You are given an array nums of n positive integers.

You can perform two types of operations on any element of the array any number of times:

The deviation of the array is the maximum difference between any two elements in the array.

Return the minimum deviation the array can have after performing some number of operations.


Test Cases

Input:

(int[]) nums = [1,2,3,4]

Output:

(int) 1

Explanation:

You can transform the array to [1,2,3,2], then to [2,2,3,2], then the deviation will be 3 - 2 = 1.

Input:

(int[]) nums = [4,1,5,20,3]

Output:

(int) 3

Explanation:

You can transform the array after two operations to [4,2,5,5,3], then the deviation will be 5 - 2 = 3.

Solution

class Solution {
    public int minimumDeviation(int[] nums) {
        PriorityQueue<Integer> pq = new PriorityQueue<>((a, b) -> b-a);
        int min = Integer.MAX_VALUE, res = Integer.MAX_VALUE;
        for(int a: nums) {
            if (a%2 == 1) a*=2;
            pq.offer(a);
            min = Math.min(min, a);
        }
        while(true) {
            int max = pq.poll();
            res = Math.min(res, max - min);
            if (max%2 == 1) break;
            pq.offer(max/2);
            min = Math.min(min, max/2);
        }
        return res;
    }
}
Time Complexity: O(n)
Space Complexity: O(n)