# K3CHKEQ - Editorial

Author: Soumy Jain
Tester: Divyank Goyal
Editorialist: Soumy Jain

MEDIUM

# PREREQUISITES:

Array and Precomputations

# EXPLANATION:

Let’s define ans_{i} to have the index of an element on left of i which is not equal to v_{i}. Obviously, ans_{1} = -1 because no element occur before first index. Now if, v_{i} != v_{i-1} we can directly mark ans_{i} to be i-1.

Now what if v_{i} == v_{i-1}? In this case we can mark ans_{i} = ans_{i-1}.

Now to answer the query, we can first check if v_{r} != k,
if it is true we can directly print “NO”,
else we will find the index of the element on the left which is not equal to k using ans_{r} and check if it lies in range or not.

# SOLUTIONS:

Setter's Solution
#include <bits/stdc++.h>
#define ll long long
using namespace std;

int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
ll n;
cin >> n;
vector<ll> v(n);
for (auto &i : v)
cin >> i;
vector<ll> ans(n);
ans[0] = -1;
for (ll i = 1; i < n; i++)
{
if (v[i] != v[i - 1])
ans[i] = i - 1;
else
ans[i] = ans[i - 1];
}
ll q;
cin >> q;
while (q--)
{
ll l, r, k;
cin >> l >> r >> k;
l--, r--;
if (v[r] == k && ans[r] < l)
cout << "YES\n";
else
cout << "NO\n";
}
return 0;
}