Introduction

In this blog post, we will explore how to solve the LeetCode problem "2087" 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 is an m x n grid, where (0, 0) is the top-left cell and (m - 1, n - 1) is the bottom-right cell. You are given an integer array startPos where startPos = [startrow, startcol] indicates that initially, a robot is at the cell (startrow, startcol). You are also given an integer array homePos where homePos = [homerow, homecol] indicates that its home is at the cell (homerow, homecol). The robot needs to go to its home. It can move one cell in four directions: left, right, up, or down, and it can not move outside the boundary. Every move incurs some cost. You are further given two 0-indexed integer arrays: rowCosts of length m and colCosts of length n. If the robot moves up or down into a cell whose row is r, then this move costs rowCosts[r]. If the robot moves left or right into a cell whose column is c, then this move costs colCosts[c]. Return the minimum total cost for this robot to return home. Example 1: Input: startPos = [1, 0], homePos = [2, 3], rowCosts = [5, 4, 3], colCosts = [8, 2, 6, 7] Output: 18 Explanation: One optimal path is that: Starting from (1, 0) -> It goes down to (2, 0). This move costs rowCosts[2] = 3. -> It goes right to (2, 1). This move costs colCosts[1] = 2. -> It goes right to (2, 2). This move costs colCosts[2] = 6. -> It goes right to (2, 3). This move costs colCosts[3] = 7. The total cost is 3 + 2 + 6 + 7 = 18 Example 2: Input: startPos = [0, 0], homePos = [0, 0], rowCosts = [5], colCosts = [26] Output: 0 Explanation: The robot is already at its home. Since no moves occur, the total cost is 0. Constraints: m == rowCosts.length n == colCosts.length 1 <= m, n <= 105 0 <= rowCosts[r], colCosts[c] <= 104 startPos.length == 2 homePos.length == 2 0 <= startrow, homerow < m 0 <= startcol, homecol < n

Explanation

Here's the solution to the problem, incorporating the requested explanations and format:

• High-Level Approach:

  • Calculate the cost of moving along the rows from the start row to the home row.
  • Calculate the cost of moving along the columns from the start column to the home column.
  • Return the sum of these two costs.

• Complexity:

  • Runtime Complexity: O(m + n), where m is the number of rows and n is the number of columns.
  • Storage Complexity: O(1) - The solution uses a constant amount of extra space.

Code

    def minCost(startPos, homePos, rowCosts, colCosts):
    """
    Calculates the minimum total cost for a robot to return home.

    Args:
        startPos (list): The starting position of the robot [startrow, startcol].
        homePos (list): The home position of the robot [homerow, homecol].
        rowCosts (list): The cost of moving into each row.
        colCosts (list): The cost of moving into each column.

    Returns:
        int: The minimum total cost for the robot to return home.
    """

    startRow, startCol = startPos
    homeRow, homeCol = homePos
    rowCost = 0
    colCost = 0

    # Calculate row cost
    if startRow < homeRow:
        for i in range(startRow + 1, homeRow + 1):
            rowCost += rowCosts[i]
    elif startRow > homeRow:
        for i in range(startRow - 1, homeRow - 1, -1):
            rowCost += rowCosts[i]

    # Calculate column cost
    if startCol < homeCol:
        for i in range(startCol + 1, homeCol + 1):
            colCost += colCosts[i]
    elif startCol > homeCol:
        for i in range(startCol - 1, homeCol - 1, -1):
            colCost += colCosts[i]

    return rowCost + colCost