Table of Contents

## Question

You are given the array `paths`

, where `paths[i] = [cityA`

means there exists a direct path going from _{i}, cityB_{i}]`cityA`

to _{i}`cityB`

. _{i}*Return the destination city, that is, the city without any path outgoing to another city.*

It is guaranteed that the graph of paths forms a line without any loop, therefore, there will be exactly one destination city.

**Example 1:**

Input:paths = [["London","New York"],["New York","Lima"],["Lima","Sao Paulo"]]Output:"Sao Paulo"Explanation:Starting at "London" city you will reach "Sao Paulo" city which is the destination city. Your trip consist of: "London" -> "New York" -> "Lima" -> "Sao Paulo".

**Example 2:**

Input:paths = [["B","C"],["D","B"],["C","A"]]Output:"A"Explanation:All possible trips are: "D" -> "B" -> "C" -> "A". "B" -> "C" -> "A". "C" -> "A". "A". Clearly the destination city is "A".

**Example 3:**

Input:paths = [["A","Z"]]Output:"Z"

**Constraints:**

`1 <= paths.length <= 100`

`paths[i].length == 2`

`1 <= cityA`

_{i}.length, cityB_{i}.length <= 10`cityA`

_{i}!= cityB_{i}- All strings consist of lowercase and uppercase English letters and the space character.

## Python Solution

class Solution: def destCity(self, paths: List[List[str]]) -> str: l=[] fn = [] for p in paths: f=0 for x in p: if f==0: fn.append(x) f=1 if x not in l: l.append(x) else: l.remove(x) for i in l: if i not in fn: return(i)