## 27 November 2011

### Given a sorted array, find k in least possible time in Java

Given a sorted array and a value k, write a program to find whether that number is present in that array or not.

The first solution that comes to the mind for this problem is to traverse through the entire array and find whether it is having the value k and return true or false. This takes O(n) running time. But then, we are presented with a hint that the input array is sorted (lets assume in ascending order). This problem can be attacked by doing a divide and rule analysis.

This can be done recursively. See whether k is inbetween the last and first element in the array. If it is, then divide the array into two and repeat the same. If its not, simply return false. Yes, its as simple as that. Following is the java code for the same.

Leave some comments if you have anything to say.

Cheers!
Braga

## 26 November 2011

### Find middle element in a circularly sorted array

Given a circularly sorted array, find its middle element in best possible time.

A circularly sorted array is an array that is obtained by shifting an array k times. For example, the following is a circularly sorted array,

[5, 6, 7, 8, 9, 1, 2, 3, 4]

If you sort this array and find the middle element it will be 5. But then, sorting is a painful operation which will take O(nlogn) timing. So it is out of the equation right away. If you closely observe this problem, it is enough to find the shifting index, the place where the sorting varies. For example, the shifting index in the above example is 5.  Once that is found, then it is a matter of applying the bias over the length.

This program can be attacked in a recursive manner. Take the first and last elements in the array and compare them. If they are sorted, then the last element is greater than the first element. If this condition is not met, we will have to divide this array into two halves and try applying the same logic. If you do this, at some point you will arrive at a situation where you are left with only two elements or lesser. The end value is the shift index. Once the shift index is found, it is easy to find the middle element(s) for odd and even cases. Java code is below.

Leave comments if something is unclear.

Cheers!
Braga

## 24 November 2011

### Queue using Stack in Java

A Queue can constructed with its underlying data structure being a stack. Typically, we will need two Stacks one for enqueue and the other for dequeue.

During every enqueue() operation over a queue, we will keep pushing value to a first stack, lets say stack1.

During every dequeue() operation, we will have to simply pop an element from stack2 if stack2 is not empty. If stack2 is empty, we will need to pop all elements from stack1 one by one and push it to stack2. And then pop an element from stack2.

Take for example, I am inserting elements from 1 to 6 to a queue. This is how the following queue and the corresponding stack will look like,

Then trying a dequeue() operation will try to immediately pop values from stack 2.

So, it will pop all values from stack1 and push them to stack2 as follows,

Now the dequeue() operation shall be performed with ease since stack2 is not empty.

Finally, more enqueue will add or keep pushing values to the stack.

Following is the Java code. Note, I have not comprehensively covered the entire methods in a Queue, but this should be well more than enough.

And a test class for this with output

Cheers!
Braga

## 18 November 2011

### Program to check whether a tree is a BST

Write a program to check whether a given tree is a binary search tree.

We all know that a binary search tree follows the rule : for any given node with value N, all the values in the left of that node is lesser than N and all the values in the right of that node will be greater than or equal to N. For example,

This is a tricky question. We may try to immediately run through all the nodes checking only the immediate left and right nodes and coming to a conclusion that it is a binary search tree. But this program may not work for this scenario.

The simplest solution to this problem is to do an inorder traversal through the entire tree and find whether it is sorted or not. Typically we can do that by pushing into any array. Even better is to calculate the minimum value at each step. Program is as follows. Please leave some comments if something is confusing. I will get back immediately.

Cheers!
Braga