C program for optimal binary search trees

Try making all keys in interval. Next Facebook Interview Set 1. Recommended Articles. Article Contributed By :. Easy Normal Medium Hard Expert. Writing code in comment? Please use ide. Load Comments. What's New. Most popular in Binary Search Tree. Most visited in Dynamic Programming. We use cookies to ensure you have the best browsing experience on our website. We add sum of frequencies from i to j see first term in the above formula , this is added because every search will go through root and one comparison will be done for every search.

Following is recursive implementation that simply follows the recursive structure mentioned above. Time complexity of the above naive recursive approach is exponential. It should be noted that the above function computes the same subproblems again and again.

We can see many subproblems being repeated in the following recursion tree for freq[ Since same suproblems are called again, this problem has Overlapping Subprolems property. So optimal BST problem has both properties of a dynamic programming problem. Cyber Security. C Programming. Control System. Data Mining.

Data Warehouse. Javatpoint Services JavaTpoint offers too many high quality services. Let's understand through an example. If the keys are 10, 20, 30, 40, 50, 60, 70 In the above tree, all the nodes on the left subtree are smaller than the value of the root node, and all the nodes on the right subtree are larger than the value of the root node.

Now we will see how many binary search trees can be made from the given number of keys. The Formula for calculating the number of trees: When we use the above formula, then it is found that total 5 number of trees can be created. The average number of comparisons can be made as: In the above tree, the average number of comparisons that can be made as: In the above tree, the average number of comparisons that can be made as: In the above tree, the total number of comparisons can be made as 3.

Therefore, the average number of comparisons that can be made as: In the above tree, the total number of comparisons can be made as 3. Therefore, the average number of comparisons that can be made as: In the third case, the number of comparisons is less because the height of the tree is less, so it's a balanced binary search tree. Let's assume that frequencies associated with the keys 10, 20, 30 are 3, 2, 5. Dynamic Approach Consider the below table, which contains the keys and frequencies. First, we will calculate the values where j-i is equal to zero.

The cost of c[0,1] is 4 The key is 10, and the cost corresponding to key 10 is 4. The cost of c[1,2] is 2 The key is 20, and the cost corresponding to key 20 is 2.

The following are the trees that can be made if 10 is considered as a root node. So, c[0,3] is equal to It is not a wise idea to keep less frequent words near root in binary search tree. Instead of storing words in binary search tree in lexicographical order, we shall arrange them according to their probabilities.

This arrangement facilitates few searches for frequent words as they would be near the root. Such tree is called Optimal Binary Search Tree. An exhaustive search for optimal binary search tree leads to huge amount of time. The goal is to construct a tree which minimizes the total search cost. Such tree is called optimal binary search tree. OBST does not claim minimum height.

It is also not necessary that parent of sub tree has higher priority than its child.