by Sairahul099        

                    

C#

// C# program to count the number of pairs
using System;
 
class GFG {
 
    static int MAX = 256;
     
    // Function to count the number of equal pairs
    static int countPairs(string s)
    {
        // Hash table
        int []cnt = new int[MAX];
     
        // Traverse the string and count occurrence
        for (int i = 0; i < s.Length; i++)
            cnt[s[i]]++;
     
        // Stores the answer
        int ans = 0;
     
        // Traverse and check the occurrence
        // of every character
        for (int i = 0; i < MAX; i++)
            ans += cnt[i] * cnt[i];
     
        return ans;
    }
     
    // Driver Code
    public static void Main ()
    {
        string s = "geeksforgeeks";
        Console.WriteLine(countPairs(s));
    }
}
 
// This code is contributed by vt_m

                    

Javascript

<script>
// JavaScript program to count the
// number of pairs
const MAX = 256
 
// Function to count the number
// of equal pairs
function countPairs(s){
 
    // Hash table
    let cnt = new Array(MAX).fill(0)
 
    // Traverse the string and count
    // occurrence
    for(let i=0;i<s.length;i++)
        cnt[s.charCodeAt(i) - 97] += 1
 
    // Stores the answer
    let ans = 0
 
    // Traverse and check the occurrence
    // of every character
    for(let i = 0; i < MAX; i++)
        ans += cnt[i] * cnt[i]
 
    return ans
}
 
// Driver code
 
let s = "geeksforgeeks"
document.write(countPairs(s))
 
// This code is contributed
// by shinjanpatra
 
</script>

                    

PHP

<?php
 
// Function to count the number of equal pairs
function countPairs($s) {
    // Hash table
    $cnt = array_fill(0, 256, 0);
 
    // Traverse the string and count occurrence
    for ($i = 0; $i < strlen($s); $i++) {
        $cnt[ord($s[$i]) - 97]++;
    }
 
    // Stores the answer
    $ans = 0;
 
    // Traverse and check the occurrence of every character
    for ($i = 0; $i < 256; $i++) {
        $ans += $cnt[$i] * $cnt[$i];
    }
 
    return $ans;
}
 
// Driver code
$s = "geeksforgeeks";
echo countPairs($s);
 
?>

                    

Output
31














Time Complexity: O(n), where n is the length of the string
Auxiliary Space: O(n), where n is the length of the string



Last Updated : 08 Dec, 2023
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