How do you solve Minimum Deletions to Make String K-Special on LeetCode?

Minimum Deletions to Make String K-Special (LeetCode #3085) 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: Understanding character frequency is crucial to solving this problem efficiently.

What is the brute force solution for Minimum Deletions to Make String K-Special?

The brute force approach for Minimum Deletions to Make String K-Special is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible combination of character deletions to see if the resulting string meets the k-special condition. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Deletions to Make String K-Special?

Minimum Deletions to Make String K-Special uses the following concepts: Hash Table, String, Greedy, Sorting, Counting. The recommended patterns to study are: Hash Map, Sliding Window.

What are the interview tips for Minimum Deletions to Make String K-Special?

Always clarify the problem constraints and edge cases before diving into coding. Explain your thought process and reasoning as you work through the problem. Practice writing clean and efficient code, as readability can be just as important as functionality.

What are common mistakes when solving Minimum Deletions to Make String K-Special?

Failing to consider edge cases where k is larger than the maximum frequency. Not properly managing the sliding window boundaries, leading to incorrect counts.

#3085

Minimum Deletions to Make String K-Special

Medium

Hash TableStringGreedySortingCountingHash MapSliding Window

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

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

The optimal approach involves counting the frequencies of characters, sorting them, and then determining the minimum deletions needed to ensure that the difference between the maximum and minimum frequencies is within k. This is efficient and leverages sorting.

⚙️

Algorithm

4 steps

1Step 1: Count the frequency of each character in the string.
2Step 2: Store the frequencies in a list and sort it.
3Step 3: Use a sliding window to find the longest subarray where the difference between the maximum and minimum frequencies is less than or equal to k.
4Step 4: Calculate the minimum deletions as the total number of characters minus the size of the longest valid subarray.

solution.py14 lines

1# Full working Python code
2from collections import Counter
3
4def min_deletions_optimal(word, k):
5    freq = Counter(word)
6    freq_list = sorted(freq.values())
7    left = 0
8    max_length = 0
9    for right in range(len(freq_list)):
10        while freq_list[right] - freq_list[left] > k:
11            left += 1
12        max_length = max(max_length, right - left + 1)
13    return len(word) - max_length
14

ℹ

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

1Understanding character frequency is crucial to solving this problem efficiently.
2Using sorting and a sliding window approach can drastically reduce the complexity from brute-force.

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