2350. Shortest Impossible Sequence of Rolls

Description

You are given an integer array rolls of length n and an integer k. You roll a k sided dice numbered from 1 to k, n times, where the result of the i^th roll is rolls[i].

Return the length of the shortest sequence of rolls that cannot be taken from rolls.

A sequence of rolls of length len is the result of rolling a k sided dice len times.

Note that the sequence taken does not have to be consecutive as long as it is in order.

Example 1:

Input: rolls = [4,2,1,2,3,3,2,4,1], k = 4
Output: 3
Explanation: Every sequence of rolls of length 1, [1], [2], [3], [4], can be taken from rolls.
Every sequence of rolls of length 2, [1, 1], [1, 2], ..., [4, 4], can be taken from rolls.
The sequence [1, 4, 2] cannot be taken from rolls, so we return 3.
Note that there are other sequences that cannot be taken from rolls.

Example 2:

Input: rolls = [1,1,2,2], k = 2
Output: 2
Explanation: Every sequence of rolls of length 1, [1], [2], can be taken from rolls.
The sequence [2, 1] cannot be taken from rolls, so we return 2.
Note that there are other sequences that cannot be taken from rolls but [2, 1] is the shortest.

Example 3:

Input: rolls = [1,1,3,2,2,2,3,3], k = 4
Output: 1
Explanation: The sequence [4] cannot be taken from rolls, so we return 1.
Note that there are other sequences that cannot be taken from rolls but [4] is the shortest.

Constraints:

n == rolls.length
1 <= n <= 10⁵
1 <= rolls[i] <= k <= 10⁵

Solution

shortest-impossible-sequence-of-rolls.py

class Solution:
    def shortestSequence(self, rolls: List[int], k: int) -> int:
        seen = set()
        curr = 1

        for x in rolls:
            seen.add(x)

            if len(seen) == k:
                curr += 1
                seen.clear()

        return curr