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Formatted question description: https://leetcode.ca/all/2350.html

# 2350. Shortest Impossible Sequence of Rolls

• Difficulty: Hard.
• Related Topics: Array, Hash Table, Greedy.
• Similar Questions: .

## Problem

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, , , , , 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, , , 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  cannot be taken from rolls, so we return 1.
Note that there are other sequences that cannot be taken from rolls but  is the shortest.


Constraints:

• n == rolls.length

• 1 <= n <= 105

• 1 <= rolls[i] <= k <= 105

## Solution

• class Solution {
public int shortestSequence(int[] rolls, int k) {
BitSet bitSet = new BitSet(k + 1);
int cnt = 0;
int res = 1;
for (int roll : rolls) {
if (!bitSet.get(roll)) {
bitSet.set(roll);
cnt++;
}
if (cnt == k) {
res++;
cnt = 0;
bitSet.clear();
}
}
return res;
}
}

############

class Solution {
public int shortestSequence(int[] rolls, int k) {
Set<Integer> s = new HashSet<>();
int ans = 1;
for (int v : rolls) {
if (s.size() == k) {
s.clear();
++ans;
}
}
return ans;
}
}

• class Solution:
def shortestSequence(self, rolls: List[int], k: int) -> int:
ans = 1
s = set()
for v in rolls:
if len(s) == k:
ans += 1
s.clear()
return ans

############

# 2350. Shortest Impossible Sequence of Rolls
# https://leetcode.com/problems/shortest-impossible-sequence-of-rolls/

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

for x in rolls:

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

return curr


• class Solution {
public:
int shortestSequence(vector<int>& rolls, int k) {
unordered_set<int> s;
int ans = 1;
for (int v : rolls) {
s.insert(v);
if (s.size() == k) {
s.clear();
++ans;
}
}
return ans;
}
};

• func shortestSequence(rolls []int, k int) int {
s := map[int]bool{}
ans := 1
for _, v := range rolls {
s[v] = true
if len(s) == k {
ans++
s = map[int]bool{}
}
}
return ans
}


Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).