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 ithcharacter 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++ 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 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. |
# 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# 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. |
<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> |
15
Time Complexity: O(N)
Auxiliary Space: O(1)