Added medium problem.

Author: das-jishu

Leetcode/medium/flower-planting-with-no-adjacent.py

@@ -0,0 +1,79 @@
+""" 
+# FLOWER PLANTING WITH NO ADJACENT
+
+You have n gardens, labeled from 1 to n, and an array paths where paths[i] = [xi, yi] describes a bidirectional path between garden xi to garden yi. In each garden, you want to plant one of 4 types of flowers.
+
+All gardens have at most 3 paths coming into or leaving it.
+
+Your task is to choose a flower type for each garden such that, for any two gardens connected by a path, they have different types of flowers.
+
+Return any such a choice as an array answer, where answer[i] is the type of flower planted in the (i+1)th garden. The flower types are denoted 1, 2, 3, or 4. It is guaranteed an answer exists.
+
+Example 1:
+
+Input: n = 3, paths = [[1,2],[2,3],[3,1]]
+Output: [1,2,3]
+Explanation:
+Gardens 1 and 2 have different types.
+Gardens 2 and 3 have different types.
+Gardens 3 and 1 have different types.
+Hence, [1,2,3] is a valid answer. Other valid answers include [1,2,4], [1,4,2], and [3,2,1].
+
+Example 2:
+
+Input: n = 4, paths = [[1,2],[3,4]]
+Output: [1,2,1,2]
+
+Example 3:
+
+Input: n = 4, paths = [[1,2],[2,3],[3,4],[4,1],[1,3],[2,4]]
+Output: [1,2,3,4]
+ 
+Constraints:
+
+1 <= n <= 104
+0 <= paths.length <= 2 * 104
+paths[i].length == 2
+1 <= xi, yi <= n
+xi != yi
+Every garden has at most 3 paths coming into or leaving it.
+"""
+
+class Solution:
+    def gardenNoAdj(self, n: int, paths):
+        graph = [[] for i in range(n)]
+        for start, end in paths:
+            graph[start - 1].append(end - 1)
+            graph[end - 1].append(start - 1)
+            
+        flowers = [None for i in range(n)]
+        
+        visited = set()
+        for garden in range(n):
+            if garden not in visited:
+                self.bfs(graph, garden, flowers, visited)
+        
+        return flowers
+    
+    def bfs(self, graph, current, flowers, visited):
+        queue = [current]
+        while len(queue) != 0:
+            tempQueue = []
+            for garden in queue:
+                visited.add(garden)
+                flowers[garden] = self.assignFlower(graph, garden, flowers)
+                for neighbor in graph[garden]:
+                    if neighbor in visited:
+                        continue
+                    tempQueue.append(neighbor)
+            queue = tempQueue
+        
+    def assignFlower(self, graph, current, flowers):
+        for flower in range(1, 5):
+            found = True
+            for neighbor in graph[current]:
+                if flowers[neighbor] == flower:
+                    found = False
+                    break
+            if found:
+                return flower