LeetCode 1675: Minimize Deviation in Array Solution

Master LeetCode problem 1675 (Minimize Deviation in Array), a hard 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.

1675. Minimize Deviation in Array

Hard

Array Greedy Heap (Priority Queue)Ordered Set

Problem Explanation

Explanation:

To minimize the deviation in the array, we need to manipulate the elements to try to bring them closer together. We can achieve this by performing the given operations - dividing even numbers by 2 and multiplying odd numbers by 2.

We first need to multiply all odd numbers by 2 to ensure that all numbers are even.
Then, we need to repeatedly divide the largest number by 2 until we cannot perform any more operations.
At each step, we update the minimum deviation whenever the difference between the current maximum and minimum is smaller than the previous minimum deviation.

Time Complexity: O(n log n) where n is the number of elements in the array
Space Complexity: O(n) for the priority queue

Solution Code

import java.util.*;

class Solution {
    public int minimumDeviation(int[] nums) {
        PriorityQueue<Integer> pq = new PriorityQueue<>(Collections.reverseOrder());
        int minDeviation = Integer.MAX_VALUE;
        
        for (int num : nums) {
            if (num % 2 == 1) {
                pq.offer(num * 2);
            } else {
                pq.offer(num);
            }
        }
        
        while (true) {
            int max = pq.poll();
            minDeviation = Math.min(minDeviation, max - pq.peek());
            if (max % 2 == 0) {
                pq.offer(max / 2);
            } else {
                break;
            }
        }
        
        return minDeviation;
    }
}

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