Problem

There is a robot starting at 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.

The move sequence is represented by a string, and the character moves[i] represents its ith move. Valid moves are R (right), L (left), U (up), and D (down). If the robot returns to the origin after it finishes all of its moves, return true. Otherwise, return false.

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: “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: “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.

Pre analysis

Simply keeping track of x-axis and y-axis co-ordinates would give heuristic solution to this proble. If after the walk, robot returns to 0,0 then there is 0 displacement.

Post analysis

I wasn’t sure whether switch case would be give better performance or if-else, after some thought, I choose to go with if-else with continue keyword used to break the loop then and there. This solution can be independent of 2-D planar and should work with n-D plane as well I guess.

Another solution

var judgeCircle = function(moves) {
var x = 0, y = 0;

for (var i = 0; i < moves.length; i ++) {
    switch (moves[i]) {
    case 'U': y ++; break;
    case 'D': y --; break;
    case 'L': x --; break;
    case 'R': x ++; break;
    default: break;
    }
}

return (!x && !y)
};