What is the time complexity of Remove Letter To Equalize Frequency?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Remove Letter To Equalize Frequency?

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

What are the key insights for solving Remove Letter To Equalize Frequency?

Insight 1: The core idea is to iterate through each character, remove it temporarily, and check if the frequencies of the remaining characters are equal. This requires efficient frequency counting. Insight 2: Using a hash map (or similar data structure like `Counter` in Python) is crucial for efficient frequency counting of characters. This allows for O(1) lookups of character counts.

Which companies ask LeetCode 2423. Remove Letter To Equalize Frequency in interviews?

This problem is frequently asked at tcs.

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Remove Letter To Equalize Frequency - LeetCode 2423 Solution

Remove Letter To Equalize Frequency - Complete Solution Guide

Remove Letter To Equalize Frequency is LeetCode problem 2423, a Easy 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 0-indexed string word , consisting of lowercase English letters. You need to select one index and remove the letter at that index from word so that the frequency of every letter present in word is equal. Return true if it is possible to remove one letter so that the frequency of all letters in word are equal, and false otherwise . Note: The frequency of a letter x is the number of times it occurs in the string. You must remove exactly one letter and cannot choose to do nothing. E

Detailed Explanation

The problem asks whether it's possible to remove exactly one character from a given string `word` such that the frequency of each remaining character is the same. The input is a string containing lowercase English letters. The output is a boolean: `true` if such a removal is possible, and `false` otherwise. Constraints limit the string length to between 2 and 100 characters.

Solution Approach

The provided solutions use a frequency counting approach. They iterate through each character in the input string. For each character, they temporarily remove it, recalculate the frequencies of the remaining characters, and check if all remaining characters have the same frequency. If they do, the function immediately returns `true`. If no such character can be removed to achieve equal frequencies, the function returns `false`.

Step-by-Step Algorithm

Step 1: Count the frequency of each character in the input string using a hash map (or array in C/Java).
Step 2: Iterate through the input string. For each character:
Step 3: Decrement its frequency in the frequency map.
Step 4: If the frequency becomes 0, remove the character entry from the map.
Step 5: Check if the remaining frequencies are all equal. This can be done by creating a set of the frequencies and checking if the set size is 1. If true, return `true`.
Step 6: If the loop completes without finding a character whose removal leads to equal frequencies, return `false`.

Key Insights

Insight 1: The core idea is to iterate through each character, remove it temporarily, and check if the frequencies of the remaining characters are equal. This requires efficient frequency counting.
Insight 2: Using a hash map (or similar data structure like `Counter` in Python) is crucial for efficient frequency counting of characters. This allows for O(1) lookups of character counts.
Insight 3: Optimizing the frequency check after removal. Instead of recounting every time, efficiently update the frequency map.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n)

Topics

This problem involves: Hash Table, String, Counting.

Companies

Asked at: tcs.

Remove Letter To Equalize Frequency | LeetCode 2423

Easy

Hash Table String Counting

Time: O(n^2)Space: O(n)

Solution Code

Available in:

import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Set;

class Solution {
    public boolean equalFrequency(String word) {
        int[] freq = new int[26];
        for (char c : word.toCharArray()) {
            freq[c - 'a']++;
        }

        for (int i = 0; i < word.length(); i++) {
            int[] tempFreq = new int[26];
            System.arraycopy(freq, 0, tempFreq, 0, 26);
            tempFreq[word.charAt(i) - 'a']--;

            Set<Integer> frequencies = new HashSet<>();
            boolean possible = true;
            for (int j = 0; j < 26; j++) {
                if (tempFreq[j] > 0) {
                    frequencies.add(tempFreq[j]);
                }
            }

            if (frequencies.size() > 1) {
                possible = false;
            }

            if (possible) {
                return true;
            }
        }

        return false;
    }
}

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Problem Info

Description

Problem 2423: Remove Letter To Equalize Frequency - Detailed explanation and solution approaches.

Topics

Hash Table String Counting

Companies

tcs

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:2423

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

6 steps

1
Step 1: Count the frequency of each character in the input string using a hash map (or array in C/Java).
2
Step 2: Iterate through the input string. For each character:
3
Step 3: Decrement its frequency in the frequency map.
4
Step 4: If the frequency becomes 0, remove the character entry from the map.
5
Step 5: Check if the remaining frequencies are all equal. This can be done by creating a set of the frequencies and checking if the set size is 1. If true, return `true`.
6
Step 6: If the loop completes without finding a character whose removal leads to equal frequencies, return `false`.

💡

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^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The frequency counting takes O(n) time, where n is the length of the string. Iterating through the string once also takes O(n) time. The frequency map uses at most 26 entries (for lowercase English letters), so its space complexity is O(1). The set of frequencies also has a maximum size of 26, which is constant. Therefore, the overall time complexity is O(n) and space complexity is O(1).

📖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

Brute-force approach

Generate all possible substrings by removing one character at a time and check the frequency for each substring. This is less efficient.

Time: O(n^2)Space: O(n)

Trade-offs: This approach is less efficient than the frequency counting approach. It's easier to understand but computationally more expensive for larger strings.

✨

Key Insights

1Insight 1: The core idea is to iterate through each character, remove it temporarily, and check if the frequencies of the remaining characters are equal. This requires efficient frequency counting.
2Insight 2: Using a hash map (or similar data structure like `Counter` in Python) is crucial for efficient frequency counting of characters. This allows for O(1) lookups of character counts.
3Insight 3: Optimizing the frequency check after removal. Instead of recounting every time, efficiently update the frequency map.

📚

Key Terms Explained

For Students

Hash Table

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

String

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

Edge Cases

Edge case 1: Empty string or string with only one character. The solutions correctly handle this by returning `false` (implicitly because the loop never executes).
Edge case 2: String where removal of any character results in unequal frequencies. The solutions accurately return `false` in such cases.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case where the frequency of a character becomes 0 after removal. The character should be removed from the frequency map to avoid incorrect results.

✗Common mistake 2: Inefficient frequency checking. Recounting frequencies after each removal is less efficient than updating the existing frequency map.

✅

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 analysis tasks involving text processing where you need to determine if a dataset is balanced after removing outlier entries or correcting typos.

🏦Real-world scenario 2: Analyzing word frequencies in natural language processing to ensure the text adheres to certain balance or pattern criteria.

💡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

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ 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: August 23, 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

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

@CodeNinjaJul 14

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

@TechLeadJul 7

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

@DevExpert92Jun 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.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 "Remove Letter To Equalize Frequency"?

For this easy-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.

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