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Moving Stones Until Consecutive II

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.

  • In particular, if the stones are at say, stones = [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).

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: stones = [7,4,9]
Output: [1,2]
Explanation: We can move 4 -> 8 for one move to finish the game.
Or, we can move 9 -> 5, 4 -> 6 for two moves to finish the game.

Example 2:

Input: stones = [6,5,4,3,10]
Output: [2,3]
Explanation: We can move 3 -> 8 then 10 -> 7 to finish the game.
Or, we can move 3 -> 7, 4 -> 8, 5 -> 9 to finish the game.
Notice we cannot move 10 -> 2 to finish the game, because that would be an illegal move.

 

Constraints:

  • 3 <= stones.length <= 104
  • 1 <= stones[i] <= 109
  • All the values of stones are unique.

Solution Explanation for Moving Stones Until Consecutive II

This problem involves finding the minimum and maximum number of moves required to make a set of stones on the x-axis consecutive. A move consists of taking an endpoint stone (the stone with the smallest or largest position) and moving it to an unoccupied position such that it's no longer an endpoint stone. The game ends when the stones occupy three consecutive positions.

Understanding the Problem

The key is to realize that the minimum number of moves depends on finding the longest consecutive sequence within the sorted stones and filling the gaps. The maximum number of moves arises from strategically moving stones to maximize the number of moves needed before achieving consecutiveness.

Approach

  1. Sort the Stones: Sort the input stones array to easily identify consecutive sequences and endpoints.

  2. Calculate Maximum Moves: The maximum number of moves occurs when the stones are spread out as much as possible. Consider the sorted stones. The maximum move is the larger of (max(stones) - min(stones) + 1) - (n-1) and (stones[n-2] - stones[0] + 1) - (n-1). This essentially calculates the difference between the maximum possible length of the consecutive sequence and the actual length, and uses the second largest to determine the difference from the smallest stone. It subtracts (n-1) because a consecutive sequence of n stones needs n-1 intervals.

  3. Calculate Minimum Moves: This is more complex. We use a two-pointer sliding window approach:

    • i and j are the pointers defining the window.
    • The window expands until the length of the subarray ( j-i+1) plus the gaps exceeds the total number of stones n.
    • If the window size is n-1 and the difference between the largest and smallest stone in the window is n-2, then it means only two moves are needed (minimum case).
    • Otherwise, the minimum moves are calculated as n - (j - i + 1), representing the number of stones to move to achieve consecutiveness given a consecutive subarray determined by the window.

Time Complexity Analysis

  • Sorting the stones takes O(n log n) time.
  • The two-pointer approach iterates through the sorted array once, taking O(n) time.
  • Calculating the maximum moves is O(1).

Therefore, the overall time complexity is dominated by the sorting step, resulting in O(n log n).

Space Complexity Analysis

The space complexity is O(1) (constant) because we are using a small number of variables to store intermediate results, irrespective of the input size.

Code Implementation (Python)

def numMovesStonesII(stones):
    stones.sort()
    n = len(stones)
    min_moves = n  # Initialize with maximum possible moves
    max_moves = max(stones[-1] - stones[1] + 1, stones[-2] - stones[0] + 1) - (n - 1)
    i = 0
    for j in range(n):
        while stones[j] - stones[i] + 1 > n:
            i += 1
        if j - i + 1 == n - 1 and stones[j] - stones[i] == n - 2:
            min_moves = min(min_moves, 2)
        else:
            min_moves = min(min_moves, n - (j - i + 1))
    return [min_moves, max_moves]

The code in other languages (Java, C++, Go, TypeScript) follows a very similar structure and logic, with only syntactic differences to reflect the language's specific features. The core algorithm remains consistent across all implementations.