657. Robot Return to Origin

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String Simulation

Time: O(n)

Space: O(1)

Problem Overview

The robot only moves on a 2D grid.

Intuition

The robot only moves on a 2D grid. Track net displacement in x (L/R) and y (U/D). After all moves, the robot returns to the origin exactly when both coordinates are zero.

Algorithm

1Initialize x = 0, y = 0.
2For each character: L decrements x, R increments x, U increments y, D decrements y.
3Return true iff x == 0 and y == 0 after the full string.

Example Walkthrough

Input: moves = "UD"

1.U: y=1. D: y=0. x stays 0.

Output: true

Common Pitfalls

•Empty moves string returns true.
•Only final displacement matters, not visit order.
•Input is always valid moves only.

657.cs

// Approach: Count horizontal (L/R) and vertical (U/D) moves separately;
// return true only if both net displacements are zero.
// Time: O(n) Space: O(1)

public class Solution
{
    public bool JudgeCircle(string moves)
    {
        int right = 0;
        int up = 0;

        foreach (char move in moves)
        {
            switch (move)
            {
                case 'R':
                    right++;
                    break;
                case 'L':
                    right--;
                    break;
                case 'U':
                    up++;
                    break;
                case 'D':
                    up--;
                    break;
            }
        }

        return right == 0 && up == 0;
    }
}

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