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Robot Return to Origin

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, or false 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.

 

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 * 104
  • moves only contains the characters 'U', 'D', 'L' and 'R'.

Solution Explanation for Robot Return to Origin

This problem asks whether a robot, starting at the origin (0, 0), returns to the origin after a sequence of moves. The moves are represented by a string containing 'U' (up), 'D' (down), 'L' (left), and 'R' (right).

Approach

The solution involves tracking the robot's x and y coordinates as it moves. We initialize x and y to 0. Then, we iterate through the moves string:

  • If the move is 'R', we increment x.
  • If the move is 'L', we decrement x.
  • If the move is 'U', we increment y.
  • If the move is 'D', we decrement y.

After processing all moves, we check if x and y are both 0. If they are, the robot has returned to the origin; otherwise, it hasn't.

Code Implementation (Python, Java, TypeScript)

The code implementations in Python, Java, and TypeScript follow the same logic. They differ slightly in syntax, but the core algorithm remains consistent.

Python3:

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

Java:

class Solution {
    public boolean judgeCircle(String moves) {
        int x = 0, y = 0;
        for (int i = 0; i < moves.length(); ++i) {
            char c = moves.charAt(i);
            if (c == 'R')
                ++x;
            else if (c == 'L')
                --x;
            else if (c == 'U')
                ++y;
            else if (c == 'D')
                --y;
        }
        return x == 0 && y == 0;
    }
}

TypeScript:

function judgeCircle(moves: string): boolean {
    let x = 0, y = 0;
    for (let i = 0; i < moves.length; i++) {
        const move = moves[i];
        if (move === 'R') x++;
        else if (move === 'L') x--;
        else if (move === 'U') y++;
        else if (move === 'D') y--;
    }
    return x === 0 && y === 0;
}

Time and Space Complexity Analysis

Time Complexity: O(n), where n is the length of the moves string. We iterate through the string once.

Space Complexity: O(1). We use a constant amount of extra space to store the variables x and y. The space used doesn't depend on the input size.