# 13.01.2025 [3223. Minimum Length of String After Operations]

Length after removing repeatings > 2 #medium

13.01.2025

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Problem TLDR

Length after removing repeatings > 2 #medium

Intuition

The takeaways are always either 1 or 2:

    // 1 -> 1
    // 2 -> 2
    // 3 -> 1
    // 4 -> 2
    // 5 -> 3 -> 1
    // 6 -> 4 -> 2
    // 7 -> 1
    // 8 -> 2

Count each char's frequency.

Approach

• we can do some arithmetics 2 - 2 * f

• careful: only count existing characters

• we can apply bitmasks instead of f[26]

Complexity

• Time complexity: $$O(n)$$

• Space complexity: $$O(1)$$

Code

    fun minimumLength(s: String) =
        s.groupBy { it }.values.sumBy {
            2 - it.size % 2
        }

    pub fn minimum_length(s: String) -> i32 {
        let mut f = vec![0; 26];
        for b in s.bytes() { f[(b - b'a') as usize] += 1 }
        (0..26).filter(|&b| f[b] > 0).map(|b| 2 - f[b] % 2).sum()
    }

    int minimumLength(string s) {
        int e = 0, f = 0;
        for (char c: s)
            f ^= 1 << (c - 'a'), e |= 1 << (c - 'a');
        return 2 * __builtin_popcount(e) - __builtin_popcount(f & e);
    }