Is Moving Stones Until Consecutive easy or hard?

Moving Stones Until Consecutive is generally considered a medium difficulty problem. The implementation is simple once the gap insight is discovered, but recognizing the special cases for the minimum moves makes it a small brainteaser.

Moving Stones Until Consecutive Python/Java solution

The Python or Java implementation sorts the three inputs and computes two values: the minimum moves using gap conditions and the maximum moves using (c − a − 2). Since the logic is constant time, the code is short and relies mainly on sorting and simple conditional checks.

How to solve Moving Stones Until Consecutive in O(1)?

Sort the three stone positions first. If both gaps (b − a and c − b) equal 1, the stones are already consecutive so the minimum moves is 0. If either gap is ≤ 2, the minimum is 1; otherwise it is 2. The maximum moves always equal (c − a − 2), which counts the empty spaces between the outer stones.

What is the best approach for Moving Stones Until Consecutive?

The optimal approach sorts the three stone positions and analyzes the gaps between them. After sorting a ≤ b ≤ c, compute gap1 = b − a and gap2 = c − b. The minimum moves depend on special cases where a gap equals 1 or 2, while the maximum moves equal (c − a − 2). This approach runs in O(1) time and O(1) space.

Is Moving Stones Until Consecutive asked at Google/Amazon/Meta?

Problems like Moving Stones Until Consecutive appear in interview prep sets for companies that test mathematical reasoning and edge‑case handling. Variants of small simulation or math brainteasers are commonly reported in interviews at companies like Amazon, Google, and Meta.

What data structure is used in Moving Stones Until Consecutive?

No advanced data structure is required. The solution only sorts three integers and performs arithmetic on the gaps between them, making it primarily a math and brainteaser problem.

What is the time complexity of Moving Stones Until Consecutive?

The optimal solution runs in O(1) time because it only sorts three numbers and performs constant arithmetic checks. Space complexity is also O(1) since no additional data structures are required.

Description Hints 1 Solution Video 8:15

1033. Moving Stones Until Consecutive

Medium52.8% acceptance1 company

There are three stones in different positions on the X-axis. You are given three integers a, b, and c, the positions of the stones.

In one move, you pick up a stone at an endpoint (i.e., either the lowest or highest position stone), and move it to an unoccupied position between those endpoints. Formally, let's say the stones are currently at positions x, y, and z with x < y < z. You pick up the stone at either position x or position z, and move that stone to an integer position k, with x < k < z and k != y.

The game ends when you cannot make any more moves (i.e., the stones are in three consecutive positions).

Return an integer array answer of length 2 where:

answer[0] is the minimum number of moves you can play, and
answer[1] is the maximum number of moves you can play.

Example 1:

Input: a = 1, b = 2, c = 5
Output: [1,2]
Explanation: Move the stone from 5 to 3, or move the stone from 5 to 4 to 3.

Example 2:

Input: a = 4, b = 3, c = 2
Output: [0,0]
Explanation: We cannot make any moves.

Example 3:

Input: a = 3, b = 5, c = 1
Output: [1,2]
Explanation: Move the stone from 1 to 4; or move the stone from 1 to 2 to 4.

Constraints:

1 <= a, b, c <= 100
a, b, and c have different values.

Asked at 1 company

Topics

Brainteaser Math

Hints open one at a time

Hint 1

For the minimum: We can always do it in at most 2 moves, by moving one stone next to another, then the third stone next to the other two. When can we do it in 1 move? 0 moves? For the maximum: Every move, the maximum position minus the minimum position must decrease by at least 1.

8:15

Still stuck?Read 2 approaches, or watch a walkthrough.

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1class Solution:
2    def numMovesStones(self, a: int, b: int, c: int) -> List[int]:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

1
2
5

Input

4
3
2

Input

3
5
1

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