### View *Robot Return to Origin* on LeetCode

## Statistics

**Time Spent Coding**

3 minutes

**Time Complexity**

O(n) - The entire input list must be iterated through, resulting in the O(n) time complexity.

**Space Complexity**

O(1) - The number of variables declared is constant and independent from the input list, resulting in the O(1) space complexity.

**Runtime Beats**

57.93% of other submissions

**Memory Beats**

69.80% of other sumbissions

## Explanation

The algorithm works similarly to a math problem; it adds and subtracts from an `x`

and `y`

variable according to the move, `U`

being `y += 1`

and so on.

Finally, it compares if the x and y values equal zero and returns that result.

## Solution

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class Solution:
def judgeCircle(self, moves: str) -> bool:
x,y = 0,0
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