1473A - Replacing Elements.cpp

/*
You have an array a1,a2,…,an. All ai are positive integers.

In one step you can choose three distinct indices i, j, and k (i?j; i?k; j?k) and assign the sum of aj and ak to ai, i. e. make ai=aj+ak.

Can you make all ai lower or equal to d using the operation above any number of times (possibly, zero)?

Input
The first line contains a single integer t (1=t=2000) — the number of test cases.

The first line of each test case contains two integers n and d (3=n=100; 1=d=100) — the number of elements in the array a and the value d.

The second line contains n integers a1,a2,…,an (1=ai=100) — the array a.

Output
For each test case, print YES, if it's possible to make all elements ai less or equal than d using the operation above. Otherwise, print NO.

You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).

Example
inputCopy
3
5 3
2 3 2 5 4
3 4
2 4 4
5 4
2 1 5 3 6
outputCopy
NO
YES
YES
Note
In the first test case, we can prove that we can't make all ai=3.

In the second test case, all ai are already less or equal than d=4.

In the third test case, we can, for example, choose i=5, j=1, k=2 and make a5=a1+a2=2+1=3. Array a will become [2,1,5,3,3].

After that we can make a3=a5+a2=3+1=4. Array will become [2,1,4,3,3] and all elements are less or equal than d=4.
*/






#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, d;
    	cin >> n >> d;
    	
    	vector <int> v(n);
    	
    	for (int i = 0; i < n; i++) cin >> v[i];
    	
    	sort(v.begin(), v.end());
    	
    	if (v[n - 1] <= d) cout << "YES" << endl;
		else {
			int tmp = v[0] + v[1];
			if (tmp <= d) cout << "YES" << endl;
			else cout << "NO" << endl;
		} 
	}
    
 
}