Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

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 <bits/stdc++.h>
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




<script>
// Javascript 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
function countValidPairs(str, N)
{
 
    // Stores the required count
    let ans = 0;
 
    // Loop to iterate the given array
    for (let 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
let str = "aaaaaa";
document.write(countValidPairs(str, str.length));
 
// This code is contributed by Samim Hossain Mondal.
</script>
Output
15

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


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!