Given string str of lower case alphabets and a non-negative integer K. The task is to find the number of pairs of characters in the given string whose ASCII value difference is exactly K.
Examples:
Input: str = “abcdab”, K = 0
Output: 2
(a, a) and (b, b) are the only valid pairs.
Input: str = “neveropen”, K = 1
Output: 8
(e, f), (e, f), (f, e), (f, e), (g, f),
(f, g), (s, r) and (r, s) are the valid pairs.
Approach: Store the frequency of each character in an array. Traverse through this frequency array to get the required answer. There exist two cases:
- If K = 0 then check if the similar character appears more than once i.e. if freq[i] > 1. If yes then add (freq[i] * (freq[i] – 1)) / 2 to the count.
- If K != 0 then check if there exist two characters with ASCII value difference as K say freq[i] and freq[j]. Then add freq[i] * freq[j] to the count.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;#define MAX 26// Function to return the count of// required pairs of charactersint countPairs(string str, int k){ // Length of the string int n = str.size(); // To store the frequency // of each character int freq[MAX]; memset(freq, 0, sizeof freq); // Update the frequency // of each character for (int i = 0; i < n; i++) freq[str[i] - 'a']++; // To store the required // count of pairs int cnt = 0; // If ascii value difference is zero if (k == 0) { // If there exists similar characters // more than once for (int i = 0; i < MAX; i++) if (freq[i] > 1) cnt += ((freq[i] * (freq[i] - 1)) / 2); } else { // If there exits characters with // ASCII value difference as k for (int i = 0; i < MAX; i++) if (freq[i] > 0 && i + k < MAX && freq[i + k] > 0) cnt += (freq[i] * freq[i + k]); ; } // Return the required count return cnt;}// Driver codeint main(){ string str = "abcdab"; int k = 0; cout << countPairs(str, k); return 0;} |
Java
// Java implementation of the approachimport java.util.Arrays;class GFG { static int MAX = 26; // Function to return the count of // required pairs of characters static int countPairs(char[] str, int k) { // Length of the string int n = str.length; // To store the frequency // of each character int[] freq = new int[MAX]; // Update the frequency // of each character for (int i = 0; i < n; i++) { freq[str[i] - 'a']++; } // To store the required // count of pairs int cnt = 0; // If ascii value difference is zero if (k == 0) { // If there exists similar characters // more than once for (int i = 0; i < MAX; i++) { if (freq[i] > 1) { cnt += ((freq[i] * (freq[i] - 1)) / 2); } } } else { // If there exits characters with // ASCII value difference as k for (int i = 0; i < MAX; i++) { if (freq[i] > 0 && i + k < MAX && freq[i + k] > 0) { cnt += (freq[i] * freq[i + k]); } } ; } // Return the required count return cnt; } // Driver code public static void main(String[] args) { String str = "abcdab"; int k = 0; System.out.println(countPairs(str.toCharArray(), k)); }}/* This code contributed by PrinciRaj1992 */ |
Python3
# Python3 implementation of the approach MAX = 26# Function to return the count of # required pairs of characters def countPairs(string, k) : # Length of the string n = len(string); # To store the frequency # of each character freq = [0] * MAX; # Update the frequency # of each character for i in range(n) : freq[ord(string[i]) - ord('a')] += 1; # To store the required # count of pairs cnt = 0; # If ascii value difference is zero if (k == 0) : # If there exists similar characters # more than once for i in range(MAX) : if (freq[i] > 1) : cnt += ((freq[i] * (freq[i] - 1)) // 2); else : # If there exits characters with # ASCII value difference as k for i in range(MAX) : if (freq[i] > 0 and i + k < MAX and freq[i + k] > 0) : cnt += (freq[i] * freq[i + k]); # Return the required count return cnt; # Driver code if __name__ == "__main__" : string = "abcdab"; k = 0; print(countPairs(string, k)); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approachusing System; class GFG { static int MAX = 26; // Function to return the count of // required pairs of characters static int countPairs(char[] str, int k) { // Length of the string int n = str.Length; // To store the frequency // of each character int[] freq = new int[MAX]; // Update the frequency // of each character for (int i = 0; i < n; i++) { freq[str[i] - 'a']++; } // To store the required // count of pairs int cnt = 0; // If ascii value difference is zero if (k == 0) { // If there exists similar characters // more than once for (int i = 0; i < MAX; i++) { if (freq[i] > 1) { cnt += ((freq[i] * (freq[i] - 1)) / 2); } } } else { // If there exits characters with // ASCII value difference as k for (int i = 0; i < MAX; i++) { if (freq[i] > 0 && i + k < MAX && freq[i + k] > 0) { cnt += (freq[i] * freq[i + k]); } } ; } // Return the required count return cnt; } // Driver code public static void Main(String[] args) { String str = "abcdab"; int k = 0; Console.WriteLine(countPairs(str.ToCharArray(), k)); }}// This code has been contributed by 29AjayKumar |
Javascript
<script> // JavaScript implementation of the approach const MAX = 26; // Function to return the count of // required pairs of characters function countPairs(str, k) { // Length of the string var n = str.length; // To store the frequency // of each character var freq = new Array(MAX).fill(0); // Update the frequency // of each character for (var i = 0; i < n; i++) { freq[str[i].charCodeAt(0) - "a".charCodeAt(0)]++; } // To store the required // count of pairs var cnt = 0; // If ascii value difference is zero if (k === 0) { // If there exists similar characters // more than once for (var i = 0; i < MAX; i++) { if (freq[i] > 1) { cnt += (freq[i] * (freq[i] - 1)) / 2; } } } else { // If there exits characters with // ASCII value difference as k for (var i = 0; i < MAX; i++) { if (freq[i] > 0 && i + k < MAX && freq[i + k] > 0) { cnt += freq[i] * freq[i + k]; } } } // Return the required count return cnt; } // Driver code var str = "abcdab"; var k = 0; document.write(countPairs(str.split(""), k)); </script> |
Output:
2
Time Complexity: O(|str| + MAX)
Auxiliary Space: O(MAX)
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