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  1. Practice
  2. Data Structures
  3. Heap
  4. Jesse and Cookies
  5. Discussions

Jesse and Cookies

  • max_au + 1 comment

    priority queue is NOT a queue. It's a heap.

    • nwicakson + 1 comment

      oh. then, use 2 queues. but, sort it before u push the inputs to the first queue.

      • dwij28 + 2 comments

        That makes no sense sir. This question requires a min heap priority queue which in reality is not a f-i-f-o queue.

        • arpadzsoltjakab + 0 comments

          You can solve this without a heap, only use pure Queue. You need to sort the input first, and then use two simple Queue. An init queue and an other to store the new made cookies. Every time a cookie is made, the time complexity will be constant time. The final time complexity will be O(nlogn + number of operations).

        • hammeramr + 1 comment

          a heap can be used to implement a priority que and thus they are synonymos.

          A[0] remains empty A[1] is the heap head (max or min your choice) and for A[k] the left node is at A[2k] and the right at A[2k + 1]

          here is my solution using a PriorityQueue

          public class Solution {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        int operations = 0;
        PriorityQueue<Integer> que = new PriorityQueue<Integer>();
        
        for(int i = 0; i < n; i++) {
            que.add(sc.nextInt());
        }
        
        
        while(que.size() > 1 && que.peek() < k) {
            int leastSweet = que.poll();
            int secondLeast = que.poll();
            
            que.add(leastSweet + 2*secondLeast);
            operations++;
        }
        
       
        if(que.peek() < k) {
            System.out.print(-1);
        } else {
            System.out.print(operations);
        }
        
    }
}
          • pc06matt + 1 comment

            My solution is exactly like this, but 2 cases are failing because of timeout. How did you resolve it?

            • vvbhandare + 1 comment

              Check your while should contain both conditions while (queue.size() > 1 && queue.peek() < K) because while (queue.peek() < K) won't suffice.

              • rakshittchopraa + 0 comments

                you might be lucky...my two test cases are also failing due to TLE.