String Range Queries to count number of distinct characters with updates
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
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 ).
- 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:
C++
#include <bits/stdc++.h> using namespace std; #include <ext/pb_ds/assoc_container.hpp> #include <ext/pb_ds/tree_policy.hpp> 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; } |
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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