LeetCode 3066: Minimum Operations to Exceed Threshold Value II Solution

Master LeetCode problem 3066 (Minimum Operations to Exceed Threshold Value II), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3066. Minimum Operations to Exceed Threshold Value II

Medium

Array Heap (Priority Queue)Simulation

Problem Explanation

Explanation:

To solve this problem, we can use a priority queue (min heap) to keep track of the smallest elements in the array. We repeatedly extract the two smallest elements from the priority queue, perform the operation described in the problem, and insert the new calculated value back into the priority queue. We continue this process until the smallest element in the priority queue is greater than or equal to the threshold value, k.

Algorithm:

Initialize a priority queue (min heap) with all elements from the input array, nums.
Initialize a variable, operations, to keep track of the number of operations performed.
While the smallest element in the priority queue is less than k:
- Extract the two smallest elements, x and y, from the priority queue.
- Calculate the new value, new_val = min(x, y) * 2 + max(x, y).
- Increment the operations count.
- Insert new_val back into the priority queue.
Return the operations count as the minimum number of operations needed.

Time Complexity:

The time complexity of this approach is O(n log n), where n is the number of elements in the input array nums.

Space Complexity:

The space complexity is O(n) to store the elements in the priority queue.

Solution Code

import java.util.PriorityQueue;

class Solution {
    public int minOperations(int[] nums, int k) {
        PriorityQueue<Integer> pq = new PriorityQueue<>();
        for (int num : nums) {
            pq.offer(num);
        }
        
        int operations = 0;
        while (pq.peek() < k) {
            int x = pq.poll();
            int y = pq.poll();
            int new_val = Math.min(x, y) * 2 + Math.max(x, y);
            pq.offer(new_val);
            operations++;
        }
        
        return operations;
    }
}

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