# Count of character pairs at same distance as in English alphabets

Given a string, the task is to count the number of pairs whose elements are at same distances as in the English alphabets.

**Note :** Absolute distance between characters is considered.

**Examples :**

Input:str = "geeksforgeeks"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 generate all pairs and compare pair characters with distance between them. If distance is same for a pair, then increment result.

**Algorithm:**

Step 1:Take an input stringStep 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 sameStep 4:If distance is same increment the counter (result).Step 5:Print the output.

**Implementation:**

## 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 pairs` `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` `(` `abs` `(str[i] - str[j]) == ` `abs` `(i - j))` ` ` `result++;` ` ` `return` `result;` `}` `// Driver code` `int` `main()` `{` ` ` `string str = ` `"geeksforgeeks"` `;` ` ` `cout << countPairs(str);` ` ` `return` `0;` `}` |

## Java

`// A Simple Java program to find pairs with distance` `// equal to English alphabet distance` `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 = ` `"geeksforgeeks"` `;` ` ` `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 ` `=` `"geeksforgeeks"` `; ` ` ` `print` `(countPairs(str1)); ` `# This code is contributed ` `# by Sairahul099` |

## C#

`// A Simple C# program to find pairs with distance` `// equal to English alphabet distance` `using` `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 = ` `"geeksforgeeks"` `;` ` ` `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 pairs` `function` `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` `= ` `"geeksforgeeks"` `;` `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 = ` `"geeksforgeeks"` `;` ` ` `document.write(countPairs(str));` ` ` ` ` `// This code is contributed by divyesh072019.` `</script>` |

**Output**

4

**Time complexity: O(n ^{2}). **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 distance` `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` `((` `abs` `(str[i + j] - str[i]) == j))` ` ` `result++;` ` ` `return` `result;` `}` `// Driver code` `int` `main()` `{` ` ` `string str = ` `"geeksforgeeks"` `;` ` ` `cout << countPairs(str);` ` ` `return` `0;` `}` |

## Java

`// An optimized Java program to find pairs with distance` `// equal to English alphabet distance` `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 = ` `"geeksforgeeks"` `;` ` ` `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 ` `=` `"geeksforgeeks"` `; ` ` ` `print` `(countPairs(str1))` `# This code is contributed` `# by Sairahul099` |

## C#

`// An optimized C# program to find pairs with distance` `// equal to English alphabet distance` `using` `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 = ` `"geeksforgeeks"` `;` ` ` `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 distance` `function` `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` `$str` `= ` `"geeksforgeeks"` `;` `echo` `countPairs(` `$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 = ` `"geeksforgeeks"` `;` ` ` `document.write(countPairs(str));` `</script>` |

**Output**

4

**Time complexity: O(n ^{2}) **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 GeeksforGeeks 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 GeeksforGeeks main page and help other Geeks.