How do you solve Moving Stones Until Consecutive II on LeetCode?

Moving Stones Until Consecutive II (LeetCode #1040) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: Understanding the concept of endpoint stones is crucial to solving the problem effectively.

What is the time complexity of Moving Stones Until Consecutive II?

The optimal solution for Moving Stones Until Consecutive II runs in O(n log n) time and uses O(1) space. The time complexity is O(n log n) due to the sorting step, which is efficient for the input size constraints.

What is the brute force solution for Moving Stones Until Consecutive II?

The brute force approach for Moving Stones Until Consecutive II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible moves for the endpoint stones and counting how many moves it takes to reach a state where the stones are in three consecutive positions. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Moving Stones Until Consecutive II?

Moving Stones Until Consecutive II uses the following concepts: Array, Math, Sliding Window, Sorting. The recommended patterns to study are: Sorting, Gap Analysis.

What are the interview tips for Moving Stones Until Consecutive II?

Always clarify the problem statement and constraints before diving into coding. Think about edge cases and how they might affect your solution. Explain your thought process clearly while coding; it helps the interviewer understand your approach.

What are common mistakes when solving Moving Stones Until Consecutive II?

Failing to sort the stones before analyzing gaps. Not considering the edge cases where stones are already in consecutive positions.

#1040

Moving Stones Until Consecutive II

Medium

ArrayMathSliding WindowSortingSortingGap Analysis

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n log n)Space O(1)

The optimal solution leverages sorting and gap analysis to quickly determine the minimum and maximum moves required to achieve three consecutive stones. This method is efficient and avoids unnecessary checks.

⚙️

Algorithm

3 steps

1Step 1: Sort the stones array.
2Step 2: Calculate the minimum moves by checking how many stones are already in consecutive positions and how many moves are needed to fill the gaps.
3Step 3: For maximum moves, consider the largest gaps at the ends of the sorted array and calculate how many stones can be moved into those gaps.

solution.py5 lines

1def numMovesStonesII(stones):
2    stones.sort()
3    min_moves = max(0, 2 - (stones[-1] - stones[0] + 1 - len(stones)))
4    max_moves = max(stones[-1] - stones[-2] - 1, stones[1] - stones[0] - 1)
5    return [min_moves, max_moves]

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step, which is efficient for the input size constraints.

1Understanding the concept of endpoint stones is crucial to solving the problem effectively.
2The relationship between gaps and the number of moves needed to fill them is key to determining both minimum and maximum moves.

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