Delete Characters to Make Fancy String - Practice Coding Problems

Delete Characters to Make Fancy String - Problem

String Easy

Make Your String Fancy!

A fancy string is one where no three consecutive characters are the same. Your mission is to transform any given string into a fancy string by removing the minimum possible number of characters.

Given a string s, you need to:
• Remove characters strategically to ensure no three consecutive characters are identical
• Minimize the number of deletions
• Return the resulting fancy string

Example: The string "aaabaaaa" becomes "aabaa" by removing 3 characters.

The solution is guaranteed to be unique - there's only one optimal way to make the string fancy!

Input & Output

example_1.py — Basic Case

$ Input: s = "leeetcode"

› Output: "leetcode"

💡 Note: Remove one 'e' from the triple 'eee' to make it fancy. The string becomes "leetcode" with no three consecutive identical characters.

example_2.py — Multiple Triplets

$ Input: s = "aaabaaaa"

› Output: "aabaa"

💡 Note: Remove one 'a' from "aaa" and two 'a's from "aaaa" at the end. This gives us "aabaa" which has no three consecutive identical characters.

example_3.py — Already Fancy

$ Input: s = "aab"

› Output: "aab"

💡 Note: The string is already fancy (no three consecutive identical characters), so no deletions are needed.

Constraints

1 ≤ s.length ≤ 10⁵
s consists only of lowercase English letters
The answer is guaranteed to be unique

Visualization

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Understanding the Visualization

Guests Arrive

Process each guest (character) in arrival order

Check Seating

Look at the last two seated guests

Seat or Wait

If the last two have the same name as the current guest, ask them to wait (skip)

Maintain Fancy

This ensures no three consecutive guests ever have the same name

Key Takeaway

🎯 Key Insight: The greedy approach works because removing the third occurrence of any consecutive triplet is always optimal - it preserves the maximum number of characters while ensuring the fancy property.

Asked in

a Amazon 45 G Google 35 ⊞ Microsoft 25 f Meta 20

The optimal solution uses a greedy one-pass approach with O(n) time complexity. Simply iterate through the string and add each character to the result unless it would create three consecutive identical characters. This works because removing the third occurrence in any triplet gives the minimum deletions needed.

Common Approaches

Approach	Time	Space	Notes
✓ One-Pass Greedy (Optimal)	O(n)	O(1)	Build result string character by character, skipping when three consecutive are same
Brute Force (Generate All Subsequences)	O(2^n * n)	O(2^n)	Generate all possible subsequences and find the longest valid fancy string

One-Pass Greedy (Optimal) — Algorithm Steps

Initialize an empty result string
Iterate through each character in the input string
Add character to result if it won't create three consecutive identical chars
Skip character if adding it would violate the fancy string rule
Return the final result string

Visualization

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Step-by-Step Walkthrough

Process Character

Look at current character from input

Check Last Two

Check if last two characters in result are same as current

Add or Skip

Add to result if safe, skip if it would create triplet

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char* makeFancyString(char* s) {
    int len = strlen(s);
    if (len < 3) {
        char* result = malloc(len + 1);
        strcpy(result, s);
        return result;
    }
    
    char* result = malloc(len + 1);
    int resultLen = 0;
    
    for (int i = 0; i < len; i++) {
        // Add character if it won't create three consecutive identical chars
        if (resultLen < 2 || !(result[resultLen-1] == result[resultLen-2] && result[resultLen-2] == s[i])) {
            result[resultLen++] = s[i];
        }
    }
    
    result[resultLen] = '\0';
    return result;
}

int main() {
    char s[100001];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    
    // Remove quotes
    char* start = s;
    if (s[0] == '"') start++;
    if (start[strlen(start)-1] == '"') start[strlen(start)-1] = '\0';
    
    char* result = makeFancyString(start);
    printf("\"%s\"", result);
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the input string, constant time operations per character

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra space (excluding the output string which is required)

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

One-Pass Greedy (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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