# [Solved] You are given the array paths, where paths[i] = [cityAi, cityBi] means there exists a direct path going from cityAi to cityBi. Return the destination city, that is, the city without any path outgoing to another city. ## Question

You are given the array `paths`, where `paths[i] = [cityAi, cityBi]` means there exists a direct path going from `cityAi` to `cityBi`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 <= cityAi.length, cityBi.length <= 10`
• `cityAi != cityBi`
• 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)```
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