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