added houseRobberIII

README.md

@@ -8,6 +8,7 @@ LeetCode
 
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|337|[House Robber III](https://leetcode.com/problems/house-robber-iii/) | [C++](./algorithms/cpp/houseRobber/houseRobberIII.cpp)|Medium|
 |334|[Increasing Triplet Subsequence](https://leetcode.com/problems/increasing-triplet-subsequence/) | [C++](./algorithms/cpp/increasingTripletSubsequence/increasingTripletSubsequence.cpp)|Medium|
 |330|[Patching Array](https://leetcode.com/problems/patching-array/) | [C++](./algorithms/cpp/patchingArray/PatchingArray.cpp)|Medium|
 |329|[Longest Increasing Path in a Matrix](https://leetcode.com/problems/longest-increasing-path-in-a-matrix/) | [C++](./algorithms/cpp/longestIncreasingPathInAMatrix/LongestIncreasingPathInAMatrix.cpp)|Medium|

algorithms/cpp/houseRobber/houseRobberIII.cpp

@@ -0,0 +1,88 @@
+// Source : https://leetcode.com/problems/house-robber-iii/
+// Author : Calinescu Valentin
+// Date   : 2016-04-29
+
+/*************************************************************************************** 
+ *
+ * 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:
+ *     3
+ *    / \
+ *   2   3
+ *    \   \ 
+ *     3   1
+ * Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
+ * Example 2:
+ *     3
+ *    / \
+ *   4   5
+ *  / \   \ 
+ * 1   3   1
+ * Maximum amount of money the thief can rob = 4 + 5 = 9.
+ * Credits:
+ * Special thanks to @dietpepsi for adding this problem and creating all test cases.
+ * 
+ ***************************************************************************************/
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
+ * };
+ */
+  /* 
+ * Solution 1 - O(N log N)
+ * =========
+ *
+ * We can use a recursive function that computes the solution for every node of the tree 
+ * using the previous solutions calculated for the left and right subtrees. At every step
+ * we have 2 options:
+ *
+ * 1) Take the value of the current node + the solution of the left and right subtrees of
+ * each of the left and right children of the current node.
+ * 2) Take the solution of the left and right subtrees of the current node, skipping over
+ * its value.
+ *
+ * This way we can make sure that we do not pick 2 adjacent nodes.
+ *
+ * If we implemented this right away we would get TLE. Thus, we need to optimize the
+ * algorithm. One key observation would be that we only need to compute the solution for
+ * a certain node once. We can use memoization to calculate every value once and then
+ * retrieve it when we get subsequent calls. As the header of the recursive function
+ * doesn't allow additional parameters we can use a map to link every node(a pointer) to
+ * its solution(an int). For every call the map lookup of an element and its insertion
+ * take logarithmic time and there are a constant number of calls for each node. Thus, the
+ * algorithm takes O(N log N) time to finish.
+ * 
+ */
+class Solution {
+public:
+    map<TreeNode*, int> dict;
+    int rob(TreeNode* root) {
+        if(root == NULL)
+            return 0;
+        else if(dict.find(root) == dict.end())
+        {
+            int lwith = rob(root->left);
+            int rwith = rob(root->right);
+            int lwithout = 0, rwithout = 0;
+            if(root->left != NULL)
+                lwithout = rob(root->left->left) + rob(root->left->right);
+            if(root->right != NULL)
+                rwithout = rob(root->right->left) + rob(root->right->right);
+            //cout << lwith << " " << rwith << " " << lwithout << " " << rwithout << '\n';
+            dict[root] = max(root->val + lwithout + rwithout, lwith + rwith);
+        }
+        return dict[root];
+    }
+};