How do you solve Least Number of Unique Integers after K Removals on LeetCode?

Least Number of Unique Integers after K Removals (LeetCode #1481) 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 least frequent integers first minimizes the unique count.

What is the time complexity of Least Number of Unique Integers after K Removals?

The optimal solution for Least Number of Unique Integers after K Removals runs in O(n log n) time and uses O(n) space. The time complexity is dominated by the sorting step, which is O(n log n), while the space complexity is O(n) due to the storage of frequency counts.

What is the brute force solution for Least Number of Unique Integers after K Removals?

The brute force approach for Least Number of Unique Integers after K Removals is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible combinations of the array after removing k elements and counting the unique integers left. This is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Least Number of Unique Integers after K Removals?

Least Number of Unique Integers after K Removals uses the following concepts: Array, Hash Table, Greedy, Sorting, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Least Number of Unique Integers after K Removals?

Always clarify if you can remove duplicates or if you need to consider their counts. Discuss your thought process and reasoning behind choosing specific elements to remove. Practice explaining your solution clearly, as communication is key in interviews.

What are common mistakes when solving Least Number of Unique Integers after K Removals?

Not considering the frequency of integers when deciding which to remove. Failing to sort the frequencies before processing them.

#1481

Least Number of Unique Integers after K Removals

Medium

ArrayHash TableGreedySortingCountingHash 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 leverages frequency counting and sorting. By removing the least frequent integers first, we minimize the number of unique integers remaining after k removals.

⚙️

Algorithm

4 steps

1Step 1: Count the frequency of each integer in the array using a HashMap.
2Step 2: Sort the integers based on their frequency in ascending order.
3Step 3: Remove integers starting from the least frequent until k removals are made.
4Step 4: Count the remaining unique integers.

solution.py14 lines

1# Full working Python code
2from collections import Counter
3
4def least_num_unique_ints(arr, k):
5    freq = Counter(arr)
6    unique_count = len(freq)
7    freq_values = sorted(freq.values())
8    for count in freq_values:
9        if k >= count:
10            k -= count
11            unique_count -= 1
12        else:
13            break
14    return unique_count

ℹ

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

1Removing the least frequent integers first minimizes the unique count.
2Using a frequency map helps efficiently track and manage counts.

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