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Leetcode #2350: Shortest Impossible Sequence of Rolls

In this guide, we solve Leetcode #2350 Shortest Impossible Sequence of Rolls in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

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].

Quick Facts

  • Difficulty: Hard
  • Premium: No
  • Tags: Greedy, Array, Hash Table

Intuition

Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.

By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.

Approach

Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.

This keeps the solution linear while remaining easy to explain in an interview setting.

Steps:

  • Initialize a hash map for seen items or counts.
  • Iterate through the input, querying/updating the map.
  • Return the first valid result or the final computed value.

Example

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.

Python Solution

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

Complexity

The time complexity is O(n). The space complexity is O(n).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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