#2870

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Minimum Number of Operations to Make Array Empty

Minimize the number of operations to make an array empty by leveraging array scanning and hash lookup.

Problem Statement

You are given a 0-indexed array nums consisting of positive integers. There are two operations that can be performed to remove elements from the array: remove two occurrences of the same number or remove one occurrence of a number and another from the array. Your goal is to determine the minimum number of operations required to make the array empty. If it's impossible, return -1.

The array may not always be reducible to an empty array, and we need to consider each possibility carefully. An efficient approach involves using a hash table to track counts of elements and scanning through the array to apply the operations strategically, while considering that only repeated elements can be removed in the same operation.

Examples

Example 1

Input: nums = [2,3,3,2,2,4,2,3,4]

Output: 4

We can apply the following operations to make the array empty:

Apply the first operation on the elements at indices 0 and 3. The resulting array is nums = [3,3,2,4,2,3,4].
Apply the first operation on the elements at indices 2 and 4. The resulting array is nums = [3,3,4,3,4].
Apply the second operation on the elements at indices 0, 1, and 3. The resulting array is nums = [4,4].
Apply the first operation on the elements at indices 0 and 1. The resulting array is nums = []. It can be shown that we cannot make the array empty in less than 4 operations.

Example 2

Input: nums = [2,1,2,2,3,3]

Output: -1

It is impossible to empty the array.

Constraints

2 <= nums.length <= 105
1 <= nums[i] <= 106

Solution Approach

Hash Table for Frequency Count

The key to solving this problem efficiently is using a hash table to count the occurrences of each number in the array. By storing these counts, we can determine which numbers can be removed in the same operation. This helps in deciding how to prioritize operations based on element frequencies.

Greedy Strategy for Minimizing Operations

A greedy approach works well by removing elements with the highest frequency first. We aim to reduce the array's size quickly by applying the most effective operations first, ensuring we minimize the total number of operations required.

Array Scanning and Hash Lookup

By scanning the array and leveraging hash lookups, we can find and apply the operations efficiently. This reduces the number of times we need to traverse the array, ensuring we meet the time complexity constraint of O(N).

Complexity Analysis

Metric	Value
Time	O(N)
Space	O(N)

The time complexity of this solution is O(N) because we only traverse the array once for counting the frequencies and once more to apply the operations. The space complexity is O(N) due to the hash table storing the frequencies of elements.

What Interviewers Usually Probe

Candidate demonstrates a solid understanding of greedy algorithms by focusing on frequency maximization.
The candidate should mention hash table lookups and how they help minimize operations.
Watch for candidates who misunderstand the problem's failure case, returning -1 when the operations aren't applicable.

Common Pitfalls or Variants

Common pitfalls

Failing to handle cases where the array cannot be emptied, such as when no elements have a frequency greater than 1.
Overcomplicating the solution by not leveraging a hash table to track frequencies.
Using inefficient algorithms that lead to excessive time complexity, especially for large arrays.

Follow-up variants

Consider cases where the array contains only one distinct element.
Handle the case when all elements are distinct, and no operations can be applied.
Introduce constraints like different operation types or modifying the frequency constraints.

FAQ

What is the pattern used in the Minimum Number of Operations to Make Array Empty?

The main pattern for this problem involves array scanning plus hash lookup to track frequencies of elements and determine the minimum number of operations.

How can I determine when it's impossible to empty the array?

If no elements have a frequency of at least two, it is impossible to apply the first operation, and the array cannot be emptied.

What happens if there are too many distinct elements in the array?

If all elements are distinct, no operations can be applied, and the result should be -1.

How do hash tables help in this problem?

Hash tables help by efficiently tracking the frequency of elements, allowing us to decide which elements can be removed in a single operation.

What is the time complexity of this solution?

The time complexity is O(N) because the solution requires two passes over the array: one for counting the frequencies and another for applying the operations.

terminal

Solution

Solution 1: Hash Table + Greedy

We use a hash table $count$ to count the number of occurrences of each element in the array. Then we traverse the hash table. For each element $x$, if it appears $c$ times, we can perform $\lfloor \frac{c+2}{3} \rfloor$ operations to delete $x$. Finally, we return the sum of the number of operations for all elements.

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class Solution:
    def minOperations(self, nums: List[int]) -> int:
        count = Counter(nums)
        ans = 0
        for c in count.values():
            if c == 1:
                return -1
            ans += (c + 2) // 3
        return ans

Continue Practicing

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