What is the time complexity of Minimum Total Distance Traveled?

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

What data structures are used to solve Minimum Total Distance Traveled?

This problem uses Array, Dynamic Programming, Sorting. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Minimum Total Distance Traveled?

Insight 1: Sorting both robots and factories by their positions simplifies the problem by allowing us to consider them in a specific order. This is because the total distance traveled will be minimized if closer robots are assigned to closer factories. Insight 2: Dynamic programming is suitable because the problem has overlapping subproblems. The minimum distance for the first 'i' robots and 'j' factories can be derived from the minimum distances of smaller subsets of robots and factories.

Which companies ask LeetCode 2463. Minimum Total Distance Traveled in interviews?

This problem is frequently asked at Infosys.

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Minimum Total Distance Traveled - LeetCode 2463 Solution

Minimum Total Distance Traveled - Complete Solution Guide

Minimum Total Distance Traveled is LeetCode problem 2463, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

There are some robots and factories on the X-axis. You are given an integer array robot where robot[i] is the position of the i th robot. You are also given a 2D integer array factory where factory[j] = [position j , limit j ] indicates that position j is the position of the j th factory and that the j th factory can repair at most limit j robots. The positions of each robot are unique . The positions of each factory are also unique . Note that a robot can be in the same position as a factory in

Detailed Explanation

The problem presents a scenario where we have robots and factories on a 1D axis. Each robot needs to be repaired by a factory. Each factory has a position and a limit on the number of robots it can repair. The goal is to minimize the total distance traveled by all robots to reach a factory for repair. We can decide the initial direction (positive or negative) for each robot's movement. The test cases guarantee a solution exists (all robots can be repaired).

Solution Approach

The solution utilizes dynamic programming to determine the minimum total distance. First, both the `robot` and `factory` arrays are sorted by their positions. A 1D DP array `dp` of size `n + 1` is initialized, where `dp[i]` represents the minimum total distance traveled to repair the first `i` robots. The algorithm iterates through each factory and updates the `dp` array. For each factory, it considers all possible numbers of robots that can be assigned to that factory, up to its limit. It calculates the cost of assigning a certain number of robots to the current factory and updates the `dp` array accordingly, choosing the minimum distance.

Step-by-Step Algorithm

Step 1: Sort the `robot` and `factory` arrays by their positions.
Step 2: Initialize a DP array `dp` of size `n + 1` with `dp[0] = 0` and all other elements set to infinity (or a large value).
Step 3: Iterate through each factory `j` from 0 to `m - 1`.
Step 4: For each factory `j`, extract its position `pos` and limit `limit`.
Step 5: Iterate backwards through the DP array from `i = n` to 1.
Step 6: Calculate the cost of assigning `k` robots to the current factory `j`, where `k` ranges from 1 to `min(i, limit)`. The cost is the sum of the absolute differences between the positions of the robots and the factory's position.
Step 7: Update `dp[i]` with the minimum of its current value and `dp[i - k] + current_cost`, where `dp[i - k]` represents the minimum distance to repair the first `i - k` robots, and `current_cost` is the cost of repairing the next `k` robots with the current factory.
Step 8: After iterating through all factories, `dp[n]` will contain the minimum total distance traveled to repair all robots.

Key Insights

Insight 1: Sorting both robots and factories by their positions simplifies the problem by allowing us to consider them in a specific order. This is because the total distance traveled will be minimized if closer robots are assigned to closer factories.
Insight 2: Dynamic programming is suitable because the problem has overlapping subproblems. The minimum distance for the first 'i' robots and 'j' factories can be derived from the minimum distances of smaller subsets of robots and factories.
Insight 3: The limit of each factory must be considered when assigning robots. We need to explore all possible assignments of robots to factories within the factory's capacity.

Complexity Analysis

Time Complexity: O(m*n*n*limit)

Space Complexity: O(n)

Topics

This problem involves: Array, Dynamic Programming, Sorting.

Companies

Asked at: Infosys.

Minimum Total Distance Traveled | LeetCode 2463

Hard

Array Dynamic Programming Sorting

Time: O(m*n*n*limit)Space: O(n)

Solution Code

Available in:

import java.util.Arrays;
import java.util.List;

class Solution {
    public long minimumTotalDistance(List<Integer> robot, int[][] factory) {
        int n = robot.size();
        int m = factory.length;

        robot.sort(null);
        Arrays.sort(factory, (a, b) -> Integer.compare(a[0], b[0]));

        long[] dp = new long[n + 1];
        Arrays.fill(dp, Long.MAX_VALUE);
        dp[0] = 0;

        for (int j = 0; j < m; j++) {
            int pos = factory[j][0];
            int limit = factory[j][1];

            for (int i = n; i > 0; i--) {
                long current_cost = 0;
                for (int k = 1; k <= Math.min(i, limit); k++) {
                    current_cost += Math.abs((long)robot.get(i - k) - pos);
                    if (dp[i - k] != Long.MAX_VALUE) {
                        dp[i] = Math.min(dp[i], dp[i - k] + current_cost);
                    }
                }
            }
        }

        return dp[n];
    }
}

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Problem Info

Description

Problem 2463: Minimum Total Distance Traveled - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Sorting

Companies

Infosys

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2463

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

8 steps

1
Step 1: Sort the `robot` and `factory` arrays by their positions.
2
Step 2: Initialize a DP array `dp` of size `n + 1` with `dp[0] = 0` and all other elements set to infinity (or a large value).
3
Step 3: Iterate through each factory `j` from 0 to `m - 1`.
4
Step 4: For each factory `j`, extract its position `pos` and limit `limit`.
5
Step 5: Iterate backwards through the DP array from `i = n` to 1.
6
Step 6: Calculate the cost of assigning `k` robots to the current factory `j`, where `k` ranges from 1 to `min(i, limit)`. The cost is the sum of the absolute differences between the positions of the robots and the factory's position.
7
Step 7: Update `dp[i]` with the minimum of its current value and `dp[i - k] + current_cost`, where `dp[i - k]` represents the minimum distance to repair the first `i - k` robots, and `current_cost` is the cost of repairing the next `k` robots with the current factory.
8
Step 8: After iterating through all factories, `dp[n]` will contain the minimum total distance traveled to repair all robots.

💡

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*n*limit)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity arises from the nested loops. The outer loop iterates 'm' times. The first inner loop iterates 'n' times. The second inner loop iterates at most 'n' times because the limit can be at most the number of robots. The space complexity is determined by the size of the DP table, which is proportional to the number of robots.

📖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

Greedy Approach with Hungarian Algorithm

A greedy approach might try to assign the closest robots to the closest factories first. This can be formulated as a min-cost max-flow problem or solved using the Hungarian algorithm. Calculate the cost matrix of assigning each robot to each factory (taking limits into account). Apply the Hungarian algorithm to find the optimal assignment.

Time: O(n^3) or O(n^2*m) depending on the exact implementation (Hungarian Algorithm). Cost matrix construction will be O(n*m).Space: O(n*m) for the cost matrix.

Trade-offs: The Hungarian algorithm provides the optimal solution but can be more complex to implement. For smaller input sizes, the DP approach may be simpler. The Hungarian algorithm has a potentially better time complexity for certain distributions of robot and factory positions, especially if factory limits are not restrictive.

✨

Key Insights

1Insight 1: Sorting both robots and factories by their positions simplifies the problem by allowing us to consider them in a specific order. This is because the total distance traveled will be minimized if closer robots are assigned to closer factories.
2Insight 2: Dynamic programming is suitable because the problem has overlapping subproblems. The minimum distance for the first 'i' robots and 'j' factories can be derived from the minimum distances of smaller subsets of robots and factories.
3Insight 3: The limit of each factory must be considered when assigning robots. We need to explore all possible assignments of robots to factories within the factory's capacity.

📚

Key Terms Explained

For Students

Array

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: When a factory's position is the same as a robot's position. The distance traveled is 0, which the `abs()` function correctly handles.
Edge case 2: When a factory's limit is reached. The inner loop's `min(i, limit)` condition prevents exceeding the factory's capacity.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to sort the `robot` and `factory` arrays. Sorting is crucial for the DP approach to work correctly as it builds up the solution based on the order of positions.

✗Common mistake 2: Not considering the factory limits properly. Incorrectly handling the `limit` parameter can lead to incorrect distance calculations and exceeding factory capacity.

✗Common mistake 3: Initializing the DP array incorrectly. All values except `dp[0]` must be initialized to infinity to ensure that only valid minimum distances are considered.

✅

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: Resource allocation - Assigning tasks (robots) to workers (factories) while minimizing the total cost (distance). For example, routing delivery trucks to distribution centers.

🏦Real-world scenario 2: Supply chain optimization - Determining the optimal location and capacity of warehouses (factories) to minimize the distance products (robots) need to travel to reach customers.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ 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: October 2, 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

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

@CodeNinjaJul 14

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

@TechLeadJul 7

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 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.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 Total Distance Traveled"?

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

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