# Count number of equal pairs in a string

• Difficulty Level : Basic
• Last Updated : 02 Jun, 2022

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. But this is not efficient enough for longer length of strings. 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 ` `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 ` `#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

 ``

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

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