LeetCode Entry

3289. The Two Sneaky Numbers of Digitville

31.10.2025 easy 2025 kotlin rust

Two extra numbers from 0..n

3289. The Two Sneaky Numbers of Digitville easy blog post substack youtube

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Problem TLDR

Two extra numbers from 0..n #easy

Intuition

Use any of:

• HashSet for visited
• bitmask for visited
• array itself for visited

The clever solution with bit manipulation:

• total xor, no extras: a^b^c – can compute as x1
• total xor for single extra: a^b^c^a
• total xor for two extras: a^b^c^a^b – can compute as x2
• xor of x1^x2: a^b^c ^ a^b^c^a^b = a^b – can compute as x1^x2

Now we have a^b, each bits is a different between a and b.

Split all given numbers by have or have-nots of this bit. xor(have_bit) = xx1 xor(have_not_bit) == xx2

Then split range numbers similarly: xor(have_bit) == yy1 xor(have_not_bit) == yy2

Then a = xx1 ^ yy1, b = xx2 ^ yy2

Approach

• just brute-force

Complexity

• Time complexity: \(O()\)

• Space complexity: \(O()\)

Code

// 17ms
    fun getSneakyNumbers(n: IntArray) =
        n.indices.filter { x -> n.count { it == x } > 1 }

// 0ms
    pub fn get_sneaky_numbers(n: Vec<i32>) -> Vec<i32> {
        let mut m = 0u128;
        n.into_iter().filter(|x| { let u = (1 << x) & m > 0; m |= 1<<x; u}).collect()
    }