How do you solve Moving Stones Until Consecutive on LeetCode?

Moving Stones Until Consecutive (LeetCode #1033) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) time complexity and O(1) space complexity. Key insight: The maximum number of moves is determined by the total distance between the stones minus 2.

What is the time complexity of Moving Stones Until Consecutive?

The optimal solution for Moving Stones Until Consecutive runs in O(1) time and uses O(1) space. The complexity remains constant as we perform a fixed number of operations regardless of the input size.

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

The brute force approach for Moving Stones Until Consecutive is Brute Force, with O(1) time complexity and O(1) space complexity. The brute-force approach involves checking all possible moves that can be made until the stones are in consecutive positions. This is straightforward but inefficient due to the potential number of moves. The optimal Optimal Solution approach improves this to O(1).

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

Moving Stones Until Consecutive uses the following concepts: Math, Brainteaser. The recommended patterns to study are: Sorting, Greedy Algorithms.

What are the interview tips for Moving Stones Until Consecutive?

Always consider edge cases, such as when stones are already consecutive. Explain your thought process clearly while solving the problem. Practice similar problems to recognize patterns and improve problem-solving speed.

What are common mistakes when solving Moving Stones Until Consecutive?

Failing to sort the positions before calculating gaps. Misunderstanding the conditions for minimum moves, especially when stones are already close.

#1033

Moving Stones Until Consecutive

Medium

MathBrainteaserSortingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(1)Space O(1)

The optimal solution leverages the gaps between the stones to determine the minimum and maximum moves efficiently without unnecessary iterations.

⚙️

Algorithm

4 steps

1Step 1: Sort the positions of the stones.
2Step 2: Calculate the gaps between the stones.
3Step 3: Determine the minimum moves based on the gaps.
4Step 4: Calculate the maximum moves based on the total distance.

solution.py11 lines

1def numMovesStones(a, b, c):
2    positions = sorted([a, b, c])
3    min_moves = 0
4    max_moves = positions[2] - positions[0] - 2
5    if max_moves == 0:
6        return [0, 0]
7    if positions[1] - positions[0] <= 2 or positions[2] - positions[1] <= 2:
8        min_moves = 1
9    else:
10        min_moves = 2
11    return [min_moves, max_moves]

ℹ

Complexity note: The complexity remains constant as we perform a fixed number of operations regardless of the input size.

1The maximum number of moves is determined by the total distance between the stones minus 2.
2The minimum number of moves depends on the gaps between the stones and can be as low as 0 or 1.

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