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# Count of index pairs (i, j) such that string after deleting ith character is equal to string after deleting jth character

• Last Updated : 03 Dec, 2021

Given a string str of N characters, the task is to calculate the count of valid unordered pairs of (i, j) such that the string after deleting ith character is equal to the string after deleting the jth character.

Examples:

Input: str = “aabb”
Output: 2
Explanation: The string after deletion of 1st element is “abb” and the string after deletion of  2nd element is “abb”. Similarly, the string after deletion of 3rd element is “aab” and the string after deletion of  4th element is “aab”. Hence, the number of valid pairs of (i, j) are 2, i.e, (1, 2) and (3, 4).

Input: str = “aaaaaa”
Output: 15

Approach: The given problem can be solved using the following observations:

• If Si = Sj, then Si = Si+1 = Si+2 … = Sj must hold true.
• Also if Si = Sj, then str[i] = str[i+1] = str[i+2] … = str[j] must hold true as well.

Therefore, using the above observations, the two-pointer technique can be used to calculate the intervals (l, r) in the string str such that str[l] = str[l+1] … = str[r], and for each valid (l, r), the count of valid pairs will be r – lC2.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to find the count of valid``// pairs (i, j) such that string after``// deleting ith character is equal to``// string after deleting jth character``int` `countValidPairs(string str, ``int` `N)``{` `    ``// Stores the required count``    ``int` `ans = 0;` `    ``// Loop to iterate the given array``    ``for` `(``int` `l = 0, r; l < N; l = r) {` `        ``// initial end point``        ``r = l;` `        ``// Loop to calculate the range``        ``// [l, r] such that all characters``        ``// from l to r are equal``        ``while` `(r < N && str[l] == str[r]) {``            ``r++;``        ``}` `        ``// Update the answer``        ``ans += ((r - l) * (r - l - 1)) / 2;``    ``}` `    ``// Return Answer``    ``return` `ans;``}` `// Driver Code``int` `main()``{``    ``string str = ``"aaaaaa"``;``    ``cout << countValidPairs(str, str.length());` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;``public` `class` `GFG``{``  ` `// Function to find the count of valid``// pairs (i, j) such that string after``// deleting ith character is equal to``// string after deleting jth character``static` `int` `countValidPairs(String str, ``int` `N)``{` `    ``// Stores the required count``    ``int` `ans = ``0``;` `    ``// Loop to iterate the given array``    ``for` `(``int` `l = ``0``, r; l < N; l = r) {` `        ``// initial end point``        ``r = l;` `        ``// Loop to calculate the range``        ``// [l, r] such that all characters``        ``// from l to r are equal``        ``Character c1 = str.charAt(l);``        ``Character c2 = str.charAt(r);``        ``while` `(r < N && c1.equals(c2)) {``            ``r++;``        ``}` `        ``// Update the answer``        ``ans += ((r - l) * (r - l - ``1``)) / ``2``;``    ``}` `    ``// Return Answer``    ``return` `ans;``}` `// Driver Code``public` `static` `void` `main(String args[])``{``    ``String str = ``"aaaaaa"``;``    ``System.out.print(countValidPairs(str, str.length()));``}``}` `// This code is contributed by Samim Hossain Mondal.`

## Python3

 `# Python program for the above approach` `# Function to find the count of valid``# pairs (i, j) such that string after``# deleting ith character is equal to``# string after deleting jth character``def` `countValidPairs(``str``, N):` `    ``# Stores the required count``    ``ans ``=` `0``;` `    ``# Loop to iterate the given array``    ``l ``=` `0``;``    ``r ``=` `None``;``    ` `    ``while``(l < N):``        ``# initial end point``        ``r ``=` `l;` `        ``# Loop to calculate the range``        ``# [l, r] such that all characters``        ``# from l to r are equal``        ``while` `(r < N ``and` `str``[l] ``=``=` `str``[r]):``            ``r ``+``=` `1` `        ``# Update the answer``        ``ans ``+``=` `((r ``-` `l) ``*` `(r ``-` `l ``-` `1``)) ``/``/` `2``;` `        ``l ``=` `r;``    ` `    ``# Return Answer``    ``return` `ans;` `# Driver Code``str` `=` `"aaaaaa"``;``print``(countValidPairs(``str``, ``len``(``str``)));` `# This code is contributed by Saurabh Jaiswal`

## C#

 `// C# program for the above approach``using` `System;``class` `GFG``{``  ` `// Function to find the count of valid``// pairs (i, j) such that string after``// deleting ith character is equal to``// string after deleting jth character``static` `int` `countValidPairs(``string` `str, ``int` `N)``{` `    ``// Stores the required count``    ``int` `ans = 0;` `    ``// Loop to iterate the given array``    ``for` `(``int` `l = 0, r; l < N; l = r) {` `        ``// initial end point``        ``r = l;` `        ``// Loop to calculate the range``        ``// [l, r] such that all characters``        ``// from l to r are equal``        ``char` `c1 = str[l];``        ``char` `c2 = str[r];``        ``while` `(r < N && c1.Equals(c2)) {``            ``r++;``        ``}` `        ``// Update the answer``        ``ans += ((r - l) * (r - l - 1)) / 2;``    ``}` `    ``// Return Answer``    ``return` `ans;``}` `// Driver Code``public` `static` `void` `Main()``{``    ``string` `str = ``"aaaaaa"``;``    ``Console.Write(countValidPairs(str, str.Length));` `}``}` `// This code is contributed by Samim Hossain Mondal.`

## Javascript

 ``

Output

`15`

Time Complexity: O(N)
Auxiliary Space: O(1)

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