Add medium problem.

Author: das-jishu

Leetcode/medium/reconstruct-itinerary.py

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+"""
+# RECONSTRUCT ITINERARY
+
+You are given a list of airline tickets where tickets[i] = [fromi, toi] represent the departure and the arrival airports of one flight. Reconstruct the itinerary in order and return it.
+
+All of the tickets belong to a man who departs from "JFK", thus, the itinerary must begin with "JFK". If there are multiple valid itineraries, you should return the itinerary that has the smallest lexical order when read as a single string.
+
+For example, the itinerary ["JFK", "LGA"] has a smaller lexical order than ["JFK", "LGB"].
+You may assume all tickets form at least one valid itinerary. You must use all the tickets once and only once.
+
+Example 1:
+
+Input: tickets = [["MUC","LHR"],["JFK","MUC"],["SFO","SJC"],["LHR","SFO"]]
+Output: ["JFK","MUC","LHR","SFO","SJC"]
+
+Example 2:
+
+Input: tickets = [["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]
+Output: ["JFK","ATL","JFK","SFO","ATL","SFO"]
+Explanation: Another possible reconstruction is ["JFK","SFO","ATL","JFK","ATL","SFO"] but it is larger in lexical order.
+
+Constraints:
+
+1 <= tickets.length <= 300
+tickets[i].length == 2
+from.length == 3
+to.length == 3
+from and to consist of uppercase English letters.
+from != to
+"""
+
+class Solution:
+    def findItinerary(self, tickets):
+        neighborsList = {}
+        remainingPlaces = {}
+        
+        for ticket in tickets:
+            start, end = ticket
+            string = self.convert(start, end)            
+            if start not in neighborsList:
+                neighborsList[start] = []
+            neighborsList[start].append(end)
+            if string not in remainingPlaces:
+                remainingPlaces[string] = 0
+            remainingPlaces[string] += 1
+        
+        for edges in neighborsList:
+            neighborsList[edges].sort()
+            
+        return self.dfs("JFK", neighborsList, ["JFK"], len(tickets), remainingPlaces)
+    
+    def dfs(self, current, neighbors, itinerary, n, remainingPlaces):
+        if len(itinerary) == n + 1:
+            return itinerary
+        
+        if current not in neighbors:
+            return False
+        
+        for destination in neighbors[current]:
+            journey = self.convert(current, destination)
+            if journey not in remainingPlaces or remainingPlaces[journey] == 0:
+                continue
+            remainingPlaces[journey] -= 1
+            itinerary.append(destination)
+            check = self.dfs(destination, neighbors, itinerary, n, remainingPlaces)
+            if check != False:
+                return check
+            remainingPlaces[journey] += 1
+            itinerary.pop()
+        return False
+            
+    def convert(self, a, b):
+        return str(a) + "-" + str(b)
+            
+                
+        
+        
+            

Leetcode/script.py

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+import os
+difficulty = input("Enter difficulty: ")
+os.system("git add .")
+os.system("git commit -m \"Add " + difficulty + " problem.\"")
+os.system("git push -u origin master")