How do you solve Minimum Number of Operations to Make Array Empty on LeetCode?

Minimum Number of Operations to Make Array Empty (LeetCode #2870) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Elements that appear an odd number of times must be handled carefully, as they can't be removed in pairs.

What is the brute force solution for Minimum Number of Operations to Make Array Empty?

The brute force approach for Minimum Number of Operations to Make Array Empty is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible pairs and triplets of elements to see if we can remove them. This method is straightforward but inefficient, as it may require checking many combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Number of Operations to Make Array Empty?

Minimum Number of Operations to Make Array Empty uses the following concepts: Array, Hash Table, Greedy, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Number of Operations to Make Array Empty?

Always think about edge cases, such as single elements or very small arrays. Explain your thought process clearly, especially how you handle odd and even counts. Practice counting frequencies and deriving operations from them, as this is a common pattern.

What are common mistakes when solving Minimum Number of Operations to Make Array Empty?

Forgetting to check if an odd count is less than 3, which leads to returning incorrect results. Not considering that pairs and triplets can overlap in their removal.

#2870

Minimum Number of Operations to Make Array Empty

Medium

ArrayHash TableGreedyCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal approach counts the frequency of each element, then calculates how many operations are needed based on pairs and triplets. This method is efficient because it processes each element only once.

⚙️

Algorithm

3 steps

1Step 1: Count the frequency of each element in the array using a hashmap.
2Step 2: For each unique element, determine how many pairs and triplets can be formed.
3Step 3: Calculate the total number of operations needed to remove all elements.

solution.py13 lines

1from collections import Counter
2
3def min_operations(nums):
4    freq = Counter(nums)
5    operations = 0
6    for count in freq.values():
7        if count % 2 == 1:
8            if count < 3:
9                return -1
10            operations += count // 3 + (count % 3) // 2
11        else:
12            operations += count // 2
13    return operations

ℹ

Complexity note: The time complexity is O(n) because we only traverse the array once to count frequencies and once more to calculate operations, making it linear in relation to the input size.

1Elements that appear an odd number of times must be handled carefully, as they can't be removed in pairs.
2If an element appears less than 2 times and is odd, it's impossible to remove it.

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