Count number of equal pairs in a string

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: airOutput: 3Explanation :3 pairs that are equal are (a, a), (i, i) and (r, r)Input : geeksforgeeksOutput : 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 ``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

 ``

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 ``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

 ``

PHP

 ``

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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