LeetCode in Kotlin

2350. Shortest Impossible Sequence of Rolls

Hard

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 ith 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

import java.util.BitSet

class Solution {
    fun shortestSequence(rolls: IntArray, k: Int): Int {
        val bitSet = BitSet(k + 1)
        var cnt = 0
        var res = 1
        for (roll in rolls) {
            if (!bitSet[roll]) {
                bitSet.set(roll)
                cnt++
            }
            if (cnt == k) {
                res++
                cnt = 0
                bitSet.clear()
            }
        }
        return res
    }
}

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