Leetcode #2731: Movement of Robots

In this guide, we solve Leetcode #2731 Movement of Robots 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.

Leetcode

Problem Statement

Some robots are standing on an infinite number line with their initial coordinates given by a 0-indexed integer array nums and will start moving once given the command to move. The robots will move a unit distance each second.

Quick Facts

Difficulty: Medium
Premium: No
Tags: Brainteaser, Array, Prefix Sum, Sorting

Intuition

Range queries become simple once we precompute cumulative sums.

We can transform subarray conditions into prefix comparisons.

Approach

Compute prefix sums and use a map to find matching prefixes.

This avoids nested loops while keeping the logic clear.

Steps:

Compute prefix sums.
Use a map to find valid ranges.
Update the answer.

Example

Input: nums = [-2,0,2], s = "RLL", d = 3
Output: 8
Explanation: 
After 1 second, the positions are [-1,-1,1]. Now, the robot at index 0 will move left, and the robot at index 1 will move right.
After 2 seconds, the positions are [-2,0,0]. Now, the robot at index 1 will move left, and the robot at index 2 will move right.
After 3 seconds, the positions are [-3,-1,1].
The distance between the robot at index 0 and 1 is abs(-3 - (-1)) = 2.
The distance between the robot at index 0 and 2 is abs(-3 - 1) = 4.
The distance between the robot at index 1 and 2 is abs(-1 - 1) = 2.
The sum of the pairs of all distances = 2 + 4 + 2 = 8.

Python Solution

class Solution:
    def sumDistance(self, nums: List[int], s: str, d: int) -> int:
        mod = 10**9 + 7
        for i, c in enumerate(s):
            nums[i] += d if c == "R" else -d
        nums.sort()
        ans = s = 0
        for i, x in enumerate(nums):
            ans += i * x - s
            s += x
        return ans % mod

Complexity

The time complexity is $O(n \times \log n)$ and the space complexity is $O(n)$ , where $n$ is the number of robots. The space complexity is $O(n)$ , where $n$ is the number of robots.

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.

Example

Input: nums = [-2,0,2], s = "RLL", d = 3
Output: 8
Explanation: 
After 1 second, the positions are [-1,-1,1]. Now, the robot at index 0 will move left, and the robot at index 1 will move right.
After 2 seconds, the positions are [-2,0,0]. Now, the robot at index 1 will move left, and the robot at index 2 will move right.
After 3 seconds, the positions are [-3,-1,1].
The distance between the robot at index 0 and 1 is abs(-3 - (-1)) = 2.
The distance between the robot at index 0 and 2 is abs(-3 - 1) = 4.
The distance between the robot at index 1 and 2 is abs(-1 - 1) = 2.
The sum of the pairs of all distances = 2 + 4 + 2 = 8.

Python Solution

class Solution:
    def sumDistance(self, nums: List[int], s: str, d: int) -> int:
        mod = 10**9 + 7
        for i, c in enumerate(s):
            nums[i] += d if c == "R" else -d
        nums.sort()
        ans = s = 0
        for i, x in enumerate(nums):
            ans += i * x - s
            s += x
        return ans % mod

Leetcode #2731: Movement of Robots

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview

Leetcode #2731: Movement of Robots

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview