Algorithm
You are given a string s consisting of lowercase Latin letters and q queries for this string.
Recall that the substring s[l;r] of the string s is the string slsl+1…sr. For example, the substrings of "codeforces" are "code", "force", "f", "for", but not "coder" and "top".
There are two types of queries:
- 1 pos c (1≤pos≤|s|, c is lowercase Latin letter): replace spos with c (set spos:=c);
- 2 l r (1≤l≤r≤|s|): calculate the number of distinct characters in the substring s[l;r].
The first line of the input contains one string s consisting of no more than 105 lowercase Latin letters.
The second line of the input contains one integer q (1≤q≤105) — the number of queries.
The next q lines contain queries, one per line. Each query is given in the format described in the problem statement. It is guaranteed that there is at least one query of the second type.
For each query of the second type print the answer for it — the number of distinct characters in the required substring in this query.
abacaba 5 2 1 4 1 4 b 1 5 b 2 4 6 2 1 7
3 1 2
dfcbbcfeeedbaea 15 1 6 e 1 4 b 2 6 14 1 7 b 1 12 c 2 6 8 2 1 6 1 7 c 1 2 f 1 10 a 2 7 9 1 10 a 1 14 b 1 1 f 2 1 11
5 2 5 2 6
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include <bits/stdc++.h>
using namespace std;
char c;
int q, ssqrt, t, pos, l, r;
string s;
vector<vector<int> > buckets;
void build() {
for(int i = 0, j; i < s.length(); i += ssqrt) {
j = i;
vector<int> fr;
for(int j = 0; j < 26; ++j)
fr.push_back(0);
while(j < s.length() && j < i + ssqrt)
++fr[s[j++] - 'a'];
buckets.push_back(fr);
}
}
void update() {
int bucket = (pos + ssqrt - 1) / ssqrt - 1;
--buckets[bucket][s[pos - 1] - 'a'];
s[pos - 1] = c;
++buckets[bucket][s[pos - 1] - 'a'];
}
int get() {
int lbucket = (l + ssqrt - 1) / ssqrt - 1 + 1;
int rbucket = (r + ssqrt - 1) / ssqrt - 1 - 1;
bool exist[26] = {false};
if(lbucket < rbucket) {
for(int i = lbucket; i <= rbucket; ++i)
for(int j = 0; j < 26; ++j)
exist[j] |= buckets[i][j] > 0;
for(int i = l - 1; i < ssqrt * lbucket; ++i)
exist[s[i] - 'a'] = true;
for(int i = ssqrt * rbucket + ssqrt; i < r; ++i)
exist[s[i] - 'a'] = true;
} else {
for(int i = l - 1; i < r; ++i)
exist[s[i] - 'a'] = true;
}
int res = 0;
for(int i = 0; i < 26; ++i)
res += exist[i];
return res;
}
int main() {
#ifndef ONLINE_JUDGE
freopen("in", "r", stdin);
#endif
cin >> s;
ssqrt = sqrt(s.length());
build();
scanf("%d", &q);
for(int i = 0; i < q; ++i) {
scanf("%d", &t);
if(t == 1) {
scanf("%d %c", &pos, &c);
update();
} else {
scanf("%d %d", &l, &r);
printf("%d\n", get());
}
}
return 0;
}Input
5
2 1 4
1 4 b
1 5 b
2 4 6
2 1 7
Output
1
2
Demonstration
Codeforcess Solution D. Distinct Characters Queries-Solution in C, C++, Java, Python,Codeforcess Solution