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.

  <ul>
  	<li>For example, if the array is <code>[1,2,3,4]</code>, then you can do this operation on the last element, and the array will be <code>[1,2,3,<u>2</u>].</code></li>
  </ul>
  </li>
  <li>If the element is <strong>odd</strong>, <strong>multiply</strong> it by <code>2</code>.
  <ul>
  	<li>For example, if the array is <code>[1,2,3,4]</code>, then you can do this operation on the first element, and the array will be <code>[<u>2</u>,2,3,4].</code></li>
  </ul>
  </li>

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

class Solution:
    def minimumDeviation(self, nums: List[int]) -> int:
        heap = []
        minimum  = float("inf")
        for num in nums:
            if num % 2 == 1:
                heap.append(-num*2) # Make all odds as even
            else:
                heap.append(-num) # Keep even same
        minimum = -max(heap) # Pick Minimum from list
        # print('Minimum', minimum)
        heapq.heapify(heap)
        diff = float("inf")
        while True:
            ele = -heapq.heappop(heap) # Get max element from heap
            diff = min(diff, ele - minimum) # Get Difference b/w min and max
            # print('Element', ele)
            if ele & 1: # If Max is odd then simply return
                return diff
            ele >>= 1 # Divide by half
            minimum = min(minimum, ele) # Get new minimum
            heapq.heappush(heap, -ele) # Put half of max back to heap
        return diff