Leetcode #2463: Minimum Total Distance Traveled
In this guide, we solve Leetcode #2463 Minimum Total Distance Traveled in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Array, Dynamic Programming, Sorting
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
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.
Python Solution
class Solution:
def minimumTotalDistance(self, robot: List[int], factory: List[List[int]]) -> int:
def dfs(i, j):
if i == len(robot):
return 0
if j == len(factory):
return inf
ans = dfs(i, j + 1)
t = 0
for k in range(factory[j][1]):
if i + k == len(robot):
break
t += abs(robot[i + k] - factory[j][0])
ans = min(ans, t + dfs(i + k + 1, j + 1))
return ans
robot.sort()
factory.sort()
ans = dfs(0, 0)
dfs.cache_clear()
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.