How do you solve Reduce Array Size to The Half on LeetCode?

Reduce Array Size to The Half (LeetCode #1338) 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: Removing the most frequent integers first is the most effective strategy.

What is the brute force solution for Reduce Array Size to The Half?

The brute force approach for Reduce Array Size to The Half is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking all possible combinations of integers to remove from the array. This is simple but inefficient, as it can take a lot of time to find the right combination. The optimal Optimal Solution approach improves this to O(n log n).

What are the interview tips for Reduce Array Size to The Half?

Always start by understanding the problem and identifying key constraints. Discuss your thought process out loud; it helps interviewers understand your approach. Practice writing clean and efficient code, as clarity is crucial in interviews.

What are common mistakes when solving Reduce Array Size to The Half?

Overthinking the combinations instead of focusing on frequency counts. Not considering edge cases where all elements are the same.

#1338

Reduce Array Size to The Half

Medium

ArrayHash TableGreedySortingHeap (Priority Queue)Hash MapArray

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 approach focuses on counting the frequency of each integer and then strategically removing the most frequent integers first. This is efficient because removing high-frequency integers reduces the array size significantly.

⚙️

Algorithm

3 steps

1Step 1: Count the frequency of each integer in the array using a HashMap.
2Step 2: Sort the frequencies in descending order.
3Step 3: Iterate through the sorted frequencies, removing integers until at least half of the array is removed, counting how many integers were removed.

solution.py15 lines

1# Full working Python code
2from collections import Counter
3
4def minSetSize(arr):
5    n = len(arr)
6    half = n // 2
7    freq = Counter(arr)
8    sorted_freq = sorted(freq.values(), reverse=True)
9    removed_count = 0
10    set_size = 0
11    for count in sorted_freq:
12        removed_count += count
13        set_size += 1
14        if removed_count >= half:
15            return set_size

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step of the frequency counts, while the space complexity is O(n) for storing the frequency counts.

1Removing the most frequent integers first is the most effective strategy.
2The problem can be efficiently solved using frequency counting and sorting.

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