LeetCode 3012: Minimize Length of Array Using Operations Solution

Master LeetCode problem 3012 (Minimize Length of Array Using Operations), 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.

3012. Minimize Length of Array Using Operations

Medium

Array Math Greedy Number Theory

Problem Explanation

Explanation:

To solve this problem, we need to iteratively perform operations on the given array nums to minimize its length. The key insight is to observe that we can reduce the length of the array by finding two elements with values greater than 0, calculating their modulo operation, and adding the result back to the array while removing the original elements.

We can achieve this by maintaining a priority queue (min heap) where we insert all positive elements from the array. We then repeatedly extract two elements from the priority queue, perform the modulo operation, and insert the result back into the priority queue until we can no longer reduce the length of the array.

Here is a step-by-step outline of the algorithm:

Create a priority queue (min heap) to store positive elements from the given array.
Iterate over the array and add all positive elements to the priority queue.
While the priority queue has at least two elements:
- Extract the top two elements from the priority queue.
- Calculate the modulo operation of the two elements and add the result back to the priority queue.
Return the final length of the priority queue as the minimum achievable length of the array.

Solution Code

import java.util.PriorityQueue;

class Solution {
    public int minimizeLength(int[] nums) {
        PriorityQueue<Integer> pq = new PriorityQueue<>();
        
        for (int num : nums) {
            if (num > 0) {
                pq.offer(num);
            }
        }
        
        while (pq.size() >= 2) {
            int num1 = pq.poll();
            int num2 = pq.poll();
            pq.offer(num1 % num2);
        }
        
        return pq.size();
    }
}

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