# String Range Queries to count number of distinct characters with updates

• Last Updated : 07 Jul, 2020

Given a string S of length N, and Q queries of the following type:

Type 1: 1 i X
Update the i-th character of the string with the given character, X.

Type 2: L R
Count number of distinct characters in the given range [L, R].

Constraint:

• 1<=N<=500000
• 1<=Q<20000
• |S|=N
• String S contains only lowercase alphabets.

Examples:

Input: S = “abcdbbd” Q = 6
2 3 6
1 5 z
2 1 1
1 4 a
1 7 d
2 1 7
Output:
3
1
5
Explanation:
For the Queries:
1. L = 3, R = 6
The different characters are:c, b, d.
ans = 3.
2. String after query updated as S=”abcdzbd”.
3. L = 1, R = 1
Only one different character.
and so on process all queries.

Input: S = “aaaaa”, Q = 2
1 2 b
2 1 4
Output:
2

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive approach:

Query type 1: Replace i-th character of the string with the given character.
Query type 2: Traverse the string from L to R and count number of distinct characters.

Time complexity: O(N2)

Efficient approach: This approach is based on the Frequency-counting algorithm.
The idea is to use a HashMap to map distinct characters of the string with an Ordered_set which stores indices of its all occurrence. Ordered_set is used because it is based on Red-Black tree, so insertion and deletion of character will taken O ( log N ).

1. Insert all characters of the string with its index into Hash-map
2. For query of type 1, erase the occurrence of character at index i and insert occurrence of character X at index i in Hash-map
3. For query of type 2, traverse all 26 characters and check if its occurrence is within the range [L, R], if yes then increment the count. After traversing print the value of count.

Below is the implementation of the above approach:

## C++

 `#include ``using` `namespace` `std;`` ` `#include ``#include ``using` `namespace` `__gnu_pbds;`` ` `#define ordered_set                 \``    ``tree<``int``, null_type, less<``int``>, \``         ``rb_tree_tag,               \``         ``tree_order_statistics_node_update>`` ` `// Function that returns the lower-``// bound of the element in ordered_set``int` `lower_bound(ordered_set set1, ``int` `x)``{``    ``// Finding the position of``    ``// the element``    ``int` `pos = set1.order_of_key(x);`` ` `    ``// If the element is not``    ``// present in the set``    ``if` `(pos == set1.size()) {``        ``return` `-1;``    ``}`` ` `    ``// Finding the element at``    ``// the position``    ``else` `{``        ``int` `element = *(set1.find_by_order(pos));`` ` `        ``return` `element;``    ``}``}`` ` `// Utility function to add the``// position of all characters``// of string into ordered set``void` `insert(``    ``unordered_map<``int``, ordered_set>& hMap,``    ``string S, ``int` `N)``{``    ``for` `(``int` `i = 0; i < N; i++) {``        ``hMap[S[i] - ``'a'``].insert(i);``    ``}``}`` ` `// Utility function for update``// the character at position P``void` `Query1(``    ``string& S,``    ``unordered_map<``int``, ordered_set>& hMap,``    ``int` `pos, ``char` `c)``{`` ` `    ``// we delete the position of the``    ``// previous character as new``    ``// character is to be replaced``    ``// at the same position.``    ``pos--;``    ``int` `previous = S[pos] - ``'a'``;``    ``int` `current = c - ``'a'``;``    ``S[pos] = c;``    ``hMap[previous].erase(pos);``    ``hMap[current].insert(pos);``}`` ` `// Utility function to determine``// number of different characters``// in given range.``void` `Query2(``    ``unordered_map<``int``, ordered_set>& hMap,``    ``int` `L, ``int` `R)``{``    ``// Iterate over all 26 alphabets``    ``// and check if it is in given``    ``// range using lower bound.``    ``int` `count = 0;``    ``L--;``    ``R--;``    ``for` `(``int` `i = 0; i < 26; i++) {``        ``int` `temp = lower_bound(hMap[i], L);``        ``if` `(temp <= R and temp != -1)``            ``count++;``    ``}``    ``cout << count << endl;``}`` ` `// Driver code``int` `main()``{``    ``string S = ``"abcdbbd"``;``    ``int` `N = S.size();`` ` `    ``unordered_map<``int``, ordered_set> hMap;`` ` `    ``// Insert all characters with its``    ``// occurrence in the hash map``    ``insert(hMap, S, N);`` ` `    ``// Queries for sample input``    ``Query2(hMap, 3, 6);``    ``Query1(S, hMap, 5, ``'z'``);``    ``Query2(hMap, 1, 1);``    ``Query1(S, hMap, 4, ``'a'``);``    ``Query1(S, hMap, 7, ``'d'``);``    ``Query2(hMap, 1, 7);`` ` `    ``return` `0;``}`
Output:
```3
1
5
```

Time complexity: O(Q * logN) where Q is number of queries and N is the size of the string.

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