# 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: 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 ` `using` `namespace` `std; ` ` `  `// Vector of vector to store ` `// position of all characters ` `// as they appear in string ` `vector > 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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