657. Robot Return to Origin

String
Simulation

TIME

O(n)

SPACE

O(1)

Intuition

Track x and y coordinates. The robot returns to origin if and only if the net displacement is zero in both directions.

Solution

Initialize x = 0, y = 0
For each move:
- 'U' → y++
- 'D' → y--
- 'R' → x++
- 'L' → x--
Return x == 0 && y == 0

Alternative: Count characters

Return count('U') == count('D') && count('L') == count('R')

Complexity

Time: O(n), single pass through moves
Space: O(1), only two variables

Edge Cases

Empty moves → true (already at origin)
All same direction → false
Alternating UD or LR → true

Code

class Solution:
    def judgeCircle(self, moves: str) -> bool:
        """
        657. Robot Return to Origin
        Time: O(n)
        Space: O(1)
        """
        x = y = 0

        for m in moves:
            if m == 'U':
                y += 1
            elif m == 'D':
                y -= 1
            elif m == 'R':
                x += 1
            else:
                x -= 1

        return x == 0 and y == 0

/**
 * 657. Robot Return to Origin
 * Time: O(n)
 * Space: O(1)
 */
class Solution {
    public boolean judgeCircle(String moves) {
        int x = 0, y = 0;

        for (char m : moves.toCharArray()) {
            if (m == 'U') y++;
            else if (m == 'D') y--;
            else if (m == 'R') x++;
            else x--;
        }

        return x == 0 && y == 0;
    }
}