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Robot Return to Origin

Difficulty: Easy

Problem Description

There is a robot starting at the position (0, 0), the origin, on a 2D plane. Given a sequence of its moves, judge if this robot ends up at (0, 0) after it completes its moves. You are given a string moves that represents the move sequence of the robot where moves[i] represents its i-th move. Valid moves are 'R' (right), 'L' (left), 'U' (up), and 'D' (down). Return true if the robot returns to the origin after it finishes all of its moves, or false otherwise.

Key Insights

  • The robot's movements can be represented as coordinate changes on a 2D plane.
  • Each 'U' increases the y-coordinate, while 'D' decreases it. Similarly, 'R' increases the x-coordinate, and 'L' decreases it.
  • To return to the origin, the number of 'U' moves must equal the number of 'D' moves, and the number of 'L' moves must equal the number of 'R' moves.

Space and Time Complexity

Time Complexity: O(n)
Space Complexity: O(1)

Solution

To solve the problem, we can maintain two counters for the movements in the x and y directions. We iterate through the string of moves and update our counters based on the type of move:

  • Increment the y counter for 'U'
  • Decrement the y counter for 'D'
  • Increment the x counter for 'R'
  • Decrement the x counter for 'L'

At the end of the iteration, if both counters are zero, it means the robot has returned to the origin.

Code Solutions

def judgeCircle(moves: str) -> bool:
    x = 0
    y = 0
    for move in moves:
        if move == 'U':
            y += 1
        elif move == 'D':
            y -= 1
        elif move == 'R':
            x += 1
        elif move == 'L':
            x -= 1
    return x == 0 and y == 0
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