What is the time complexity of Minimum Deletions to Make Character Frequencies Unique?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Minimum Deletions to Make Character Frequencies Unique?

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

What are the key insights for solving Minimum Deletions to Make Character Frequencies Unique?

Insight 1: The problem can be solved using a greedy approach. We can iterate through the frequencies of each character and reduce the frequencies until they are unique or zero. Insight 2: We need to keep track of the frequencies we've already used to avoid creating duplicate frequencies.

Which companies ask LeetCode 1647. Minimum Deletions to Make Character Frequencies Unique in interviews?

This problem is frequently asked at American Express, ConsultAdd, smartnews.

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Minimum Deletions to Make Character Frequencies Unique - LeetCode 1647 Solution

Minimum Deletions to Make Character Frequencies Unique - Complete Solution Guide

Minimum Deletions to Make Character Frequencies Unique is LeetCode problem 1647, 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

A string s is called good if there are no two different characters in s that have the same frequency . Given a string s , return the minimum number of characters you need to delete to make s good . The frequency of a character in a string is the number of times it appears in the string. For example, in the string "aab" , the frequency of 'a' is 2 , while the frequency of 'b' is 1 . Example 1: Input: s = "aab" Output: 0 Explanation: s is already good. Example 2: Input: s = "aaabbbcc" Output: 2 Ex

Detailed Explanation

The problem asks us to find the minimum number of characters we need to delete from a given string 's' such that no two distinct characters in the resulting string have the same frequency. A string that satisfies this condition is called a 'good' string. We need to return the minimum number of deletions required to transform the input string 's' into a 'good' string. The input string 's' consists of lowercase English letters, and its length is between 1 and 10^5.

Solution Approach

The solution uses a greedy approach. First, it calculates the frequency of each character in the string 's'. Then, it iterates through the frequencies. For each frequency, it checks if that frequency has already been used. If it has, the algorithm decrements the frequency and increments the number of deletions until the frequency is unique or zero. A set is used to store the frequencies that have already been used.

Step-by-Step Algorithm

Step 1: Create an array `freq_counts` of size 26 to store the frequency of each lowercase English letter in the string 's'. Initialize all elements of the array to 0.
Step 2: Iterate through the string 's' and increment the corresponding element in `freq_counts` for each character.
Step 3: Initialize a variable `deletions` to 0 to store the number of deletions required.
Step 4: Create a set `used_frequencies` to store the frequencies that have already been used.
Step 5: Iterate through the `freq_counts` array.
Step 6: For each frequency `freq` in `freq_counts`, while `freq` is greater than 0 and `freq` is already in `used_frequencies`, decrement `freq` and increment `deletions`.
Step 7: If `freq` is greater than 0 after the while loop, add `freq` to `used_frequencies`.
Step 8: Return the value of `deletions`.

Key Insights

Insight 1: The problem can be solved using a greedy approach. We can iterate through the frequencies of each character and reduce the frequencies until they are unique or zero.
Insight 2: We need to keep track of the frequencies we've already used to avoid creating duplicate frequencies.
Insight 3: Deleting characters with higher frequency first may not always be optimal. It's best to reduce frequencies one by one until uniqueness is achieved.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

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

Companies

Asked at: American Express, ConsultAdd, smartnews.

Minimum Deletions to Make Character Frequencies Unique | LeetCode 1647

Medium

Hash Table String Greedy Sorting

Time: O(n)Space: O(1)

Solution Code

Available in:

import java.util.HashSet;
import java.util.Set;

class Solution {
    public int minDeletions(String s) {
        int[] freqCounts = new int[26];
        for (char c : s.toCharArray()) {
            freqCounts[c - 'a']++;
        }

        int deletions = 0;
        Set<Integer> usedFrequencies = new HashSet<>();

        for (int freq : freqCounts) {
            while (freq > 0 && usedFrequencies.contains(freq)) {
                freq--;
                deletions++;
            }

            if (freq > 0) {
                usedFrequencies.add(freq);
            }
        }

        return deletions;
    }
}

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Problem Info

Description

Problem 1647: Minimum Deletions to Make Character Frequencies Unique - Detailed explanation and solution approaches.

Topics

Hash Table String Greedy Sorting

Companies

American ExpressConsultAddsmartnews

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:1647

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

8 steps

1
Step 1: Create an array `freq_counts` of size 26 to store the frequency of each lowercase English letter in the string 's'. Initialize all elements of the array to 0.
2
Step 2: Iterate through the string 's' and increment the corresponding element in `freq_counts` for each character.
3
Step 3: Initialize a variable `deletions` to 0 to store the number of deletions required.
4
Step 4: Create a set `used_frequencies` to store the frequencies that have already been used.
5
Step 5: Iterate through the `freq_counts` array.
6
Step 6: For each frequency `freq` in `freq_counts`, while `freq` is greater than 0 and `freq` is already in `used_frequencies`, decrement `freq` and increment `deletions`.
7
Step 7: If `freq` is greater than 0 after the while loop, add `freq` to `used_frequencies`.
8
Step 8: Return the value of `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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is dominated by the initial frequency calculation and the while loop which may decrement frequency upto n times. Space complexity is constant since only 26 characters are considered even with a frequency upto n.

📖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

Sorting Frequencies

Instead of using a set, we could sort the frequencies in descending order and then iterate through the sorted frequencies, decrementing frequencies to ensure uniqueness. If a frequency is equal to or greater than the previous frequency, we would decrement it.

Time: O(n log n) due to sortingSpace: O(1) in place sort, O(n) if requires additional space for sorting

Trade-offs: The sorting approach has a higher time complexity of O(n log n) due to sorting. The current approach is faster O(n) but sorting could still be beneficial if we expected to receive strings where the set operations would have similar cost to sorting.

✨

Key Insights

1Insight 1: The problem can be solved using a greedy approach. We can iterate through the frequencies of each character and reduce the frequencies until they are unique or zero.
2Insight 2: We need to keep track of the frequencies we've already used to avoid creating duplicate frequencies.
3Insight 3: Deleting characters with higher frequency first may not always be optimal. It's best to reduce frequencies one by one until uniqueness is achieved.

📚

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 will correctly return 0 deletions since the frequencies array will be all zeros, and the loop won't execute.
Edge case 2: String with all same characters (e.g., "aaaaaa"). The code will iteratively decrement the frequency until it becomes 0, resulting in the correct number of deletions.
Edge case 3: String with unique characters. The code will correctly return 0 deletions as no frequencies are duplicated.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the case where a frequency needs to be decremented to 0. If you simply skip duplicate frequencies without decrementing, you might miss opportunities to make the string good.

✗Common mistake 2: Optimizing prematurely by deleting higher frequency characters assuming it's optimal. Greedy approach of decrementing by 1 until uniqueness is reached is best.

✗Common mistake 3: Forgetting to handle the condition where the decreased frequency becomes zero, as this will require adding to the deletion count

✅

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. Reducing redundancy in data by minimizing the repetition of characters, which relates to the problem of frequency optimization and making character frequencies unique.

🏦Real-world scenario 2: Database indexing. Optimizing the distribution of index values to improve query performance. This involves managing the frequencies of different index values to avoid skewed distributions that can lead to inefficient queries.

🎮Real-world scenario 3: Resource allocation. Allocating limited resources such as bandwidth or processing power among different users or applications. This often involves balancing the frequency with which different resources are requested to minimize conflicts and ensure fair usage.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

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

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 28, 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

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

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 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 Character Frequencies Unique"?

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?