LeetCode Entry

1957. Delete Characters to Make Fancy String

21.07.2025 easy 2025 kotlin rust

No three chars repeats

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Problem TLDR

No three chars repeats #easy

Intuition

Scan, count, filter.

Approach

  • make separate decisions for counter and for appending
  • there are Regex way, chunks way, dedup way
  • leetcode has itertools available

Complexity

  • Time complexity: \(O(n)\)

  • Space complexity: \(O(n)\)

Code


// 193ms
    fun makeFancyString(s: String) = s
    .replace(Regex("(.)\\1{2,}"), "$1$1")



// 24ms
    fun makeFancyString(s: String) = s
    .filterIndexed { i, c -> i < 2 || s[i - 2] != c || s[i - 1] != c }



// 17ms
    fun makeFancyString(s: String) = buildString {
        var p = '.'; var pc = 0
        for (c in s) {
            if (c == p) ++pc else pc = 1
            if (pc < 3) append(c)
            p = c
        }
    }



// 53ms
    pub fn make_fancy_string(mut s: String) -> String {
        s.chars().dedup_with_count().into_iter()
        .map(|(cnt, c)| c.to_string().repeat(2.min(cnt))).collect()
    }



// 11ms
    pub fn make_fancy_string(s: String) -> String {
        String::from_utf8_lossy(
        &s.bytes().enumerate().filter(|&(i, b)|
        i < 2 || s.as_bytes()[i - 2] != b || s.as_bytes()[i - 1] != b)
        .map(|(i, b)| b).collect::<Vec<_>>()).into()
    }



// 5ms
    pub fn make_fancy_string(mut s: String) -> String {
        let (mut a, mut b) = ('.', '.');
        s.retain(|c| { let r = a != c || b != c; a = b; b = c; r }); s
    }



// 2826ms
    string makeFancyString(string s) {
        return regex_replace(s, regex("(.)\\1\\1+"), "$1$1");
    }