What is the time complexity of Delete Characters to Make Fancy String?

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

What data structures are used to solve Delete Characters to Make Fancy String?

This problem uses String. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Delete Characters to Make Fancy String?

Insight 1: The problem can be solved efficiently by iterating through the string only once. No need for recursion or complex data structures. Insight 2: A simple counter is sufficient to track consecutive character repetitions. When the count of consecutive identical characters exceeds 2, characters need to be removed.

Which companies ask LeetCode 1957. Delete Characters to Make Fancy String in interviews?

This problem is frequently asked at Wayfair.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Delete Characters to Make Fancy String - LeetCode 1957 Solution

Delete Characters to Make Fancy String - Complete Solution Guide

Delete Characters to Make Fancy String is LeetCode problem 1957, 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

A fancy string is a string where no three consecutive characters are equal. Given a string s , delete the minimum possible number of characters from s to make it fancy . Return the final string after the deletion . It can be shown that the answer will always be unique . Example 1: Input: s = "le e etcode" Output: "leetcode" Explanation: Remove an 'e' from the first group of 'e's to create "leetcode". No three consecutive characters are equal, so return "leetcode". Example 2: Input: s = " a aab a

Detailed Explanation

The problem asks you to take a string as input and modify it such that no three consecutive characters are the same. The goal is to remove the minimum number of characters to achieve this. The input is a string consisting of lowercase English letters. The output is a modified string that satisfies the 'fancy string' condition. Constraints limit the string length to 10<sup>5</sup> characters.

Solution Approach

The provided solutions employ a greedy approach. They iterate through the input string, maintaining a count of consecutive identical characters. If the count exceeds 2, the extra characters are skipped. This ensures that no more than two consecutive identical characters remain in the resulting string. The approach works because it locally optimizes at each character; removing extra repetitions immediately prevents future violations of the 'fancy string' condition.

Step-by-Step Algorithm

Step 1: Initialize an empty string (or StringBuilder in Java) to store the result.
Step 2: Initialize a counter to track consecutive identical characters.
Step 3: Iterate through the input string.
Step 4: If the current character is the same as the previous character, increment the counter. Otherwise, reset the counter to 1.
Step 5: If the counter is less than 3, append the current character to the result string. Otherwise, skip the current character.
Step 6: Repeat steps 4 and 5 until all characters have been processed.
Step 7: Return the resulting string.

Key Insights

Insight 1: The problem can be solved efficiently by iterating through the string only once. No need for recursion or complex data structures.
Insight 2: A simple counter is sufficient to track consecutive character repetitions. When the count of consecutive identical characters exceeds 2, characters need to be removed.
Insight 3: The solution needs to handle edge cases such as empty strings, strings with less than 3 characters, and strings with long sequences of repeating characters.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: String.

Companies

Asked at: Wayfair.

Delete Characters to Make Fancy String | LeetCode 1957

Easy

String

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

Solution Code

Available in:

class Solution {
    public String makeFancyString(String s) {
        StringBuilder sb = new StringBuilder();
        int count = 0;
        for (int i = 0; i < s.length(); i++) {
            if (sb.length() > 0 && sb.charAt(sb.length() - 1) == s.charAt(i)) {
                count++;
                if (count < 2) {
                    sb.append(s.charAt(i));
                }
            } else {
                sb.append(s.charAt(i));
                count = 0;
            }
        }
        return sb.toString();
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 1957: Delete Characters to Make Fancy String - Detailed explanation and solution approaches.

Topics

String

Companies

Wayfair

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1957

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

7 steps

1
Step 1: Initialize an empty string (or StringBuilder in Java) to store the result.
2
Step 2: Initialize a counter to track consecutive identical characters.
3
Step 3: Iterate through the input string.
4
Step 4: If the current character is the same as the previous character, increment the counter. Otherwise, reset the counter to 1.
5
Step 5: If the counter is less than 3, append the current character to the result string. Otherwise, skip the current character.
6
Step 6: Repeat steps 4 and 5 until all characters have been processed.
7
Step 7: Return the resulting string.

💡

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(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The algorithm iterates through the input string once. The space used is proportional to the length of the output string, which can be at most the same length as the input string.

📖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

Using a stack

A stack can be used to track the last two characters. When processing a new character, compare it to the top of the stack and push it accordingly. If the new character is the same as the top two, it's skipped.

Time: O(n)Space: O(1) (since the stack's size is limited to 2)

Trade-offs: Using a stack would have slightly higher constant factors due to stack operations. However, the space complexity is reduced to constant, which is better for very long strings

✨

Key Insights

1Insight 1: The problem can be solved efficiently by iterating through the string only once. No need for recursion or complex data structures.
2Insight 2: A simple counter is sufficient to track consecutive character repetitions. When the count of consecutive identical characters exceeds 2, characters need to be removed.
3Insight 3: The solution needs to handle edge cases such as empty strings, strings with less than 3 characters, and strings with long sequences of repeating characters.

📚

Key Terms Explained

For Students

String

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

Edge Cases

Edge case 1: Empty string. The solutions handle this correctly by either implicitly (Python) or explicitly (Java, C++) handling the case where the input length is 0. The result will be an empty string.
Edge case 2: String with less than 3 characters. The solutions correctly handle this case; no characters are removed as the condition is already satisfied.
Edge case 3: String with long sequences of repeating characters (e.g., 'aaaaaaaaa'). The solutions effectively reduce the consecutive repetitions to a maximum of two ('aa').

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling the counter. Failing to reset the counter when a different character is encountered can lead to incorrect results.

✗Common mistake 2: Not considering edge cases, especially empty strings or strings with fewer than three characters. This could result in errors or unexpected behavior.

✅

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 sanitization or data cleaning. This algorithm could be used to remove consecutive duplicate entries in data streams.

🏦Real-world scenario 2: Text processing or natural language processing (NLP). It could be part of a larger system for cleaning up text data to improve its quality or for specialized text formatting.

💡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

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

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: November 5, 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

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 "Delete Characters to Make Fancy String"?

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

🤖 Need help with this problem?