{ "data": { "question": { "questionId": "1113", "questionFrontendId": "1040", "boundTopicId": null, "title": "Moving Stones Until Consecutive II", "titleSlug": "moving-stones-until-consecutive-ii", "content": "
There are some stones in different positions on the X-axis. You are given an integer array stones
, the positions of the stones.
Call a stone an endpoint stone if it has the smallest or largest position. In one move, you pick up an endpoint stone and move it to an unoccupied position so that it is no longer an endpoint stone.
\n\nstones = [1,2,5]
, you cannot move the endpoint stone at position 5
, since moving it to any position (such as 0
, or 3
) will still keep that stone as an endpoint stone.The game ends when you cannot make any more moves (i.e., the stones are in three consecutive positions).
\n\nReturn an integer array answer
of length 2
where:
answer[0]
is the minimum number of moves you can play, andanswer[1]
is the maximum number of moves you can play.\n
Example 1:
\n\n\nInput: stones = [7,4,9]\nOutput: [1,2]\nExplanation: We can move 4 -> 8 for one move to finish the game.\nOr, we can move 9 -> 5, 4 -> 6 for two moves to finish the game.\n\n\n
Example 2:
\n\n\nInput: stones = [6,5,4,3,10]\nOutput: [2,3]\nExplanation: We can move 3 -> 8 then 10 -> 7 to finish the game.\nOr, we can move 3 -> 7, 4 -> 8, 5 -> 9 to finish the game.\nNotice we cannot move 10 -> 2 to finish the game, because that would be an illegal move.\n\n\n
\n
Constraints:
\n\n3 <= stones.length <= 104
1 <= stones[i] <= 109
stones
are unique.Compiled with clang 11
using the latest C++ 20 standard.
Your code is compiled with level two optimization (-O2
). AddressSanitizer is also enabled to help detect out-of-bounds and use-after-free bugs.
Most standard library headers are already included automatically for your convenience.
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Python 2.7.12
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Node.js 16.13.2
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