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 <= 10^5
1 <= nums[i] <= 10^9

Solution

Max Heap

Time complexity : O(nlogn)
Space complexity : O(n)

class Solution {
public:
    int minimumDeviation(vector<int>& nums) {
        priority_queue<int> pq;
        int min_num = INT_MAX;
        for (int n: nums) {
            int x = n & 1 ? n*2 : n;
            pq.push(x);
            min_num = min(min_num, x);
        }
        
        int ans = pq.top() - min_num;
        while (pq.top() % 2 == 0) {
            int x = pq.top()/2; pq.pop();
            pq.push(x);
            min_num = min(min_num, x);
            ans = min(ans, pq.top()-min_num);
        }
        return ans;
    }
};

先做前處理，將所有奇數乘 2，之後只要考慮偶數的除 2 即可。
小數奇數已做乘 2 往上拉，接下來只考慮最大值且為偶數可以除 2。
做到最大值為奇數為止，因奇數只能乘 2，max-min 只會更大。

Ordered Set

Time complexity : O(nlogn)
Space comlexity : O(n)

class Solution {
public:
    int minimumDeviation(vector<int>& nums) {
        set<int> s;
        for (int n: nums)
            s.insert(n & 1 ? n*2 : n);
        
        int ans = *s.rbegin() - *s.begin();
        while (*s.rbegin() % 2 == 0) {
            s.insert(*s.rbegin()/2);
            s.erase(*s.rbegin());
            ans = min(ans, *s.rbegin() - *s.begin());
        }
        return ans;
    }
};

概念與上相同，只是用 set 來做。