How do you solve Shortest Impossible Sequence of Rolls on LeetCode?

Shortest Impossible Sequence of Rolls (LeetCode #2350) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding how to efficiently check for subsequences is crucial.

What is the brute force solution for Shortest Impossible Sequence of Rolls?

The brute force approach for Shortest Impossible Sequence of Rolls is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks all possible sequences of rolls of increasing lengths until we find one that cannot be formed from the given rolls. This is straightforward but inefficient as it involves generating combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Shortest Impossible Sequence of Rolls?

Shortest Impossible Sequence of Rolls uses the following concepts: Array, Hash Table, Greedy. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Shortest Impossible Sequence of Rolls?

Always start with a brute-force solution to understand the problem better. Explain your thought process clearly; it shows your understanding. Practice generating combinations and sequences, as they are common in DSA problems.

What are common mistakes when solving Shortest Impossible Sequence of Rolls?

Failing to account for all possible sequences when checking lengths. Overlooking the need to optimize the solution instead of brute-forcing through combinations.

#2350

Shortest Impossible Sequence of Rolls

Hard

ArrayHash TableGreedyHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach uses a greedy method to determine the shortest impossible sequence by checking the maximum number of unique rolls that can be formed at each length. This is efficient because it avoids generating all combinations.

⚙️

Algorithm

3 steps

1Step 1: Create a frequency map to count occurrences of each number in rolls.
2Step 2: For each length starting from 1, check if we can form all sequences of that length using the frequency map.
3Step 3: If we cannot form a sequence, return the current length.

solution.py11 lines

1def shortest_impossible_sequence(rolls, k):
2    from collections import Counter
3    freq = Counter(rolls)
4    length = 1
5    while True:
6        if length > len(freq):
7            return length
8        length += 1
9
10# Example usage
11print(shortest_impossible_sequence([4,2,1,2,3,3,2,4,1], 4))

ℹ

Complexity note: This complexity is linear because we only traverse the rolls array once to create the frequency map and then check the lengths, which is efficient.

1Understanding how to efficiently check for subsequences is crucial.
2Using frequency maps can significantly reduce the complexity of checking combinations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.