Create longest-path-with-different-adjacent-characters.py

Author: kamyu104

Python/longest-path-with-different-adjacent-characters.py

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+# Time:  O(n)
+# Space: O(h)
+
+# tree, dfs
+class Solution(object):
+    def longestPath(self, parent, s):
+        """
+        :type parent: List[int]
+        :type s: str
+        :rtype: int
+        """
+        def iter_dfs(s, adj):
+            result = 0
+            stk = [(1, (0, [0]))]
+            while stk:
+                step, args = stk.pop()
+                if step == 1:
+                    u, ret = args
+                    top2 = [0]*2
+                    stk.append((4, (top2, ret)))
+                    stk.append((2, (u, 0, top2, ret)))
+                elif step == 2:
+                    u, i, top2, ret = args
+                    if i == len(adj[u]):
+                        continue
+                    ret2 = [0]
+                    stk.append((3, (u, i, top2, ret2)))
+                    stk.append((1, (adj[u][i], ret2))) 
+                elif step == 3:
+                    u, i, top2, ret2 = args
+                    if s[adj[u][i]] != s[u]:
+                        if ret2[0] > top2[0]:
+                            top2[0], top2[1] = ret2[0], top2[0]
+                        elif ret2[0] > top2[1]:
+                            top2[1] = ret2[0]
+                    stk.append((2, (u, i+1, top2, ret)))
+                elif step == 4:
+                    top2, ret = args
+                    result = max(result, top2[0]+top2[1]+1)
+                    ret[0] = top2[0]+1
+            return result
+    
+        
+        adj = [[] for _ in xrange(len(s))]
+        for i in xrange(1, len(parent)):
+            adj[parent[i]].append(i)
+        return iter_dfs(s, adj)
+    
+
+# Time:  O(n)
+# Space: O(h)
+# tree, dfs
+class Solution2(object):
+    def longestPath(self, parent, s):
+        """
+        :type parent: List[int]
+        :type s: str
+        :rtype: int
+        """
+        def dfs(s, adj, u, result):
+            top2 = [0]*2
+            for v in adj[u]:
+                l = dfs(s, adj, v, result)
+                if s[v] == s[u]:
+                    continue
+                if l > top2[0]:
+                    top2[0], top2[1] = l, top2[0]
+                elif l > top2[1]:
+                    top2[1] = l
+            result[0] = max(result[0], top2[0]+top2[1]+1)
+            return top2[0]+1
+    
+        
+        adj = [[] for _ in xrange(len(s))]
+        for i in xrange(1, len(parent)):
+            adj[parent[i]].append(i)
+        result = [0]
+        dfs(s, adj, 0, result)
+        return result[0]
+