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

Difficulty Level : Medium
Last Updated : 24 Feb, 2021

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

// 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;
}

Python3

# Python implementation to Find the total
# number of duplicate character in a
# range L to R for Q number of queries in a string S
 
import bisect
 
# Vector of vector to store
# position of all characters
# as they appear in string
v = [[] for _ in range(26)]
 
# Function to store position of each character
def calculate(s: str) -> None:
 
    for i in range(len(s)):
        # Inserting position of each
        # character as they appear
        v[ord(s[i]) - ord('a')].append(i)
 
 
# Function to calculate duplicate
# characters for Q queries
def query(L: int, R: int) -> None:
 
    # Variable to count duplicates
    duplicates = 0
 
    # Iterate over all 26 characters
    for i in range(26):
 
        # Finding the first element which
        # is less than or equal to L
        first = bisect.bisect_left(v[i], L - 1)
 
        # Check if first pointer exists
        # and is less than R
        if (first < len(v[i]) and v[i][first] < R):
            # Incrementing first pointer to check
            # if the next duplicate element exists
            first += 1
 
            # Check if the next element exists
            # and is less than R
            if (first < len(v[i]) and v[i][first] < R):
                duplicates += 1
 
    print(duplicates)
 
 
# Driver Code
if __name__ == "__main__":
 
    s = "geeksforgeeks"
 
    Q = 2
 
    l1 = 1
    r1 = 5
    l2 = 4
    r2 = 8
 
    calculate(s)
 
    query(l1, r1)
    query(l2, r2)
 
# This code is contributed by sanjeev2552

Output:

1
0

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

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