Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Count pairs of characters in a string whose ASCII value difference is K

  • Last Updated : 17 May, 2021

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:
(a, a) and (b, b) are the only valid pairs.
Input: str = “geeksforgeeks”, K = 1 
Output:
(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: 
 

  1. 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.
  2. 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 characters
int 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 code
int main()
{
    string str = "abcdab";
    int k = 0;
 
    cout << countPairs(str, k);
 
    return 0;
}

Java




// Java implementation of the approach
import 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 approach
using 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

 


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!