How do you solve Minimize Deviation in Array on LeetCode?

Minimize Deviation in Array (LeetCode #1675) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: The operations allowed on the array elements can significantly change their values, thus affecting the deviation.

What is the brute force solution for Minimize Deviation in Array?

The brute force approach for Minimize Deviation in Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible variations of the array by applying the allowed operations and then calculating the deviation for each variation. This is simple but inefficient as it explores all combinations. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimize Deviation in Array?

Minimize Deviation in Array uses the following concepts: Array, Greedy, Heap (Priority Queue), Ordered Set. The recommended patterns to study are: Heap (Priority Queue), Greedy Algorithms.

What are the interview tips for Minimize Deviation in Array?

Always clarify the operations allowed on the elements before diving into the solution. Think about how to efficiently track the maximum and minimum values as you modify the array. Practice explaining your thought process clearly, as this is important in interviews.

What are common mistakes when solving Minimize Deviation in Array?

Not considering the case where odd numbers need to be multiplied by 2 before processing. Failing to update the minimum value correctly after each operation.

#1675

Minimize Deviation in Array

Hard

ArrayGreedyHeap (Priority Queue)Ordered SetHeap (Priority Queue)Greedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

The optimal solution uses a max-heap to efficiently track the maximum value while minimizing the deviation. By always halving the largest even number, we can ensure that we are reducing the maximum value while keeping track of the minimum value.

⚙️

Algorithm

4 steps

1Step 1: Convert all numbers to even by multiplying odd numbers by 2.
2Step 2: Use a max-heap to keep track of the maximum value in the array.
3Step 3: Continuously remove the maximum value, halve it, and reinsert it into the heap until it becomes odd.
4Step 4: Calculate the current deviation and update the minimum deviation found.

solution.py20 lines

1import heapq
2
3def min_deviation(nums):
4    max_heap = []
5    min_val = float('inf')
6    for num in nums:
7        if num % 2 != 0:
8            num *= 2
9        heapq.heappush(max_heap, -num)
10        min_val = min(min_val, num)
11    min_deviation = float('inf')
12    while max_heap:
13        max_val = -heapq.heappop(max_heap)
14        min_deviation = min(min_deviation, max_val - min_val)
15        if max_val % 2 == 0:
16            heapq.heappush(max_heap, -max_val // 2)
17            min_val = min(min_val, max_val // 2)
18        else:
19            break
20    return min_deviation

ℹ

Complexity note: The time complexity is O(n log n) due to the operations on the max-heap, where n is the number of elements in the array. The space complexity is O(n) for storing the elements in the heap.

1The operations allowed on the array elements can significantly change their values, thus affecting the deviation.
2Using a max-heap allows us to efficiently manage the maximum value while minimizing the deviation.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.