What is the time complexity of Minimum Cost Homecoming of a Robot in a Grid?

The optimal solution has a time complexity of O(m+n).

What data structures are used to solve Minimum Cost Homecoming of a Robot in a Grid?

This problem uses Array, Greedy. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Cost Homecoming of a Robot in a Grid?

Insight 1: The problem can be solved greedily because the cost is fixed for each row and column, and we want to minimize the total cost. Therefore, the optimal path will always be moving along the rows and columns directly towards the destination, without unnecessary backtracking or detours. Insight 2: The solution involves iterating through the rows and columns between the start and home positions, summing the corresponding costs in the `rowCosts` and `colCosts` arrays.

Which companies ask LeetCode 2087. Minimum Cost Homecoming of a Robot in a Grid in interviews?

This problem is frequently asked at Goldman Sachs, HP.

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Minimum Cost Homecoming of a Robot in a Grid - LeetCode 2087 Solution

Minimum Cost Homecoming of a Robot in a Grid - Complete Solution Guide

Minimum Cost Homecoming of a Robot in a Grid is LeetCode problem 2087, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 = [start row , start col ] indicates that initially , a robot is at the cell (start row , start col ) . You are also given an integer array homePos where homePos = [home row , home col ] indicates that its home is at the cell (home row , home col ) . The robot needs to go to its home. It can move one cell in four directions: left , right ,

Detailed Explanation

The problem asks us to find the minimum cost for a robot to travel from its starting position to its home position in a grid. The robot can only move up, down, left, or right one cell at a time. Moving up or down incurs a cost based on the `rowCosts` array, where `rowCosts[r]` is the cost of moving into row `r`. Moving left or right incurs a cost based on the `colCosts` array, where `colCosts[c]` is the cost of moving into column `c`. We need to calculate the total cost of the optimal path.

Solution Approach

The solution calculates the cost by iterating from the starting row to the home row and from the starting column to the home column. The row and column indices are incremented or decremented based on whether the home row/column is greater or smaller than the start row/column. In each iteration, the cost of moving to the new row or column is added to the total cost.

Step-by-Step Algorithm

Step 1: Initialize the starting row `sr`, starting column `sc`, home row `hr`, and home column `hc` with the provided input.
Step 2: Initialize a variable `cost` to 0 to store the total cost.
Step 3: Iterate from `sr` to `hr`. If `sr` is less than `hr`, increment `sr` by 1 in each iteration. Otherwise, decrement `sr` by 1 in each iteration. Add the cost of moving to the new row `rowCosts[sr]` to the `cost` variable.
Step 4: Iterate from `sc` to `hc`. If `sc` is less than `hc`, increment `sc` by 1 in each iteration. Otherwise, decrement `sc` by 1 in each iteration. Add the cost of moving to the new column `colCosts[sc]` to the `cost` variable.
Step 5: Return the `cost` variable, which represents the minimum total cost.

Key Insights

Insight 1: The problem can be solved greedily because the cost is fixed for each row and column, and we want to minimize the total cost. Therefore, the optimal path will always be moving along the rows and columns directly towards the destination, without unnecessary backtracking or detours.
Insight 2: The solution involves iterating through the rows and columns between the start and home positions, summing the corresponding costs in the `rowCosts` and `colCosts` arrays.
Insight 3: An edge case to consider is when the start and home positions are the same. In this case, the cost is zero, as no movement is required.

Complexity Analysis

Time Complexity: O(m+n)

Space Complexity: O(1)

Topics

This problem involves: Array, Greedy.

Companies

Asked at: Goldman Sachs, HP.

Minimum Cost Homecoming of a Robot in a Grid | LeetCode 2087

Medium

Array Greedy

Time: O(m+n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int minCost(int[] startPos, int[] homePos, int[] rowCosts, int[] colCosts) {
        int sr = startPos[0];
        int sc = startPos[1];
        int hr = homePos[0];
        int hc = homePos[1];

        int cost = 0;
        
        while (sr != hr) {
            if (sr < hr) {
                sr += 1;
            } else {
                sr -= 1;
            }
            cost += rowCosts[sr];
        }
        
        while (sc != hc) {
            if (sc < hc) {
                sc += 1;
            } else {
                sc -= 1;
            }
            cost += colCosts[sc];
        }
        
        return cost;
    }
}

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Problem Info

Description

Problem 2087: Minimum Cost Homecoming of a Robot in a Grid - Detailed explanation and solution approaches.

Topics

Array Greedy

Companies

Goldman SachsHP

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:2087

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

5 steps

1
Step 1: Initialize the starting row `sr`, starting column `sc`, home row `hr`, and home column `hc` with the provided input.
2
Step 2: Initialize a variable `cost` to 0 to store the total cost.
3
Step 3: Iterate from `sr` to `hr`. If `sr` is less than `hr`, increment `sr` by 1 in each iteration. Otherwise, decrement `sr` by 1 in each iteration. Add the cost of moving to the new row `rowCosts[sr]` to the `cost` variable.
4
Step 4: Iterate from `sc` to `hc`. If `sc` is less than `hc`, increment `sc` by 1 in each iteration. Otherwise, decrement `sc` by 1 in each iteration. Add the cost of moving to the new column `colCosts[sc]` to the `cost` variable.
5
Step 5: Return the `cost` variable, which represents the minimum total cost.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(m+n)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the number of iterations in the two while loops. The first loop iterates a maximum of |startRow - homeRow| times, and the second loop iterates a maximum of |startCol - homeCol| times. The space complexity is constant because we only use a fixed number of variables regardless of the input grid size.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

No alternative approach significantly better than the given greedy solution.

Due to the fixed cost of each row and column, and the constraint of only moving horizontally and vertically, dynamic programming or graph search algorithms like Dijkstra's or A* are not required. They would add unnecessary complexity and overhead without improving performance.

Time: O(m+n) if implemented reasonably, but with higher constant factorsSpace: O(m*n) or O(m+n), depending on the DP approach

Trade-offs: The greedy approach is the most efficient and straightforward solution. DP or graph algorithms would be less efficient for this problem.

✨

Key Insights

1Insight 1: The problem can be solved greedily because the cost is fixed for each row and column, and we want to minimize the total cost. Therefore, the optimal path will always be moving along the rows and columns directly towards the destination, without unnecessary backtracking or detours.
2Insight 2: The solution involves iterating through the rows and columns between the start and home positions, summing the corresponding costs in the `rowCosts` and `colCosts` arrays.
3Insight 3: An edge case to consider is when the start and home positions are the same. In this case, the cost is zero, as no movement is required.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: The start and home positions are the same (startPos == homePos). The solution correctly handles this case because the while loops will not execute, and the initial cost of 0 will be returned.
Edge case 2: Large differences between start and home rows/columns. The algorithm efficiently handles this because it iterates only through the necessary rows and columns, summing the corresponding costs.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the direction of movement (incrementing when should decrement or vice versa). This would lead to incorrect cost calculation.

✗Common mistake 2: Off-by-one errors in the loop conditions (e.g., not including the home row/column in the cost calculation).

✗Common mistake 3: Assuming there are obstacles in the path, which is not specified in the problem, thus attempting to find a path around obstacles, leading to unnecessary complexity.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Path planning for robots in a warehouse or factory, where each cell has a cost associated with traversing it. The algorithm can be used to find the minimum cost path for the robot to reach its destination.

🏦Real-world scenario 2: Route optimization in a city grid where different streets have different costs (e.g., toll roads, traffic congestion). The algorithm can be used to find the cheapest route between two locations.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 16, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

What are common greedy problem patterns?

Activity selection (sort by end time), interval scheduling (sort by start/end), Huffman coding (always combine smallest), fractional knapsack (sort by value/weight ratio), and minimum spanning tree. Look for problems where local decisions don't affect future options.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Minimum Cost Homecoming of a Robot in a Grid"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether greedy techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?