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].
- String S contains only lowercase alphabets.
Input: S = “abcdbbd” Q = 6
2 3 6
1 5 z
2 1 1
1 4 a
1 7 d
2 1 7
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
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 ).
- Insert all characters of the string with its index into Hash-map
- 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
- 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:
3 1 5
Time complexity: O(Q * logN) where Q is number of queries and N is the size of the string.