CRELECTIONS - Editorial

PROBLEM LINK:

Practice
Contest Source

Author, Tester and Editorialist: Jitin

DIFFICULTY:

EASY

EXPLANATION:

Consider the case when the i_{th} contestant competes with the $i-1$th contestant:
\bullet The score of the i_{th} contestant is given as A_iA_i+1...A_{i+k-2}A_{i+k-1}
\bullet The score of the i-1_{th} contestant is given as A_{i-1}A_i...A_{i+k-3}A_{i+k-2}
On Careful Observation we can notice that the product of A_iA_{i+1}...A_{i+k-2} is common in both the products. So we can directly compare the elements A_{i-1} and A_{i+k-1} to determine whether the i_{th} contestant will win or not.

SOLUTIONS:

C++ Solution
	#include <bits/stdc++.h>
	using namespace std;
	
	int main()
	{
		int n, k;
		cin >> n >> k;
		int a[n + 1];
		for (int i = 0; i < n; i++)
		{
			cin >> a[i];
		}
		for (int i = k; i < n; i++)
		{
			if (a[i] > a[i - k])
			{
				cout << "Yes\n";
			}
			else
			{
				cout << "No\n";
			}
		}
		return 0;
	}