Queries to find total number of duplicate character in range L to R in the string S

Given a string S of size N consisting of lower case alphabets and an integer Q which represents the number of queries for S. Our task is to print the number of duplicate characters in the substring L to R for all the queries Q.

Note: 1 ≤N ≤ 10⁶ and 1 ≤ Q≤ 10⁶

Examples:

Input :
S = “geeksforgeeks”, Q = 2
L = 1 R = 5
L = 4 R = 8
Output :
1
0
Explanation:
For the first query ‘e’ is the only duplicate character in S from range 1 to 5.
For the second query theres is no duplicate character in S.

Input :
S = “Geekyy”, Q = 1
L = 1 R = 6
Output :
2
Explanation:
For the first query ‘e’ and ‘y’ are duplicate characters in S from range 1 to 6.

Naive Approach:

The naive approach would be to maintain a frequency array of size 26, to store the count of each character. For each query, given a range [L, R] we will traverse substring S[L] to S[R] and keep counting the occurrence of each character. Now, if the frequency of any character is greater than 1 then we would add 1 to answer.

Efficient Approach:

To solve the above problem in an efficient way we will store the position of each character as it appears in the string in a dynamic array. For each given query we will iterate over all the 26 lower case alphabets. If the current letter is in the substring S[L: R] then the next element of the first element which is greater than or equal L to in the corresponding vector should exist and be less than or equal to R.

Diagram below shows how we store characters in the dynamic array:

Picture explaining the above approach

Below is the implementation of the above approach:

// CPP implementation to Find the total 
// number of duplicate character in a 
// range L to R for Q number of queries in a string S 
  
#include <bits/stdc++.h> 
using namespace std; 
  
// Vector of vector to store 
// position of all characters 
// as they appear in string 
vector<vector<int> > v(26); 
  
// Function to store position of each character 
void calculate(string s) 
{ 
    for (int i = 0; i < s.size(); i++) { 
        // Inserting position of each 
        // character as they appear 
        v[s[i] - 'a'].push_back(i); 
    } 
} 
  
// Function to calculate duplicate 
// characters for Q queries 
void query(int L, int R) 
{ 
    // Variable to count duplicates 
    int duplicates = 0; 
  
    // Iterate over all 26 characters 
    for (int i = 0; i < 26; i++) { 
  
        // Finding the first element which 
        // is less than or equal to L 
        auto first = lower_bound(v[i].begin(), 
                                 v[i].end(), L - 1); 
  
        // Check if first pointer exists 
        // and is less than R 
        if (first != v[i].end() && *first < R) { 
            // Incrementing first pointer to check 
            // if the next duplicate element exists 
            first++; 
  
            // Check if the next element exists 
            // and is less than R 
            if (first != v[i].end() && *first < R) 
                duplicates++; 
        } 
    } 
  
    cout << duplicates << endl; 
} 
  
// Driver Code 
int main() 
{ 
    string s = "geeksforgeeks"; 
  
    int Q = 2; 
  
    int l1 = 1, r1 = 5; 
    int l2 = 4, r2 = 8; 
  
    calculate(s); 
  
    query(l1, r1); 
    query(l2, r2); 
  
    return 0; 
} 

Output:

1
0

`
Time complexity: O( Q * 26 * log N)

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