How do you solve Robot Return to Origin on LeetCode?

Robot Return to Origin (LeetCode #657) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The robot's movements can be balanced by counting opposing moves.

What is the brute force solution for Robot Return to Origin?

The brute force approach for Robot Return to Origin is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves simulating each move of the robot one by one and tracking its position. This is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Robot Return to Origin?

Robot Return to Origin uses the following concepts: String, Simulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Robot Return to Origin?

Always think about the simplest way to solve the problem first. Consider edge cases, such as empty strings or very long strings. Explain your thought process clearly while coding.

What are common mistakes when solving Robot Return to Origin?

Not considering the case where moves are not balanced. Confusing the directions and their effects on the position.

#657

Robot Return to Origin

Easy

StringSimulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

Instead of tracking the robot's position step by step, we can count the occurrences of each move. The robot returns to the origin if the number of 'U's equals 'D's and 'L's equals 'R's.

⚙️

Algorithm

4 steps

1Step 1: Initialize counters for U, D, L, and R to zero.
2Step 2: Count each move in the string.
3Step 3: Check if the counts of U and D are equal and the counts of L and R are equal.
4Step 4: Return true if both conditions are met, otherwise return false.

solution.py2 lines

1def judgeCircle(moves):
2    return moves.count('U') == moves.count('D') and moves.count('L') == moves.count('R')

ℹ

Complexity note: The time complexity is O(n) because we traverse the string a constant number of times (4 counts). The space complexity is O(1) since we only use a few counters.

1The robot's movements can be balanced by counting opposing moves.
2The problem can be solved efficiently without simulating each move.

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