Given a string, the task is to count the number of pairs whose elements are at the same distances as in the English alphabet.
Note : Absolute distance between characters is considered.
Examples :
Input: str = "neveropen" Output: 4 Explanation: In this (g, s), (e, g), (e, k), (e, g) are the pairs that are at same distances as in English alphabets. Input: str = "observation" Output: 4 Explanation: (b, i), (s, v), (o, n), (v, t) are at same distances as in English alphabets.
A simple solution is to consider generating all pairs and comparing pair characters with the distance between them. If distance is same for a pair, then increment result.
Algorithm:
- Step 1: Take an input string
- Step 2: Initialize result equals to 0 and n to the length of the string.
- Step 3: Using nested for loops check if the distance between characters is same
- Step 4: If distance is same increment the counter (result).
- Step 5: Print the output.
Below is the implementation of the above algorithm:
C++
// A Simple C++ program to find pairs with distance// equal to English alphabet distance#include <bits/stdc++.h>using namespace std;// Function to count pairsint countPairs(string str){ int result = 0; int n = str.length(); for (int i = 0; i < n; i++) for (int j = i + 1; j < n; j++) // Increment count if characters are at // same distance if (abs(str[i] - str[j]) == abs(i - j)) result++; return result;}// Driver codeint main(){ string str = "neveropen"; cout << countPairs(str); return 0;} |
Java
// A Simple Java program to find pairs with distance// equal to English alphabet distanceimport java.io.*;class Test { // Method to count pairs static int countPairs(String str) { int result = 0; int n = str.length(); for (int i = 0; i < n; i++) for (int j = i + 1; j < n; j++) // Increment count if characters // are at same distance if (Math.abs(str.charAt(i) - str.charAt(j)) == Math.abs(i - j)) result++; return result; } // Driver method public static void main(String args[]) { String str = "neveropen"; System.out.println(countPairs(str)); }} |
Python 3
# Simple Python3 program to find pairs with # distance equal to English alphabet distance # Function to count pairs def countPairs(str1): result = 0; n = len(str1) for i in range(0, n): for j in range(i + 1, n): # Increment count if characters # are at same distance if (abs(ord(str1[i]) - ord(str1[j])) == abs(i - j)): result += 1; return result; # Driver code if __name__ == "__main__": str1 = "neveropen"; print(countPairs(str1)); # This code is contributed # by Sairahul099 |
C#
// A Simple C# program to find pairs with distance// equal to English alphabet distanceusing System;class Test { // Method to count pairs static int countPairs(string str) { int result = 0; int n = str.Length; for (int i = 0; i < n; i++) for (int j = i + 1; j < n; j++) // Increment count if characters // are at same distance if (Math.Abs(str[i] - str[j]) == Math.Abs(i - j)) result++; return result; } // Driver method public static void Main() { string str = "neveropen"; Console.WriteLine(countPairs(str)); }}// This Code is contributed by vt_m. |
PHP
<?php// PHP program for Count // of character pairs at // same distance as in // English alphabets// Function to count pairsfunction countPairs($str){ $result = 0; $n = strlen($str); for ($i = 0; $i < $n; $i++) for ($j = $i + 1; $j < $n; $j++) // Increment count if // characters are at // same distance if (abs(ord($str[$i]) - ord($str[$j])) == abs($i - $j)) $result++; return $result;}// Driver code$str = "neveropen";echo countPairs($str);// This code is contributed by Sam007?> |
Javascript
<script> // A Simple Javascript program to find pairs with distance // equal to English alphabet distance // Method to count pairs function countPairs(str) { let result = 0; let n = str.length; for (let i = 0; i < n; i++) for (let j = i + 1; j < n; j++) // Increment count if characters // are at same distance if (Math.abs(str[i].charCodeAt() - str[j].charCodeAt()) == Math.abs(i - j)) result++; return result; } let str = "neveropen"; document.write(countPairs(str)); // This code is contributed by divyesh072019.</script> |
Output
4
Time complexity: O(n2). The above method can be optimized by using the fact that there can be only 26 alphabets i.e. instead of checking an element up to length of string, check only from current index to 26th index.
Auxiliary Space: O(n), where n is the length of string. This is because when string is passed to any function it is passed by value and creates a copy of itself in stack.
C++
// An optimized C++ program to find pairs with distance// equal to English alphabet distance#include <bits/stdc++.h>using namespace std;const int MAX_CHAR = 26;// Function to count pairs with distance// equal to English alphabet distanceint countPairs(string str){ int result = 0; int n = str.length(); for (int i = 0; i < n; i++) // This loop runs at most 26 times for (int j = 1; (i + j) < n && j <= MAX_CHAR; j++) if ((abs(str[i + j] - str[i]) == j)) result++; return result;}// Driver codeint main(){ string str = "neveropen"; cout << countPairs(str); return 0;} |
Java
// An optimized Java program to find pairs with distance// equal to English alphabet distanceimport java.io.*;class Test { static final int MAX_CHAR = 26; // Method to count pairs with distance // equal to English alphabet distance static int countPairs(String str) { int result = 0; int n = str.length(); for (int i = 0; i < n; i++) // This loop runs at most 26 times for (int j = 1; (i + j) < n && j <= MAX_CHAR; j++) if ((Math.abs(str.charAt(i + j) - str.charAt(i)) == j)) result++; return result; } // Driver method public static void main(String args[]) { String str = "neveropen"; System.out.println(countPairs(str)); }} |
Python 3
# An optimized C++ program to find pairs with# distance equal to English alphabet distance MAX_CHAR = 26# Function to count pairs with distance # equal to English alphabet distance def countPairs(str1): result = 0; n = len(str1) for i in range(0, n): # This loop runs at most 26 times for j in range(1, MAX_CHAR + 1): if((i + j) < n): if ((abs(ord(str1[i + j]) - ord(str1[i])) == j)): result += 1; return result # Driver code if __name__ == "__main__": str1 = "neveropen"; print(countPairs(str1))# This code is contributed# by Sairahul099 |
C#
// An optimized C# program to find pairs with distance// equal to English alphabet distanceusing System;class Test { static int MAX_CHAR = 26; // Method to count pairs with distance // equal to English alphabet distance static int countPairs(string str) { int result = 0; int n = str.Length; for (int i = 0; i < n; i++) // This loop runs at most 26 times for (int j = 1; (i + j) < n && j <= MAX_CHAR; j++) if ((Math.Abs(str[i + j] - str[i]) == j)) result++; return result; } // Driver method public static void Main() { string str = "neveropen"; Console.WriteLine(countPairs(str)); }}// This Code is contributed by vt_m. |
PHP
<?php// An optimized PHP program // to find pairs with distance// equal to English alphabet distance// Function to count pairs // with distance equal to // English alphabet distancefunction countPairs($str){ $result = 0; $n = strlen($str); for ($i = 0; $i < $n; $i++) // This loop runs at // most 26 times for ($j = 1; ($i + $j) < $n && $j <= 26; $j++) if ((abs(ord($str[$i + $j]) - ord($str[$i]) ) == $j)) $result++; return $result;}// Driver code$str = "neveropen";echo countPairs($str);// This code is contributed by Sam007?> |
Javascript
<script> // An optimized Javascript program to find pairs with distance // equal to English alphabet distance let MAX_CHAR = 26; // Method to count pairs with distance // equal to English alphabet distance function countPairs(str) { let result = 0; let n = str.length; for (let i = 0; i < n; i++) // This loop runs at most 26 times for (let j = 1; (i + j) < n && j <= MAX_CHAR; j++) if ((Math.abs(str[i + j].charCodeAt() - str[i].charCodeAt()) == j)) result++; return result; } let str = "neveropen"; document.write(countPairs(str));</script> |
Output
4
Time complexity: O(n2) under the assumption that alphabet size is constant.
Auxiliary Space: O(n), where n is the length of string. This is because when string is passed to any function it is passed by value and creates a copy of itself in stack.
This article is contributed by Sahil Chhabra (akku). If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
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