### View *Destination City* on LeetCode

## Statistics

**Time Spent Coding**

2 minutes

**Time Complexity**

O(n) - In the worst case, every element in the input list will be visited twice, but constant multiples of n do not increase the growth rate, resulting in the O(n) time complexity.

**Space Complexity**

O(n) - In the worst case, every element in the input list will be stored twice, but constant multiples of n do not increase the growth rate, resulting in the O(n) space complexity.

**Runtime Beats**

99.93% of other submissions

**Memory Beats**

89.62% of other sumbissions

## Explanation

*The comments explain the program*

## Data Structures Used

**Hash Table (Set) -** An unordered container of non-repeating values.

**Visual Examples**

An array being transformed into a set, click to view

## Solution

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class Solution:
def destCity(self, paths: List[List[str]]) -> str:
# Initialize sets to store all starting and ending cities
startings = set()
endings = set()
# Fill sets with corresponding elements from the input list
for s, e in paths:
startings.add(s)
endings.add(e)
# Iterate through all ending cities until one is not found within the starting cities set
for e in endings:
if e not in startings:
return e