How do you solve Movement of Robots on LeetCode?

Movement of Robots (LeetCode #2731) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Ignoring collisions simplifies the problem.

What is the time complexity of Movement of Robots?

The optimal solution for Movement of Robots runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to the sorting step. The space complexity is O(n) for storing the positions of the robots.

What is the brute force solution for Movement of Robots?

The brute force approach for Movement of Robots is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute-force approach, we simulate the movement of each robot for 'd' seconds and track their positions. After that, we calculate the distances between all pairs of robots. This method is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Movement of Robots?

Movement of Robots uses the following concepts: Array, Brainteaser, Sorting, Prefix Sum. The recommended patterns to study are: Sorting, Prefix Sum.

What are the interview tips for Movement of Robots?

Explain your thought process clearly when approaching the problem. Discuss edge cases, such as robots starting at the same position. Practice similar problems to get comfortable with movement and distance calculations.

What are common mistakes when solving Movement of Robots?

Not considering the modulo operation for large sums. Confusing the direction of movement when calculating final positions.

#2731

Movement of Robots

Medium

ArrayBrainteaserSortingPrefix SumSortingPrefix Sum

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

In the optimal solution, we can ignore the collisions and directly calculate the final positions of the robots after 'd' seconds. By sorting these positions, we can efficiently compute the sum of distances using a prefix sum approach.

⚙️

Algorithm

3 steps

1Step 1: Calculate the final position of each robot after 'd' seconds based on its direction.
2Step 2: Sort the final positions.
3Step 3: Use a prefix sum to calculate the total distance efficiently.

solution.py14 lines

1# Full working Python code
2
3def sum_of_distances(nums, s, d):
4    n = len(nums)
5    positions = [0] * n
6    for i in range(n):
7        positions[i] = nums[i] + d if s[i] == 'R' else nums[i] - d
8    positions.sort()
9    total_distance = 0
10    prefix_sum = 0
11    for i in range(n):
12        total_distance += positions[i] * i - prefix_sum
13        prefix_sum += positions[i]
14    return total_distance % (10**9 + 7)

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step. The space complexity is O(n) for storing the positions of the robots.

1Ignoring collisions simplifies the problem.
2Sorting helps in efficiently calculating distances.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.