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Robot Return to Origin

EasyUpdated: Aug 2, 2025
Practice on:
LeetCode

Problem

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 ith 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, orfalse otherwise.

Note : The way that the robot is "facing" is irrelevant. 'R' will always make the robot move to the right once, 'L' will always make it move left, etc. Also, assume that the magnitude of the robot's movement is the same for each move.

Examples

Example 1

Input: moves = "UD"
Output: true
Explanation: The robot moves up once, and then down once. All moves have the same magnitude, so it ended up at the origin where it started. Therefore, we return true.

Example 2

Input: moves = "LL"
Output: false
Explanation: The robot moves left twice. It ends up two "moves" to the left of the origin. We return false because it is not at the origin at the end of its moves.

Constraints

  • 1 <= moves.length <= 2 * 10^4
  • moves only contains the characters 'U', 'D', 'L' and 'R'.

Solution

Method 1 - Count Moves

Intuition

The robot returns to the origin if the number of moves up equals down, and left equals right. We can count each direction and check if the net movement is zero.

Approach

Iterate through the string, incrementing or decrementing counters for vertical and horizontal moves. At the end, if both counters are zero, the robot is back at the origin.

Code

C++
class Solution {
public:
    bool judgeCircle(string moves) {
        int x = 0, y = 0;
        for (char c : moves) {
            if (c == 'U') y++;
            else if (c == 'D') y--;
            else if (c == 'L') x--;
            else if (c == 'R') x++;
        }
        return x == 0 && y == 0;
    }
};
Go
func judgeCircle(moves string) bool {
    x, y := 0, 0
    for _, c := range moves {
        switch c {
        case 'U': y++
        case 'D': y--
        case 'L': x--
        case 'R': x++
        }
    }
    return x == 0 && y == 0
}
Java
class Solution {
    public boolean judgeCircle(String moves) {
        int x = 0, y = 0;
        for (char c : moves.toCharArray()) {
            if (c == 'U') y++;
            else if (c == 'D') y--;
            else if (c == 'L') x--;
            else if (c == 'R') x++;
        }
        return x == 0 && y == 0;
    }
}
Kotlin
class Solution {
    fun judgeCircle(moves: String): Boolean {
        var x = 0
        var y = 0
        for (c in moves) {
            when (c) {
                'U' -> y++
                'D' -> y--
                'L' -> x--
                'R' -> x++
            }
        }
        return x == 0 && y == 0
    }
}
Python
class Solution:
    def judgeCircle(self, moves: str) -> bool:
        x = y = 0
        for c in moves:
            if c == 'U': y += 1
            elif c == 'D': y -= 1
            elif c == 'L': x -= 1
            elif c == 'R': x += 1
        return x == 0 and y == 0
Rust
impl Solution {
    pub fn judge_circle(moves: String) -> bool {
        let (mut x, mut y) = (0, 0);
        for c in moves.chars() {
            match c {
                'U' => y += 1,
                'D' => y -= 1,
                'L' => x -= 1,
                'R' => x += 1,
                _ => {}
            }
        }
        x == 0 && y == 0
    }
}
TypeScript
function judgeCircle(moves: string): boolean {
    let x = 0, y = 0;
    for (const c of moves) {
        if (c === 'U') y++;
        else if (c === 'D') y--;
        else if (c === 'L') x--;
        else if (c === 'R') x++;
    }
    return x === 0 && y === 0;
}

Complexity

  • ⏰ Time complexity: O(n) — where n is the length of moves.
  • 🧺 Space complexity: O(1) — only a few counters are used.

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