I am using min heap to solve this problem

#include <bits/stdc++.h>
#define lli long long int
using namespace std;

struct greaters {
    bool operator() (const long&a, const long&b) const {
        return a > b;
    }
};

int main (int argc, char *argv[]) {

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    vector<int> v;
    lli ans = 0;

    lli n, k;
    cin >> n >> k;
    while(n--) {
        lli x;
        cin >> x;
        v.push_back(x);
    }
    // min heap
    make_heap(v.begin(), v.end(), greaters());
		// check if there's a minimum element < k
    if (v.front() < k) {
        while(v.size() > 1) {
            int firstMinVal = v.front();
            // delete min heap
            pop_heap(v.begin(), v.end(), greaters());
            v.pop_back();

            int secondMinVal = v.front();
            // delete min heap
            pop_heap(v.begin(), v.end(), greaters());
            v.pop_back();

            int formula = firstMinVal + (2*secondMinVal);
						// push heap
            v.push_back(formula);
            push_heap(v.begin(), v.end(), greaters());

            ans++;

            int minVal = v.front();
            if (minVal >= k) {
                break;
            }
        }
    }

    if (v.size() == 1 && v.front() < k) {
        cout << -1 << "\n";
    } else {
        cout << ans << endl;
    }


    return 0;
}

The problem state that the sweetness have to be greater than k But there's a test where the last sweet is exactly k and it still consider valid, not return -1 What do you even mean?

def cookies(k,A): # convert list A to priority queue heapq heapq.heapify(A)

iteration = 0
while any(c < k for c in A) and len(A) > 1:
    mix = heapq.heappop(A) + 2 * heapq.heappop(A)
    heapq.heappush(A,mix)
    iteration += 1


if not all(c >= k for c in A):
    return -1

return iteration

def cookies(k, A): A.sort(reverse = True) queue = [] mixes = 0 while True: if len(A) <= 1 and queue: A = queue[::-1] + A queue = [] if not A and not queue: return mixes if queue: if A[-1] <= queue[0]: last_num = A.pop() else: last_num = queue.pop(0) else: last_num = A.pop() if last_num >= k: next elif len(A) == 0 and last_num < k: return -1 else: if not queue: sec_num = A.pop() else: if A[-1] <= queue[0]: sec_num = A.pop() else: sec_num = queue.pop(0) new_num = last_num + 2 * sec_num queue.append(new_num) mixes += 1

Is it a good idea to use it for my content based website of cookieclickerunblocked, I'm running this website on WordPress right now, Please give me a honest advice. Thanks