Given a string S of length N, and M queries of the following type:
Type 1: 1 L x,
Indicates update Lth index of string S by character ‘x’.
Type 2: 2 L R str
Find the number of subsets in range L to R
which is equal to the string str modulo 1000000007.
|S| <= 100000,
M <= 100000,
1 <= L, R <= |S|,
|str| <= 26,
All characters in str are unique,
S & str consisting of lowercase latin letters.
Input : N = 16, M = 3, S = “geeekkksgskeegks”
type = 2: L = 1, R = 7, str = “gek”
type = 1: index = 2, character = ‘x’
type = 2: L = 1, R = 7, str = “gek”
Output : 9 6 4
query2 : No of subsets in string S i range [1…7] that is equal to gek is 9.
query1 : string S is changed to gxeekkksgskeegks after second query.
query2 : No of subsets in string S i range [1…7] that is equal to gek is 6.
Naive Approach :
Query type 1: We will calculate the frequency of each character of the query string in the range [L…R] and then multiply all the calculated frequencies to get the desired result.
Query type 2: We will replace the i’th character of the string with the given character.
Time complexity : 0(m*n)
Efficient Approach :
- Using Segment Tree we can perform both operations in log(n) time. Every node of segment tree will contain frequency of characters in range [L..R].
- The function build takes n*log(n) time to create a segment tree with every node containing the frequency of characters of some segment of the string.
- The function get returns a vector containing frequency of all characters. Multiplication of all frequency of the given query string modulo 1e9+7 gives the desired result.
- The function update decreases the frequency of character placed earlier and increases the frequency of new character present in the nodes of the segment tree by one.
- The function add_two_result adds two vector and returns their result.
Below is the C++ implementation of the above approach:
Time complexity : 0(m*log(n)+n*log(n))
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