1380A - Three Indices.cpp

/*
You are given a permutation p1,p2,…,pn. Recall that sequence of n integers is called a permutation if it contains all integers from 1 to n exactly once.

Find three indices i, j and k such that:

1=i<j<k=n;
pi<pj and pj>pk.
Or say that there are no such indices.
Input
The first line contains a single integer T (1=T=200) — the number of test cases.

Next 2T lines contain test cases — two lines per test case. The first line of each test case contains the single integer n (3=n=1000) — the length of the permutation p.

The second line contains n integers p1,p2,…,pn (1=pi=n; pi?pj if i?j) — the permutation p.

Output
For each test case:

if there are such indices i, j and k, print YES (case insensitive) and the indices themselves;
if there are no such indices, print NO (case insensitive).
If there are multiple valid triples of indices, print any of them.

Example
inputCopy
3
4
2 1 4 3
6
4 6 1 2 5 3
5
5 3 1 2 4
outputCopy
YES
2 3 4
YES
3 5 6
NO
*/




#include<bits/stdc++.h>
using namespace std;

int main(){
    ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	
	int t;
	cin >> t;
	
	while (t--) {
		int n;
		cin >> n;
		
		vector <int> v(n);
		
		for (auto &x: v) cin >> x;
		
		bool res = false;
		
		for (int i = 1; i <  n - 1; i++) {
			if (v[i - 1] < v[i] && v[i] > v[i + 1]) {
				cout << "YES\n";
				cout << i << " " << i + 1 << " " << i + 2 << "\n";
				res = true;
				break;
			}
		}
		
		if (!res) cout << "NO\n";
	}
	
    return 0;
}