The goal is to minimize deletions to make a string k-special by ensuring the frequency difference between any two characters is at most k. A greedy approach works by counting character frequencies and deleting the least frequent ones to satisfy the condition.
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.
How GhostInterview Helps
- GhostInterview helps you focus on counting character frequencies, a crucial step in solving the problem.
- It guides you through the greedy approach, ensuring you delete characters in the most efficient way possible.
- GhostInterview helps you practice the validation of the k-special condition to ensure correctness in your solution.
Topic Pages
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- Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
- How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.
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.
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