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 * 104moves only contains the characters 'U', 'D', 'L' and 'R'.This approach involves counting the number of moves in each direction and checking if the horizontal movements ('L' and 'R') balance each other and the vertical movements ('U' and 'D') balance each other to return to the origin.
We initialize two integer variables, x and y, to track the robot's position. We iterate through the moves string, adjusting the position according to each move. If 'U' is found, y is incremented; if 'D' is found, y is decremented; 'L' decrements x; and 'R' increments x. Finally, we check if both x and y are zero, indicating a return to the origin, and return accordingly.
C++
Java
Python
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Time Complexity: O(n), where n is the length of the moves string.
Space Complexity: O(1), constant space used for variables x and y.
Instead of tracking coordinates, this approach counts the occurrences of each move character to determine if the robot returns to the origin. The idea is that the number of 'U' moves should equal 'D' moves and 'L' moves should equal 'R'.
We use four counters to track the number of each move type in the input string. The robot returns to the origin if 'U' matches 'D' and 'L' matches 'R'. Thus, we count the occurrences of each direction and compare. The function returns true if these conditions are met.
C++
Java
Python
C#
JavaScript
Time Complexity: O(n), where n is the length of the input moves.
Space Complexity: O(1), for direct counting with integer variables.
| Approach | Complexity |
|---|---|
| Balancing Moves | Time Complexity: O(n), where n is the length of the moves string. |
| Net Balance Using Counts | Time Complexity: O(n), where n is the length of the input |
Robot Return to Origin • Kevin Naughton Jr. • 34,200 views views
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