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# Day 30 - Minimize Deviation in Array

## The Problem

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`

## Tests

``````import pytest
from .Day30_MinimizeDeviationInArray import Solution

s = Solution()

@pytest.mark.parametrize(
"nums,expected",
[
([1, 2, 3, 4], 1),
([4, 1, 5, 20, 3], 3),
([2, 10, 8], 3),
],
)
def test_minimum_deviation(nums, expected):
assert s.minimumDeviation(nums) == expected

``````

## Solution

``````from typing import List
import heapq
import math

class Solution:
def minimumDeviation(self, nums: List[int]) -> int:
vals = []
minimum = math.inf
minimum_deviation = math.inf

for n in nums:
if n % 2 == 0:
vals.append(-n)
minimum = min(minimum, n)
else:
evened = n * 2
vals.append(-evened)
minimum = min(minimum, evened)

heapq.heapify(vals)

while vals:
e = -heapq.heappop(vals)
minimum_deviation = min(minimum_deviation, e - minimum)

if e % 2 != 0:
return minimum_deviation
e = e // 2
minimum = min(minimum, e)
heapq.heappush(vals, -e)

return minimum_deviation
``````