LeetCode #2463 — HARD

Minimum Total Distance Traveled

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

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 ith robot. You are also given a 2D integer array factory where factory[j] = [positionj, limitj] indicates that positionj is the position of the jth factory and that the jth factory can repair at most limitj 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 initially.

All the robots are initially broken; they keep moving in one direction. The direction could be the negative or the positive direction of the X-axis. When a robot reaches a factory that did not reach its limit, the factory repairs the robot, and it stops moving.

At any moment, you can set the initial direction of moving for some robot. Your target is to minimize the total distance traveled by all the robots.

Return the minimum total distance traveled by all the robots. The test cases are generated such that all the robots can be repaired.

Note that

  • All robots move at the same speed.
  • If two robots move in the same direction, they will never collide.
  • If two robots move in opposite directions and they meet at some point, they do not collide. They cross each other.
  • If a robot passes by a factory that reached its limits, it crosses it as if it does not exist.
  • If the robot moved from a position x to a position y, the distance it moved is |y - x|.

Example 1:

Input: robot = [0,4,6], factory = [[2,2],[6,2]]
Output: 4
Explanation: As shown in the figure:
- The first robot at position 0 moves in the positive direction. It will be repaired at the first factory.
- The second robot at position 4 moves in the negative direction. It will be repaired at the first factory.
- The third robot at position 6 will be repaired at the second factory. It does not need to move.
The limit of the first factory is 2, and it fixed 2 robots.
The limit of the second factory is 2, and it fixed 1 robot.
The total distance is |2 - 0| + |2 - 4| + |6 - 6| = 4. It can be shown that we cannot achieve a better total distance than 4.

Example 2:

Input: robot = [1,-1], factory = [[-2,1],[2,1]]
Output: 2
Explanation: As shown in the figure:
- The first robot at position 1 moves in the positive direction. It will be repaired at the second factory.
- The second robot at position -1 moves in the negative direction. It will be repaired at the first factory.
The limit of the first factory is 1, and it fixed 1 robot.
The limit of the second factory is 1, and it fixed 1 robot.
The total distance is |2 - 1| + |(-2) - (-1)| = 2. It can be shown that we cannot achieve a better total distance than 2.

Constraints:

  • 1 <= robot.length, factory.length <= 100
  • factory[j].length == 2
  • -109 <= robot[i], positionj <= 109
  • 0 <= limitj <= robot.length
  • The input will be generated such that it is always possible to repair every robot.
Patterns Used
📊Dynamic Programming→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: 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 ith robot. You are also given a 2D integer array factory where factory[j] = [positionj, limitj] indicates that positionj is the position of the jth factory and that the jth factory can repair at most limitj 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 initially. All the robots are initially broken; they keep moving in one direction. The direction could be the negative or the positive direction of the X-axis. When a robot reaches a factory that did not reach its limit, the factory repairs the robot, and it stops moving. At any moment, you can set the initial direction of moving for some robot. Your target is to minimize the total distance traveled by all

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming

Example 1

[0,4,6]
[[2,2],[6,2]]

Example 2

[1,-1]
[[-2,1],[2,1]]

Related Problems

  • Capacity To Ship Packages Within D Days (capacity-to-ship-packages-within-d-days)
  • Number of Ways to Earn Points (number-of-ways-to-earn-points)
Step 02

Core Insight

What unlocks the optimal approach

  • Sort robots and factories by their positions.
  • After sorting, notice that each factory should repair some subsegment of robots.
  • Find the minimum total distance to repair first i robots with first j factories.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2463: Minimum Total Distance Traveled
class Solution {
    private long[][] f;
    private List<Integer> robot;
    private int[][] factory;

    public long minimumTotalDistance(List<Integer> robot, int[][] factory) {
        Collections.sort(robot);
        Arrays.sort(factory, (a, b) -> a[0] - b[0]);
        this.robot = robot;
        this.factory = factory;
        f = new long[robot.size()][factory.length];
        return dfs(0, 0);
    }

    private long dfs(int i, int j) {
        if (i == robot.size()) {
            return 0;
        }
        if (j == factory.length) {
            return Long.MAX_VALUE / 1000;
        }
        if (f[i][j] != 0) {
            return f[i][j];
        }
        long ans = dfs(i, j + 1);
        long t = 0;
        for (int k = 0; k < factory[j][1]; ++k) {
            if (i + k == robot.size()) {
                break;
            }
            t += Math.abs(robot.get(i + k) - factory[j][0]);
            ans = Math.min(ans, t + dfs(i + k + 1, j + 1));
        }
        f[i][j] = ans;
        return ans;
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(m^2 × n)
Space
O(m × n)

Approach Breakdown

RECURSIVE
O(2ⁿ) time
O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING
O(n × m) time
O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

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