LeetCode 2350 - Shortest Impossible Sequence of Rolls

Difficulty: 🔴 Hard
Topics: Array, Hash Table, Greedy

Solution

Here is a complete, detailed technical solution guide for LeetCode 2350 - Shortest Impossible Sequence of Rolls, following your requested format precisely.

Problem Understanding

The problem asks us to determine the length of the shortest sequence of dice rolls that cannot be found as a subsequence within a given list of dice rolls. We are given an array rolls of length n representing the outcomes of rolling a k-sided die n times, and an integer k representing the number of sides of the die.

In simpler terms, imagine all possible sequences of dice rolls of increasing length. We want to find the first length len for which there exists at least one sequence of rolls of length len that does not appear anywhere in rolls. That length is the answer.

The constraints indicate that n and k can be as large as 10^5. Therefore, generating all possible sequences of rolls is not feasible, because the number of sequences grows exponentially with length (k^len). This requires a clever approach that does not rely on brute-force enumeration.

Important edge cases include:

When rolls does not contain every number from 1 to k, the answer is 1, because a sequence of length 1 is already impossible.
When rolls contains multiple occurrences of each number, the algorithm should count how many complete sets of 1..k can be formed and derive the minimum impossible length from that.
Large k values relative to n may limit the number of complete sequences that can appear.

Approaches

Brute-Force Approach

A naive approach would attempt to generate all sequences of length 1, then 2, and so on, and check whether each sequence appears as a subsequence in rolls. For each length len, we would need to enumerate all k^len sequences and verify their presence. While this would eventually give the correct answer, it is computationally infeasible for large k and n because the number of sequences grows exponentially.

Optimal Approach

The key insight is that the shortest impossible sequence length is determined by how many complete sets of 1..k appear in rolls. Every time we see all numbers from 1 to k at least once, we know that all sequences of the current length exist. Once we cannot form a complete set, the next sequence length becomes impossible.

This means we can iterate through rolls while maintaining a set of numbers seen in the current incomplete set. Each time the set reaches size k, we know one full set of 1..k exists, so we reset the set and increment the count of complete sets seen. At the end, the answer is the number of complete sets plus 1.

This approach is linear in n and uses at most O(k) extra space.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(k^len * n)	O(k^len)	Enumerate all sequences, infeasible for large inputs
Optimal	O(n)	O(k)	Track sets of numbers in rolls to find complete sequences

Algorithm Walkthrough

Initialize an empty set current_set to track which numbers from 1..k have been seen in the current incomplete set.
Initialize a counter complete_sets to zero, representing the number of full sets of 1..k seen.
Iterate over each roll r in rolls:
Add r to current_set.
If current_set contains all k numbers, increment complete_sets and reset current_set to empty.
After processing all rolls, the length of the shortest impossible sequence is complete_sets + 1.

Why it works: Each complete set guarantees that all sequences of the current length exist. Once we cannot form another complete set, a sequence of length complete_sets + 1 is impossible. This invariant ensures correctness without enumerating sequences.

Python Solution

from typing import List

class Solution:
    def shortestSequence(self, rolls: List[int], k: int) -> int:
        current_set = set()
        complete_sets = 0
        
        for r in rolls:
            current_set.add(r)
            if len(current_set) == k:
                complete_sets += 1
                current_set.clear()
        
        return complete_sets + 1

Explanation: The current_set tracks unique numbers until we see all numbers 1..k. When a full set is detected, we increment complete_sets and reset the set. At the end, the answer is complete_sets + 1, representing the first sequence length that cannot be formed.

Go Solution

func shortestSequence(rolls []int, k int) int {
    currentSet := make(map[int]bool)
    completeSets := 0
    
    for _, r := range rolls {
        currentSet[r] = true
        if len(currentSet) == k {
            completeSets++
            currentSet = make(map[int]bool)
        }
    }
    
    return completeSets + 1
}

Explanation: The Go solution mirrors the Python logic, using a map to simulate a set. Each time the map reaches size k, we increment completeSets and reset the map.

Worked Examples

Example 1

rolls = [4,2,1,2,3,3,2,4,1], k = 4

Step through rolls:

roll	current_set	complete_sets
4	{4}	0
2	{2,4}	0
1	{1,2,4}	0
2	{1,2,4}	0
3	{1,2,3,4}	1
3	{ }	1
2	{2}	1
4	{2,4}	1
1	{1,2,4}	1

Return complete_sets + 1 = 2 + 1 = 3.

Example 2

rolls = [1,1,2,2], k = 2

roll	current_set	complete_sets
1	{1}	0
1	{1}	0
2	{1,2}	1
2	{ }	1

Return complete_sets + 1 = 2.

Example 3

rolls = [1,1,3,2,2,2,3,3], k = 4

We never see 4, so no complete set is formed. Return 1.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	We iterate through `rolls` once, adding elements to a set and checking its size in O(1) amortized time.
Space	O(k)	The set `current_set` stores at most `k` unique elements at a time.

The solution is efficient for n, k <= 10^5.

Test Cases

# Provided examples
assert Solution().shortestSequence([4,2,1,2,3,3,2,4,1], 4) == 3  # Example 1
assert Solution().shortestSequence([1,1,2,2], 2) == 2            # Example 2
assert Solution().shortestSequence([1,1,3,2,2,2,3,3], 4) == 1    # Example 3

# Edge cases
assert Solution().shortestSequence([1,2,3], 5) == 1               # Missing numbers 4,5
assert Solution().shortestSequence([1]*100000, 1) == 2           # All same number, k=1
assert Solution().shortestSequence(list(range(1, 100001)), 100000) == 2 # Each number once, largest k
assert Solution().shortestSequence([2,3,1,2,1,3,2,1,3], 3) == 2  # Multiple full sets of 1..3

Test	Why
Example 1	Standard case with multiple complete sets
Example 2	Small `k` and repeated numbers
Example 3	Missing number in rolls
Missing numbers	Sequence length 1 impossible
All same number	Only one number appears repeatedly
Large k	Tests algorithm on very large `k`
Multiple sets	Confirms algorithm correctly counts multiple full sets

Edge Cases

Missing numbers entirely: If some numbers in 1..k never appear in rolls, the answer is immediately 1. This could trip naive implementations that assume every number appears at least once. The implementation handles