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: 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 <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; } |
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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 |
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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 |
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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 |
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Output :
31
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