Search in Sorted but Rotated Array

Given a sorted array that is sorted by rotated, find a given number. For example take an array: 1 3 8 10 12 56 and rotate it so you have 10 12 56 1 3 8 and then find a candidate e.g. 3 in it.


Input/Ouput:

Array:
10 12 56 2 3 8 9
3 is present at index: 4

Get Sentence from raw text

String getSentence(String text, Set<String> dictionary);

// text is a string without spaces, you need to insert spaces into text, so each word seperated by the space in the resulting string exists in the dictionary, return the resulting string

// getSentence("iamastudentfromwaterloo", {"from, "waterloo", "hi", "am", "yes", "i", "a", "student"}) -> "i am a student from waterloo"


Approach:
String [] tokens = {"from", "waterloo", "hi", "am", "as", "stud", "yes", "i", "a", "student"};
String text = "iamastudentfromwaterloo";


Lets create a hashtable named tokenMap which will contain all the token of tokens array.

We will also have 2 stacks, lets say indexStack to keep track of starting index of all the matched tokens in text (this will become clear below .. keep on reading :))
And another stack resultStack which will store all the tokens of the final expected sentence.

We will also have a StringBuffer sb which will keep track of substring from text which would be a candinate token (we will verify that by checking in hashtable tokenMap)

Now,
We will read one character at a time from "text"
StringBuffer sb = "i";
Our start index = 0
Lets push index onto indextStack
sb = "i" is present in the hashtable tokenMap, so now we have following
indexStack = 0 -> null
resultStack = "i" -> null

we now reset sb = "";
Next we see sb = "a"
"a" is in tokenMap so now we have following
indexStack = 1 -> 0 -> null;
resultStack = "a" -> "i" -> null

Now things get interesting as we keep on reading the characters from text and appending to sb but the string is not there in hashtable tokenMap
like sb = "mastudent" ...
So we need to decide to backtrack in such case (logic can be if sb length goes beyond max length of any of the given tokens in tokens array)

So when we backtrack we pop the stacks and reset StringBuffer sb
indexStack = 0 -> null
resultStack = "i" -> null
sb = "";

now again we start appending characters to sb till sb has a match in tokenMap HashTable

We keep repeating the above steps till we have constructed our sentence or till we decide its NOT possible to do so.

Input:

String [] tokens = {"from", "waterloo", "hi", "am", "as", "stud", "yes", "i", "a", "student"};
String text = "iamastudentfromwaterloo";

Output:

Max Token Length: 8
Sentence is: i am a student from waterloo

Binary Search Tree: Any Path Sum

You are given a binary tree in which each node contains a value. Design an algorithm to print all paths which sum up to that value. Note that it can be any path in the tree - it does not have to start at the root.

Input/Output:

printing inorder: 

1 3 4 5 6 7 8 9 10 12 13 15 17 

printing preorder: 

7 5 3 1 4 6 12 9 8 10 15 13 17 

printing path(s):

7 -> 5 -> null

5 -> 3 -> 4 -> null

12 -> null

Binary Tree: Is SubTree ?

You have two very large binary trees: T1, with millions of nodes, and T2, with hundreds of nodes. Create an algorithm to decide if T2 is a subtree of T1.

Approach 1:
T1 has 10 million nodes—this means that the data alone is about 40 mb. We could create a string representing the inorder and preorder traversals. If T2’s preorder traversal is a substring of T1’s preorder traversal, and T2’s inorder traversal is a substring of T1’s inorder traversal, then T2 is a substring of T1. We can check this using a suffix tree. However, we may hit memory limitations because suffix trees are extremely memory intensive. If this become an issue, we can use an alternative approach.

Approach 2:
Find all the occurances of  T2's root node in T1 and then check recursively if the binary tree matches or not.

Binary Search Tree: Least Common Ancestor

Least common ancestor of 2 nodes in a binary search tree:

Approach:
The main idea of the solution is — While traversing Binary Search Tree from top to bottom, the first node n we encounter with value between n1 and n2, i.e., n1 < n < n2 is the Lowest or Least Common Ancestor(LCA) of n1 and n2 (where n1 < n2). 
So just traverse the BST in pre-order, if you find a node with value in between n1 and n2 then n is the LCA,
 if it's value is greater than both n1 and n2 then our LCA lies on left side of the node,
 if it's value is smaller than both n1 and n2 then LCA lies on right side.

Binary Tree: Least Common Ancestor

Design an algorithm and write code to find the first common ancestor of two nodes in a binary tree. NOTE: This is not necessarily a binary search tree.

Approach:
Do the inorder traversal and store it in an array I
Then do the preorder traversal and store it in an array P

Inorder: D B F G E A C
PreOrder: A B D E F G C


Suppose we need to find LCA for nodes D and F
mark the positions in inorder array and find first preorder ptr which seprates it into two
different parts of Inorder array

for above example first preorder ptr is "B" which seprates inorder into "D" and "F G E A C"
so "B" is first commom ancester of "D" and "F".

Binary Tree: Inorder Predecessor

Write an algorithm to find the ‘previous’ node (e.g., in-order predecessor) of a given node in a binary search tree where each node has a link to its parent.


To Find the inorder predecessor of node u:
If u has a left child, l, then pred(u) is the rightmost descendent of l
Otherwise, pred(u) is the closest ancestor, v, of u (if any) such that u is de-
scended from the right child of v.
If there is no such ancestor, then pred(u) is undefined.

The running time of TREE-SUCCESSOR on a tree of height his O(h), since we either follow a path up the tree or follow a path down the tree. The procedure TREE-PREDECESSOR, which is symmetric to TREE-SUCCESSOR, also runs in time O(h).

Taking the nodes one at a time and applying the rule:
node C: Does not have a left child. Closest ancestor such that node-C is de-
scended from the right child is node-A. Therefore, the predecessor of node-C
is node-A.
node A: Has a left child, node-B. Rightmost descendent of node-B is node-E.
node E: Has a left child, node-F. Rightmost descendent of node-F is node-G.
node G: Does not have a left child. Closest ancestor such that node-G is de-
scended from the right child is node-F.
node F: Does not have a left child. Closest ancestor such that node-F is de-
scended from the right child is node-B.
node B: Has a left child, node-D. Rightmost descendent of node-D is node-D
itself.
node D: Does not have a left child. There is no ancestor such that node-D is
descended from the right child. Therefore, the predecessor of node-D is
undefined.