Introduction

In this blog post, we will explore how to solve the LeetCode problem "1040" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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.

Explanation

Here's the breakdown of the approach and the Python code:

• Sort the Stones: Sorting allows us to easily identify endpoints and gaps between stones.
• Calculate Maximum Moves: The maximum moves are determined by the total gaps at either end of the sorted array that need to be filled to achieve consecutive positions.

• Calculate Minimum Moves: The minimum moves involve finding the longest consecutive sequence within the stones. We minimize moves by sliding a window and looking for the densest packing within the window.

• Runtime Complexity: O(n log n) due to sorting.

• Storage Complexity: O(1) excluding the input array.

Code

    def numMovesStones(stones):
    stones.sort()
    n = len(stones)

    max_moves = max(stones[n - 1] - stones[1] - n + 2, stones[n - 2] - stones[0] - n + 2)

    min_moves = n
    j = 0
    for i in range(n):
        while stones[j] - stones[i] + 1 <= n:
            j += 1
        length = j - i - 1
        if length == n - 2 and stones[j - 1] - stones[i] + 1 == n - 1:
            min_moves = min(min_moves, 2)
        else:
            min_moves = min(min_moves, n - length - 1)

    return [min_moves, max_moves]