Add House Robber 3 Problem and rename 0213_House_robber.py -> 0213_House_robber2.py

LeetCode/0337_House_Robber_III.py

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+'''
+337. House Robber III
+The thief has found himself a new place for his thievery again.
+There is only one entrance to this area, called the "root." Besides the root, each house has one and only one parent house.
+After a tour, the smart thief realized that "all houses in this place forms a binary tree".
+It will automatically contact the police if two directly-linked houses were broken into on the same night.
+Determine the maximum amount of money the thief can rob tonight without alerting the police.
+Example 1:
+            Input: [3,2,3,null,3,null,1]
+                 3
+                / \
+               2   3
+                \   \ 
+                 3   1
+
+Output: 7 
+            Explanation: Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
+
+Time Complexity:- O(N) 
+Space Complexity:- O(N), as we are using dictionary to memoize the solution
+'''
+
+def rob(self, root: TreeNode) -> int:
+    # this dp dictionary will help us reduce the time complexity by memoizing the solutions
+    dp = {}
+
+    # this function will return the max profit that we can get 
+    def rob_helper(root):
+        # base cases
+        if(root is None):
+            return 0
+        # if we have the value for root in dp that we means we have calculated the value
+        # previously so we can simply return the saved value
+        if(root in dp.keys()):
+            return dp[root]
+        '''
+        In this problem our constraints are:-
+            1. If we add/rob the profit of parent then we can't add/rob profit of children
+               as the police will be alerted
+            2. If we don't rob the parent then we can rob its child nodes 
+
+             Example:-
+
+            lvl 1            3
+                            / \
+            lvl 2          2   3
+                            \   \ 
+            lvl 3            3   1
+            In this if we add the profit for 3 then we can't add 2 and 3 from level 2 but add 3 and 1 from level 3
+            If we add 2 and 3 from level 2 then we can't add profit from level 1 and 3
+            Therefore, in a nutshell WE CAN'T ADD THE PROFIT OF 2 ADJACENT LEVEL/HOUSES
+
+        '''
+
+        # It is the total profit if we exclude the parent and add the left and right child
+        profit1 = rob_helper(root.left) + rob_helper(root.right)
+
+        # In this case we left child nodes and added the profit for parent node
+        profit2 = root.val
+        # As we robbed the parent node so we can't rob children
+        #  but we can rob children of left and right child of parent
+
+        # If root.left has left and right children then add there profit
+        if(root.left is not None):
+            profit2 += rob_helper(root.left.left) + rob_helper(root.left.right)
+        # If root.right has left and right children then add there profit
+        if(root.right is not None):
+            profit2 += rob_helper(root.right.left) + rob_helper(root.right.right)
+
+        # save the max value in DP to memoize the solution
+        dp[root] = max(profit1, profit2)
+
+        return dp[root]
+
+    return rob_helper(root)