1451B - Non-Substring Subsequence.cpp

/*
Hr0d1y has q queries on a binary string s of length n. A binary string is a string containing only characters '0' and '1'.

A query is described by a pair of integers li, ri (1=li<ri=n).

For each query, he has to determine whether there exists a good subsequence in s that is equal to the substring s[li…ri].

A substring s[i…j] of a string s is the string formed by characters sisi+1…sj.
String a is said to be a subsequence of string b if a can be obtained from b by deleting some characters without changing the order of the remaining characters.
A subsequence is said to be good if it is not contiguous and has length =2. For example, if s is "1100110", then the subsequences s1s2s4 ("1100110") and s1s5s7 ("1100110") are good, while s1s2s3 ("1100110") is not good.
Can you help Hr0d1y answer each query?

Input
The first line of the input contains a single integer t (1=t=100) — the number of test cases. The description of each test case is as follows.

The first line contains two integers n (2=n=100) and q (1=q=100) — the length of the string and the number of queries.

The second line contains the string s.

The i-th of the next q lines contains two integers li and ri (1=li<ri=n).

Output
For each test case, output q lines. The i-th line of the output of each test case should contain "YES" if there exists a good subsequence equal to the substring s[li...ri], and "NO" otherwise.

You may print each letter in any case (upper or lower).

Example
inputCopy
2
6 3
001000
2 4
1 3
3 5
4 2
1111
1 4
2 3
outputCopy
YES
NO
YES
NO
YES
Note
In the first test case,

s[2…4]= "010". In this case s1s3s5 ("001000") and s2s3s6 ("001000") are good suitable subsequences, while s2s3s4 ("001000") is not good.
s[1…3]= "001". No suitable good subsequence exists.
s[3…5]= "100". Here s3s5s6 ("001000") is a suitable good subsequence.
*/




#include<bits/stdc++.h>
using namespace std;
 
void solve() {
	int n, q;
	cin >> n >> q;
	
	string s;
	cin >> s;
	
	while (q--) {
		int l, r;
		cin >> l >> r;
		
		string t = s.substr(l - 1, r - l + 1);
		
		int conti = 0, maxConti = t.length() - 1;
		
		int k = 0;
		for (int i = 0; i < n; i++) {
			if (t[k] == s[i]) {
				conti++;
				k++;
			} else conti = 0;
			
			if (conti == maxConti) {
				i++; // extra
				conti = 0;
			}
		}
		
		if (k == t.length()) cout << "YES\n";
		else cout << "NO\n";
	}
}
 
int main(){
    ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	
	int t = 1;
	cin >> t;
	
	while (t--)  solve();
	
    return 0;
}