Tuesday, May 20, 2025

Discussion on "Remove Consecutive Duplicates" Problem

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Problem: Remove Consecutive Duplicates

Given a string s, remove all consecutive duplicate characters and return the resulting string.

Input Format

A single line containing a string s (1 ≤ length of s ≤ 1000)

The string will contain only lowercase English letters.

Output Format

Print the string after removing consecutive duplicate characters.

Sample Input 1

aabbccddeee

Sample Output 1

abcde

Sample Input 2

aabaaabb

Sample Output 2

abab

~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~

Can you ask me a (that may be simple at rating of about 3/10) coding problem that tests problem solving skills of the candidate?

Sure! Here's a simple (3/10 difficulty) coding problem that tests basic problem-solving, string handling, and logic skills:


✅ Problem: Remove Consecutive Duplicates

Given a string s, remove all consecutive duplicate characters and return the resulting string.


🔹 Input Format

  • A single line containing a string s (1 ≤ length of s ≤ 1000)

  • The string will contain only lowercase English letters.


🔹 Output Format

  • Print the string after removing consecutive duplicate characters.


🔹 Sample Input 1

nginx
aabbccddeee

🔹 Sample Output 1

nginx
abcde

🔹 Sample Input 2

nginx
aabaaabb

🔹 Sample Output 2

nginx
abab

🧠 What it tests

  • Understanding of string traversal

  • Use of conditionals

  • Handling edge cases like repeated or single characters


Let me know when you're ready for the solution or hints!

Can you critique this rough algo:

s = "afafasgggewrtdg"
t = ""
last_seen_character = ''
streak_found = False
for c in s: 
    if c not in t: 
        t += c
        last_seen_character = c
    elif c in t and last_seen_character == c: 
        streak_found = True
        last_seen_character = c
        continue
    else (c in t) and last_seen_character != c and not streak_found:
        t += c
        last_seen_character = c
        streak_found = False

Thanks for sharing your rough algorithm! You're clearly thinking through the logic and edge cases, which is great. Let’s break down the approach and evaluate it step by step.


🔍 Your Goal:

You want to remove consecutive duplicate characters from a string.


✅ What’s good in your approach:

  1. Tracking state: You're maintaining last_seen_character — that’s useful for detecting consecutive repeats.

  2. Use of flags: You introduced streak_found to handle streaks, though it's a bit more complex than needed for this problem.

  3. Good intent: You’re trying to avoid adding characters that have already been added if they’re part of a consecutive streak.


❌ Issues in the Logic:

  1. Wrong check for character presence in t:

    • You’re checking if c not in t, but this checks if the character has ever occurred in t, not if it's the same as the previous one.

    • This can cause issues with strings like "aabaaabb" — you'll incorrectly skip valid characters.

  2. Incorrect else condition:

    • Python syntax error: else (c in t) and ... is invalid.

    • else doesn’t take a condition. You’d need to use elif again.

  3. Unnecessary use of streak_found:

    • You don’t need a streak_found flag. You only need to compare each character to the one immediately before it.

  4. Adding characters that are not duplicates but previously seen:

    • You're preventing characters from being added again even if they're not consecutive — which is incorrect.


✅ Correct Logic Should Be:

Keep the character only if it’s not equal to the previous character.


✔️ Cleaned Up Version of the Algorithm:

python
s = "afafasgggewrtdg" t = "" last_seen_character = None for c in s: if c != last_seen_character: t += c last_seen_character = c print(t)

🧪 Example:

Input: "aabaaabb"
Output: "abab"


⏱️ Time Complexity: O(n)

You visit each character once.

📦 Space Complexity: O(n)

In the worst case, the output string could be the same length as input.


🧠 Final Thoughts:

You're on the right track with tracking characters, but overcomplicating it with multiple conditions and flags. Stick to comparing only the current and previous characters — that’s sufficient here.

Let me know if you want to walk through more test cases or edge conditions!


So The Solution Is:

s = "afafasgggewrtdg"
t = ""
last_seen_character = None

for c in s:
    if c != last_seen_character:
        t += c
    last_seen_character = c

print(t)

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