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

The optimal solution has a time complexity of O(N*M).

What data structures are used to solve Minimum Deletions to Make String K-Special?

This problem uses Hash Table, String, Greedy, Sorting, Counting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Deletions to Make String K-Special?

Insight 1: The problem can be reframed as finding a suitable range of frequencies [f_min, f_min + k] such that the number of deletions needed to bring all character frequencies within that range is minimized. Insight 2: The frequencies of the characters after deletion will be in the range [f_min, f_min+k]. For any character with frequency less than f_min, we should delete the frequency entirely. For a character with frequency more than f_max, we should reduce the frequency to f_max.

Which companies ask LeetCode 3085. Minimum Deletions to Make String K-Special in interviews?

This problem is frequently asked at DE Shaw.

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Minimum Deletions to Make String K-Special - LeetCode 3085 Solution

Q: How to solve Minimum Deletions to Make String K-Special?

The solution calculates the frequencies of each character in the input string. Then, it iterates through possible values for the minimum frequency (`f_min`). For each `f_min`, it calculates the maximum allowed frequency (`f_max = f_min + k`). The algorithm then iterates through the frequencies of each character. If a frequency is less than `f_min`, all occurrences of that character must be deleted, so the number of deletions is added to `current_deletions`. If a frequency is greater than `f_max`, the number of characters to be deleted from that character will be `f - f_max`. The minimum number of deletions across all `f_min` values is then returned.

Minimum Deletions to Make String K-Special - Complete Solution Guide

Minimum Deletions to Make String K-Special is LeetCode problem 3085, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a string word and an integer k . We consider word to be k-special if |freq(word[i]) - freq(word[j])| <= k for all indices i and j in the string. Here, freq(x) denotes the frequency of the character x in word , and |y| denotes the absolute value of y . Return the minimum number of characters you need to delete to make word k-special . Example 1: Input: word = "aabcaba", k = 0 Output: 3 Explanation: We can make word 0 -special by deleting 2 occurrences of "a" and 1 occurrence of "c"

Detailed Explanation

The problem asks us to find the minimum number of characters we need to delete from a given string `word` to make it `k-special`. A string is considered `k-special` if the absolute difference between the frequencies of any two characters in the string is less than or equal to `k`. Essentially, we need to adjust the frequencies of the characters such that the difference between the minimum and maximum frequency is at most `k`, while minimizing the number of deletions.

Solution Approach

The solution calculates the frequencies of each character in the input string. Then, it iterates through possible values for the minimum frequency (`f_min`). For each `f_min`, it calculates the maximum allowed frequency (`f_max = f_min + k`). The algorithm then iterates through the frequencies of each character. If a frequency is less than `f_min`, all occurrences of that character must be deleted, so the number of deletions is added to `current_deletions`. If a frequency is greater than `f_max`, the number of characters to be deleted from that character will be `f - f_max`. The minimum number of deletions across all `f_min` values is then returned.

Step-by-Step Algorithm

Step 1: Count the frequency of each character in the input string `word` using a hash map (or Counter in Python).
Step 2: Extract the frequencies from the hash map into a list.
Step 3: Find the unique frequencies and sort them. This will be used to generate potential values for `f_min`
Step 4: Create a list of candidate `f_min` values. This list contains 0 and each of the unique frequencies from previous step.
Step 5: Iterate through each candidate `f_min` value.
Step 6: For each `f_min`, calculate `f_max = f_min + k`.
Step 7: Iterate through all the character frequencies. Calculate the number of deletions needed to make the frequency `k-special`.
Step 8: If a character's frequency `f` is less than `f_min`, the number of deletions needed for that character will be `f`.
Step 9: If a character's frequency `f` is greater than `f_max`, the number of deletions needed for that character will be `f - f_max`.
Step 10: Sum the required deletions for each frequency to compute `current_deletions`.
Step 11: Keep track of the minimum `current_deletions` across all tested `f_min` values. Update `min_deletions` if needed.
Step 12: Return the overall `min_deletions`.

Key Insights

Insight 1: The problem can be reframed as finding a suitable range of frequencies [f_min, f_min + k] such that the number of deletions needed to bring all character frequencies within that range is minimized.
Insight 2: The frequencies of the characters after deletion will be in the range [f_min, f_min+k]. For any character with frequency less than f_min, we should delete the frequency entirely. For a character with frequency more than f_max, we should reduce the frequency to f_max.
Insight 3: It's sufficient to consider only frequencies that actually occur in the word as possible values for f_min. This optimization helps to reduce the number of candidate ranges we need to evaluate.

Complexity Analysis

Time Complexity: O(N*M)

Space Complexity: O(M)

Topics

This problem involves: Hash Table, String, Greedy, Sorting, Counting.

Companies

Asked at: DE Shaw.

Minimum Deletions to Make String K-Special | LeetCode 3085

Medium

Hash Table String Greedy Sorting Counting

Time: O(N*M)Space: O(M)

Solution Code

Available in:

import java.util.HashMap;
import java.util.List;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Set;
import java.util.HashSet;

class Solution {
    public int minimumDeletions(String word, int k) {
        HashMap<Character, Integer> counts = new HashMap<>();
        for (char c : word.toCharArray()) {
            counts.put(c, counts.getOrDefault(c, 0) + 1);
        }
        if (counts.isEmpty()) {
            return 0;
        }
        
        List<Integer> freqs = new ArrayList<>(counts.values());
        Set<Integer> uniqueFreqSet = new HashSet<>(freqs);
        List<Integer> unique_freqs = new ArrayList<>(uniqueFreqSet);
        Collections.sort(unique_freqs);
        
        int min_deletions = Integer.MAX_VALUE;
        
        List<Integer> candidate_f_mins = new ArrayList<>();
        candidate_f_mins.add(0);
        candidate_f_mins.addAll(unique_freqs);
        
        for (int f_min : candidate_f_mins) {
            int f_max = f_min + k;
            int current_deletions = 0;
            for (int f : freqs) {
                if (f < f_min) {
                    current_deletions += f;
                } else if (f > f_max) {
                    current_deletions += f - f_max;
                }
            }
            min_deletions = Math.min(min_deletions, current_deletions);
        }
        
        return min_deletions;
    }
}

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Problem Info

Description

Problem 3085: Minimum Deletions to Make String K-Special - Detailed explanation and solution approaches.

Topics

Hash Table String Greedy Sorting Counting

Companies

DE Shaw

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3085

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

12 steps

1
Step 1: Count the frequency of each character in the input string `word` using a hash map (or Counter in Python).
2
Step 2: Extract the frequencies from the hash map into a list.
3
Step 3: Find the unique frequencies and sort them. This will be used to generate potential values for `f_min`
4
Step 4: Create a list of candidate `f_min` values. This list contains 0 and each of the unique frequencies from previous step.
5
Step 5: Iterate through each candidate `f_min` value.
6
Step 6: For each `f_min`, calculate `f_max = f_min + k`.
7
Step 7: Iterate through all the character frequencies. Calculate the number of deletions needed to make the frequency `k-special`.
8
Step 8: If a character's frequency `f` is less than `f_min`, the number of deletions needed for that character will be `f`.
9
Step 9: If a character's frequency `f` is greater than `f_max`, the number of deletions needed for that character will be `f - f_max`.
10
Step 10: Sum the required deletions for each frequency to compute `current_deletions`.
11
Step 11: Keep track of the minimum `current_deletions` across all tested `f_min` values. Update `min_deletions` if needed.
12
Step 12: Return the overall `min_deletions`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(N*M)

💾

Space Complexity

O(M)

📝

Why?

The time complexity is determined by the nested loops. The outer loop iterates through potential `f_min` values, which are bounded by the number of unique character frequencies (M). The inner loop iterates through each character's frequency (also bounded by M). The space complexity is determined by the number of unique characters in the word, which needs to be stored in the hash map to count frequencies.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Dynamic Programming

While less intuitive, a dynamic programming approach might be possible. We can define dp[i][j] as the minimum deletions to make the substring word[0...i] k-special with a maximum frequency of j. However, the complexity would be high and likely not more efficient.

Time: O(N*M^2) (N = word length, M = max frequency)Space: O(N*M)

Trade-offs: DP could work but might not be as efficient or easy to implement as the greedy approach. It requires more space complexity and is harder to conceptualize.

✨

Key Insights

1Insight 1: The problem can be reframed as finding a suitable range of frequencies [f_min, f_min + k] such that the number of deletions needed to bring all character frequencies within that range is minimized.
2Insight 2: The frequencies of the characters after deletion will be in the range [f_min, f_min+k]. For any character with frequency less than f_min, we should delete the frequency entirely. For a character with frequency more than f_max, we should reduce the frequency to f_max.
3Insight 3: It's sufficient to consider only frequencies that actually occur in the word as possible values for f_min. This optimization helps to reduce the number of candidate ranges we need to evaluate.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Greedy Algorithm

Making the best choice at each step without looking ahead.

String

A sequence of characters. Like a word or sentence in programming!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty string: The code correctly handles the case when the input string `word` is empty by returning 0 deletions.
Edge case 2: All characters are the same: If all characters are the same, any `f_min` will result in a 0 deletion case, as expected.
Edge case 3: k = 0: When k = 0, all characters must have equal frequency, so algorithm will test candidate frequencies and return min deletions to equate them.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering 0 as a candidate f_min. If all characters have higher frequencies than `k`, setting `f_min` to zero will result in just reducing all frequencies to `k`.

✗Common mistake 2: Incorrectly calculating the number of deletions for frequencies greater than `f_max`. The correct formula is `f - f_max`, not just `f`.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Data compression: Minimizing the variations in character frequencies can be useful in data compression algorithms to improve efficiency.

🏦Real-world scenario 2: Resource allocation: Imagine allocating resources (represented by character frequencies) with a tolerance (`k`) for variation. This algorithm helps minimize the changes needed to meet a balanced resource allocation requirement.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 26, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 30

❓

Frequently Asked Questions

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Minimum Deletions to Make String K-Special"?

For this medium-level Hash Table problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?