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Day 30 - Minimize Deviation in Array

ruarfff profile image Ruairí O'Brien ・2 min read

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.
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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.
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Example 3:

Input: nums = [2,10,8]
Output: 3
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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

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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
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Analysis

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