# 08.02.2025 [2349. Design a Number Container System]

Smallest running index of number in map #medium #treeset

08.02.2025

2349. Design a Number Container System medium blog post substack youtube

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Problem TLDR

Smallest running index of number in map #medium #treeset

Intuition

To keep all indices for the number in a sorted order use a TreeSet. Store (index, number) map to remove the old number from index.

Approach

• one small optimization is to remove old number lazily: keep removing if m[find(n)] != n

Complexity

• Time complexity: $$O(nlog(n))$$ for all operation, log(n) for change, O(1) for find, reverse for lazy.

• Space complexity: $$O(n)$$ indices & numbers are never erased

Code

class NumberContainers() {
    val iin = HashMap<Int, Int>()
    val nii = HashMap<Int, TreeSet<Int>>()
    fun change(i: Int, n: Int) {
        iin[i]?.let { nii[it]!! -= i }; iin[i] = n
        nii.getOrPut(n, ::TreeSet) += i
    }
    fun find(n: Int) = nii[n]?.firstOrNull() ?: -1
}

#[derive(Default)] struct NumberContainers(HashMap<i32, i32>, HashMap<i32, BTreeSet<i32>>);
impl NumberContainers {
    fn new() -> Self { Self::default() }
    fn change(&mut self, i: i32, n: i32) {
        self.0.insert(i, n).inspect(|j| { self.1.get_mut(&j).unwrap().remove(&i);});
        self.1.entry(n).or_default().insert(i);
    }
    fn find(&self, n: i32) -> i32 
        { *self.1.get(&n).and_then(|s| s.first()).unwrap_or(&-1) }
}

class NumberContainers {
    unordered_map<int, int> in; map<int, set<int>> ni;
public:
    void change(int i, int n) {
        if (in.count(i)) ni[in[i]].erase(i);
        in[i] = n, ni[n].insert(i);
    }
    int find(int n) { return size(ni[n]) ? *begin(ni[n]) : -1; }
};