Delete Characters to Make Fancy String

Delete Characters to Make Fancy String - Problem

String Easy

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.

Input & Output

Example 1 — Basic Case

$ Input: s = "leeetcode"

› Output: "leetcode"

💡 Note: Remove one 'e' from the three consecutive 'e's. The result is "leetcode" with no three consecutive characters equal.

Example 2 — Multiple Groups

$ Input: s = "aaabaaaa"

› Output: "aabaa"

💡 Note: Remove one 'a' from "aaa" at start and two 'a's from "aaaa" at end. Result: "aabaa".

Example 3 — Already Fancy

$ Input: s = "aab"

› Output: "aab"

💡 Note: No three consecutive characters are equal, so no deletion needed.

Constraints

1 ≤ s.length ≤ 10⁵
s consists only of lowercase English letters

Visualization

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Asked in

a Amazon 15 M Microsoft 12

The key insight is to process the string character by character, keeping track of the last two characters added to the result. When we encounter a character that would create three consecutive identical characters, we simply skip it. The optimal approach uses a single pass with O(n) time and O(n) space complexity.

Common Approaches

✓ Brute Force Character by Character

⏱️ Time: O(n) Space: O(n)

Iterate through each character and check if adding it would create three consecutive identical characters. If so, skip it; otherwise, add it to the result.

Stack-like Approach

⏱️ Time: O(n) Space: O(n)

Build the result using a list/array that acts like a stack. For each character, check the last two elements and decide whether to append or skip.

Brute Force Character by Character — Algorithm Steps

Initialize empty result string
For each character, check if it matches the last two characters in result
Skip if it would create three consecutive, otherwise append

Visualization

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Step-by-Step Walkthrough

Check Character

Compare current char with last 2 in result

Decision

Skip if 3 consecutive, otherwise append

Continue

Process next character

Code -

solution.c — C

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

char* solution(char* s) {
    int len = strlen(s);
    char* result = (char*)malloc((len + 1) * sizeof(char));
    int pos = 0;
    
    for (int i = 0; i < len; i++) {
        if (pos >= 2 && result[pos - 1] == s[i] && result[pos - 2] == s[i]) {
            continue;
        }
        result[pos++] = s[i];
    }
    result[pos] = '\0';
    return result;
}

int main() {
    char s[100001];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    
    char* result = solution(s);
    printf("%s\n", result);
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the string with constant time operations

✓ Linear Growth

Space Complexity

O(n)

Result string stores up to n characters

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force Character by Character — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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