Introduction

In this blog post, we will explore how to solve the LeetCode problem "1033" 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 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.

Explanation

Here's a breakdown of the solution:

• Sort the positions: The initial step is to sort the given positions a, b, and c to easily identify the minimum, middle, and maximum values. This simplifies the logic for calculating moves.

• Calculate minimum moves: The minimum number of moves depends on how close the stones are initially. If the gap between the smallest and middle stone or the middle and largest stone is at most 2, the minimum moves is either 0 or 1. Otherwise it is 2.

• Calculate maximum moves: The maximum number of moves is the sum of the gaps between the stones, reduced by 2 (i.e., (max - mid - 1) + (mid - min - 1)).

• Complexity Analysis:

  • Runtime Complexity: O(1) - Sorting takes constant time since we are only sorting 3 elements.
  • Storage Complexity: O(1) - We are using constant extra space.

Code

    def num_moves_stones(a: int, b: int, c: int) -> list[int]:
    """
    Calculates the minimum and maximum moves to arrange stones in consecutive positions.

    Args:
        a: The position of the first stone.
        b: The position of the second stone.
        c: The position of the third stone.

    Returns:
        A list containing the minimum and maximum number of moves.
    """

    stones = sorted([a, b, c])
    min_val = stones[0]
    mid_val = stones[1]
    max_val = stones[2]

    max_moves = (max_val - mid_val - 1) + (mid_val - min_val - 1)

    if max_val - min_val == 2:
        min_moves = 0
    elif max_val - mid_val <= 2 or mid_val - min_val <= 2:
        min_moves = 1
    else:
        min_moves = 2

    return [min_moves, max_moves]