In this guide, you will get 1436. Destination City LeetCode Solution with the best time and space complexity. The solution to Destination City problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
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.
class Solution:
def destCity(self, paths: list[list[str]]) -> str:
count = collections.Counter()
for a, b in paths:
count[a] += 1
for a, b in paths:
if b in count:
count[b] -= 1
if count[b] == 0:
del count[b]
else:
return b /* code provided by PROGIEZ */
1436. Destination City LeetCode Solution in Java
N/A // code provided by PROGIEZ
1436. Destination City LeetCode Solution in Python
