How do you solve Minimum Cost Homecoming of a Robot in a Grid on LeetCode?

Minimum Cost Homecoming of a Robot in a Grid (LeetCode #2087) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The robot must traverse all rows and columns between start and home positions, making the path choice irrelevant.

What is the brute force solution for Minimum Cost Homecoming of a Robot in a Grid?

The brute force approach for Minimum Cost Homecoming of a Robot in a Grid is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we will simulate every possible path the robot can take to reach its home position. This involves calculating the cost for each possible route and selecting the minimum cost. While this method guarantees finding the correct answer, it can be inefficient for larger grids. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Cost Homecoming of a Robot in a Grid?

Minimum Cost Homecoming of a Robot in a Grid uses the following concepts: Array, Greedy. The recommended patterns to study are: Greedy, Dynamic Programming.

What are the interview tips for Minimum Cost Homecoming of a Robot in a Grid?

Always consider the simplest approach first before optimizing. Look for patterns in the problem that can simplify your calculations. Communicate your thought process clearly, especially when transitioning from brute force to optimal solutions.

What are common mistakes when solving Minimum Cost Homecoming of a Robot in a Grid?

Students often try to implement complex pathfinding algorithms instead of recognizing the direct cost calculation. Forgetting to include the costs of the starting row and column can lead to incorrect results.

#2087

Minimum Cost Homecoming of a Robot in a Grid

Medium

ArrayGreedyGreedyDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal solution leverages the observation that the robot must traverse all rows and columns between its starting position and home position. By calculating the total cost based on the row and column costs directly, we can avoid unnecessary path exploration and achieve a more efficient solution.

⚙️

Algorithm

3 steps

1Step 1: Calculate the total cost for all rows between startRow and homeRow by summing the rowCosts.
2Step 2: Calculate the total cost for all columns between startCol and homeCol by summing the colCosts.
3Step 3: Return the total cost as the minimum cost to reach home.

solution.py7 lines

1def minCost(startPos, homePos, rowCosts, colCosts):
2    total_cost = 0
3    for r in range(min(startPos[0], homePos[0]), max(startPos[0], homePos[0]) + 1):
4        total_cost += rowCosts[r]
5    for c in range(min(startPos[1], homePos[1]), max(startPos[1], homePos[1]) + 1):
6        total_cost += colCosts[c]
7    return total_cost

ℹ

Complexity note: The time complexity is O(n) because we are only iterating through the rows and columns between the starting and home positions, which is linear in relation to the distance traveled.

1The robot must traverse all rows and columns between start and home positions, making the path choice irrelevant.
2Summing costs directly based on the range of rows and columns is more efficient than exploring all paths.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.