#3085

Medium

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Minimum Deletions to Make String K-Special

Minimize deletions to make a string k-special by adjusting character frequencies.

hash table string greedy sorting counting

Problem Statement

You are given a string word and an integer k. The string is considered k-special if for all pairs of indices i and j, the absolute difference between the frequencies of characters at those indices is less than or equal to k. In other words, the difference in frequency between any two characters should not exceed k.

Your task is to determine the minimum number of deletions required to make the string k-special. The deletions should be performed by removing characters from the string, and you must ensure that the final string meets the k-special condition.

Examples

Example 1

Input: word = "aabcaba", k = 0

Output: 3

We can make word 0 -special by deleting 2 occurrences of "a" and 1 occurrence of "c" . Therefore, word becomes equal to "baba" where freq('a') == freq('b') == 2 .

Example 2

Input: word = "dabdcbdcdcd", k = 2

Output: 2

We can make word 2 -special by deleting 1 occurrence of "a" and 1 occurrence of "d" . Therefore, word becomes equal to "bdcbdcdcd" where freq('b') == 2 , freq('c') == 3 , and freq('d') == 4 .

Example 3

Input: word = "aaabaaa", k = 2

Output: 1

We can make word 2 -special by deleting 1 occurrence of "b" . Therefore, word becomes equal to "aaaaaa" where each letter's frequency is now uniformly 6 .

Constraints

1 <= word.length <= 105
0 <= k <= 105
word consists only of lowercase English letters.

Solution Approach

Frequency Counting

Start by counting the frequency of each character in the string. This is done using a hash table to track the occurrences of each character. Sorting the frequencies helps in easily identifying which characters need deletion to balance the string.

Greedy Deletions

After sorting the frequencies, the greedy approach begins. Delete characters from the most frequent ones to ensure that the absolute difference in frequencies between any two characters is at most k. This ensures minimal deletions are made.

Invariant Validation

Continuously validate the condition for k-special by checking the frequency differences between characters. If the difference exceeds k, additional deletions are made. The goal is to reduce the string to the smallest form that satisfies the k-special property.

Complexity Analysis

Metric	Value
Time	O(n + C^2)
Space	O(C)

The time complexity is O(n + C^2), where n is the length of the string and C is the number of distinct characters. The space complexity is O(C) due to the storage of character frequencies in a hash table.

What Interviewers Usually Probe

Understand frequency counting and its importance for this problem.
Demonstrate ability to apply a greedy strategy to optimize deletions.
Check for a valid approach by maintaining the invariant while making deletions.

Common Pitfalls or Variants

Common pitfalls

Not using a hash table for efficient frequency counting can lead to slower solutions.
Forgetting to sort the frequencies might result in suboptimal deletions.
Incorrectly applying the greedy approach might lead to an invalid final string.

Follow-up variants

Instead of minimizing deletions, the problem could be framed as maximizing the number of characters that remain in the string.
The problem could involve different limits on k or the size of the string, testing the solution's scalability.
Instead of just deletions, the problem could also involve transforming characters to meet the k-special condition.

FAQ

What is a k-special string?

A k-special string is one where the absolute frequency difference between any two characters is at most k.

How do I solve the 'Minimum Deletions to Make String K-Special' problem?

The problem is solved by counting the frequencies of characters, sorting them, and using a greedy strategy to delete characters and meet the k-special condition.

What pattern is used in solving the 'Minimum Deletions to Make String K-Special' problem?

The problem uses a greedy approach combined with invariant validation, focusing on minimizing deletions based on character frequencies.

What is the time complexity of the 'Minimum Deletions to Make String K-Special' problem?

The time complexity is O(n + C^2), where n is the length of the string and C is the number of distinct characters.

Can the 'Minimum Deletions to Make String K-Special' problem be solved using dynamic programming?

No, the problem is best solved with a greedy strategy, using frequency counting and validation rather than dynamic programming.

terminal

Solution

Solution 1: Counting + Enumeration

First, we can count the occurrence of each character in the string and put all the counts into an array $nums$. Since the string only contains lowercase letters, the length of the array $nums$ will not exceed $26$.

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class Solution:
    def minimumDeletions(self, word: str, k: int) -> int:
        def f(v: int) -> int:
            ans = 0
            for x in nums:
                if x < v:
                    ans += x
                elif x > v + k:
                    ans += x - v - k
            return ans

        nums = Counter(word).values()
        return min(f(v) for v in range(len(word) + 1))

Continue Practicing

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