LeetCode Entry

3020. Find the Maximum Number of Elements in Subset

27.06.2026 medium 2026 kotlin rust

Powers of two symmetrical exponentially growing longest sequence length

3020. Find the Maximum Number of Elements in Subset medium substack youtube

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Problem TLDR

Powers of two symmetrical exponentially growing longest sequence length

Intuition

    // the base can be anything, not just 2

Compute frequencies. The length of the sequence is very small, just check every number if it has continuation by brute force, do +2 at every step, frequencey should be at least 2.

Approach

• ones 1111 is the corner case, and it can be odd or even

Complexity

• Time complexity: \(O(nlog(log(max)))\)

• Space complexity: \(O(n)\)

Code

    fun maximumLength(n: IntArray) = n.groupBy{1L*it}.run {
        max((get(1)?.size?:0).let{it-1+it%2},
            (keys-1).maxOfOrNull { x ->
                var (n,c) = x to 0
                while ((get(n)?.size?:0)>1) { c += 2; n *= n}
                c + if (n in this) 1 else -1
        }?:1 ) }

    pub fn maximum_length(n: Vec<i32>) -> i32 {
        let mut f = HashMap::new();
        for x in n { *f.entry(x as i64).or_default() += 1 }
        let o = f.remove(&1).unwrap_or(0);
        f.keys().fold(1.max(o - 1 | 1), |r, &k| {
            let (mut x, mut c) = (k, 0);
            while f.get(&x) > Some(&1) { c += 2; x *= x }
            r.max(c + (f.get(&x) > None) as i32 * 2 - 1)
        })
    }