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package g1601_1700.s1675_minimize_deviation_in_array;

// #Hard #Array #Greedy #Heap_Priority_Queue #Ordered_Set
// #2022_04_19_Time_104_ms_(88.83%)_Space_58.3_MB_(91.85%)

import java.util.PriorityQueue;

/**
 * 1675 - Minimize Deviation in Array\.
 *
 * Hard
 *
 * 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:
 *
 * *   If the element is **even** , **divide** it by `2`.
 *     *   For example, if the array is `[1,2,3,4]`, then you can do this operation on the last element, and the array will be `[1,2,3,2].`
 * *   If the element is **odd** , **multiply** it by `2`.
 *     *   For example, if the array is `[1,2,3,4]`, then you can do this operation on the first element, and the array will be `[2,2,3,4].`
 *
 * 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._
 *
 * **Example 1:**
 *
 * **Input:** nums = [1,2,3,4]
 *
 * **Output:** 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.
 *
 * **Example 2:**
 *
 * **Input:** nums = [4,1,5,20,3]
 *
 * **Output:** 3
 *
 * **Explanation:** You can transform the array after two operations to [4,2,5,5,3], then the deviation will be 5 - 2 = 3.
 *
 * **Example 3:**
 *
 * **Input:** nums = [2,10,8]
 *
 * **Output:** 3
 *
 * **Constraints:**
 *
 * *   `n == nums.length`
 * *   2 <= n <= 10⁵
 * *   1 <= nums[i] <= 10⁹
**/
public class Solution {
    public int minimumDeviation(int[] nums) {
        PriorityQueue pq = new PriorityQueue<>((a, b) -> b - a);
        int min = Integer.MAX_VALUE;
        for (int num : nums) {
            if (num % 2 == 1) {
                num *= 2;
            }
            min = Math.min(min, num);
            pq.offer(num);
        }
        int diff = Integer.MAX_VALUE;
        while (pq.peek() % 2 == 0) {
            int max = pq.poll();
            diff = Math.min(diff, max - min);
            min = Math.min(max / 2, min);
            pq.offer(max / 2);
        }
        return Math.min(diff, pq.peek() - min);
    }
}