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Ans. Simplest implementation we can have is a List wherein we can place ordered words and hence can perform Binary Search. Other implementation with better search performance is to use HashMap with key as first character of the word and value as a LinkedList. Further level up, we can have linked Hashmaps like , hashmap { a ( key ) > hashmap (keyaa , value (hashmap(keyaaa,value) b ( key ) > hashmap (keyba , value (hashmap(keybaa,value) .................................................................................... z( key ) > hashmap (keyza , value (hashmap(keyzaa,value) } upto n levels ( where n is the average size of the word in dictionary.  
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Ans. public class LinkedList { Node start = null; Node head = null; class Node{ Integer body; Node nextNode; Node(Integer value){ body = value; } } private void insertInMiddle(Integer value){ head = start; if(start == null) { start = new Node(value); head = start; head.nextNode = null; return; } while(head.body < value){ if(head.nextNode == null  head.nextNode.body >= value){ Node newNode = new Node(value); newNode.nextNode = head.nextNode; head.nextNode = newNode; break; } head = head.nextNode; } } private void traverse(){ head = start; while(head != null){ System.out.println(head.body); head = head.nextNode; } } public static void main(String[] args){ LinkedList ll = new LinkedList(); ll.insertInMiddle(5); ll.insertInMiddle(10); ll.insertInMiddle(15); ll.insertInMiddle(7); ll.traverse(); } }  
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Ans. In Breadth first algorithm, all the adjacent nodes of the starting node is visited first and then the same rule is followed while moving inwards whereas In Depth first algorithm, all the nodes of a single traversal path are visited first till a cycle or an end is found. For example , given the following entries of adjacent nodes 1,2 1,3 1,6 2,4 2,5 3,6 The Breadth first path would be 1,2,3,6,4,5 and Depth first path would be 1,2,4,5,3,6  
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Ans. 1. Collection should have an index for random access. 2. Collection should have ordered elements.  
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Ans. int duplicateArray[] = { 1, 2, 2, 3, 4, 5, 6, 8, 9} Set unique = new HashSet(); for (int i = 0; i < duplicateArray.length; i) { if (unique.contains(duplicateArray[i])) { System.out.println(duplicateArray[i]); } else { unique.add(duplicateArray[i]); } } Complexity O(n) = nHashSet contains and add has O(n) = 1  
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Ans. Binary tree is a tree in which each node has up to two children.Tree is a data structure composed of nodes.Each tree has a root node(not necessary in graph theory). The root node has zero or more child nodes.Each child node has zero or more child nodes, and so on.The tree cannot contain cycles.  
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Ans. public class LinkedList { Node start = null; Node head = null; class Node { Integer body; Node nextNode; Node(Integer value) { body = value; } } private void addNodeToEnd(Integer value) { if (start == null) { start = new Node(value); head = start; head.nextNode = null; return; } while (head.nextNode != null) { head = head.nextNode; } head.nextNode = new Node(value); } private void traverse() { head = start; while (head != null) { System.out.println(head.body); head = head.nextNode; } } public static void main(String[] args) { LinkedList ll = new LinkedList(); ll.addNodeToEnd(5); ll.addNodeToEnd(10); ll.addNodeToEnd(15); ll.traverse(); } }  
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Ans. The sorting algorithm is a modified mergesort. This algorithm offers guaranteed n log(n) performance.  
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Ans. http://stackoverflow.com/questions/494830/howtodetermineifalinkedlisthasacycleusingonlytwomemorylocations  
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Ans. http://www.geeksforgeeks.org/findduplicatesinontimeandconstantextraspace/  
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Ans. https://en.wikipedia.org/wiki/Travelling_salesman_problem  
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Ans. Please not that all such questions can be easily answered through recursion. Simple recursive implementation could be traverse(root); void traverse(Element element){ if(element.hasNext()){ traverse(element.next()); } else { System.out.println(element); } } but this algo / code lead to endless loop if there is a loop in graph traversal. So you can keep a collection to keep track of which elements have laready been traversed static List<Elements> listOfAlreadyTraversedElements = new ArrayList<Elements>(); main(){ traverse(root); } void traverse(Element element){ if(element.hasNext()){ traverse(element.next()); } else { listOfAlreadyTraversedElements.add(element); System.out.println(element); } }  
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Ans. https://www.geeksforgeeks.org/flattenabinarytreeintolinkedlist/  
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Ans. https://www.geeksforgeeks.org/heapsort/  
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Ans. https://javasearch.buggybread.com/InterviewQuestions/questionSearch.php?searchOption=label'  
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Ans. Override the behavior of collection class to sort the list upon each element addition. Though it's not recommended as list are sorting heavy data structures. List<MyType> list = new ArrayList<MyType>() { public boolean add(MyType mt) { super.add(mt); Collections.sort(list, comparator); return true; } };  
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Ans. import java.util.*; class LinkedListSolution{ protected LinkedList list; public LinkedListSolution(){ list = new LinkedList(); } public Object pop() throws NoSuchElementException{ if(list.isEmpty()) throw new NoSuchElementException(); else return list.removeFirst(); } public void push(Object obj){ list.addFirst(obj); } public Object peek() throws NoSuchElementException{ if(list.isEmpty()) throw new NoSuchElementException(); else return list.getFirst(); } public boolean isEmpty(){ return list.isEmpty(); } public String toString(){ return list.toString(); } } class TestStack{ public static void main(String args[]){ LinkedListSolution s = new LinkedListSolution(); s.push("First"); s.push("Second"); s.push("Third"); System.out.println("Top: " s.peek()); s.push("Fourth"); while(!(s.isEmpty())) System.out.println(s.pop()); } }  
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Ans. array is traversed from first element to last element. Here current element is compared with next element. If current element is greater than next element it is swapped.  
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Ans. We can simply find it by checking the criteria of a tree. A tree will not contain a cycle, so if there is any cycle in the graph, it is not a tree. We can check it using another approach, if the graph is connected and it has V1 edges, it could be a tree. Here V is the number of vertices in the graph  
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Ans. 1 > 2 > 3 > 6 > 4  > 5  
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