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Ans. O(1) for ArrayList O(n) for LinkedList | ||||
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Ans. Yes we can. But as the underlying structure of the collection class is double linked list, it will eventually have to traverse to the element linearlly and hence would result in bad performance. | ||||
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Ans. Arrays allows elements to be accessed directly using the index. As Array elements are stored in continuous memory locations it's very easy to find the memory address of any element using the formula as following Memory Address of Array start or index 0 + ( Size of array element * Index ) | ||||
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This question was recently asked at 'Spillman Technologies,Motorola Solutions'.This question is still unanswered. Can you please provide an answer. | ||||
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Ans. Use hashCode() which returns an integer value, generated by a hashing algorithm | ||||
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Ans. public class StackUsingArrayList { public static void main(String[] args) { List list = new ArrayList(); list.add("A"); list.add("B"); list.add("C"); Stack stack = new Stack(); list.forEach(a -> stack.add(a)); System.out.println(stack); } } | ||||
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Ans. Create one extra field called MAX O(1) when you an element to the stack check these two condition 1. Stack is empty, then MAX = element 2. Stack is not empty then check if the element is greater than MAX then MAX = element when getMax fuction is called, then return MAX | ||||
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Ans. list.get(index); | ||||
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Ans. Array List works on Array and when we add an element in middle of the list, Array List need to update the index of all subsequent elements. I the capacity is full, it even may need to move the whole list to a new memory location . Linked List works on Double linked list algorithm and all it has to do is to adjust the address of the previous and next elements. | ||||
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This question was recently asked at 'Amazon'.This question is still unanswered. Can you please provide an answer. | ||||
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Ans. public class BinarySearchTree { //Represent the node of binary tree public static class Node { int data; Node left; Node right; public Node(int data) { //Assign data to the new node, set left and right children to null this.data = data; this.left = null; this.right = null; } } //Represent the root of binary tree public Node root; public BinarySearchTree() { root = null; } //factorial() will calculate the factorial of given number public int factorial(int num) { int fact = 1; if (num == 0) return 1; else { while (num > 1) { fact = fact * num; num--; } return fact; } } //numOfBST() will calculate the total number of possible BST by calculating Catalan Number for given key public int numOfBST(int key) { int catalanNumber = factorial(2 * key) / (factorial(key 1) * factorial(key)); return catalanNumber; } public static void main(String[] args) { BinarySearchTree bt = new BinarySearchTree(); //Display total number of possible binary search tree with key 5 System.out.println("Total number of possible Binary Search Trees with given key: " bt.numOfBST(5)); } } | ||||
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Like Discuss Correct / Improve  Arraylist  linkedlist  queue  collections | ||||
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This question was recently asked at 'Amazon'.This question is still unanswered. Can you please provide an answer. | ||||
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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 deleteNode(Integer value) { head = start; while (head.nextNode != null) { if(head.nextNode.body == value){ head.nextNode = head.nextNode.nextNode; } 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.addNodeToEnd(5); ll.addNodeToEnd(10); ll.addNodeToEnd(15); ll.traverse(); ll.deleteNode(10); ll.traverse(); } } | ||||
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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. 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. Graph contain cycles whereas Trees cannot. | ||||
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Ans. Stack is based on Last in First out (LIFO) principle while a queue is based on FIFO (First In First Out) principle. | ||||
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This question was recently asked at 'Caprus IT'.This question is still unanswered. Can you please provide an answer. | ||||
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This question was recently asked at 'MarkMonitor'.This question is still unanswered. Can you please provide an answer. | ||||
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This question was recently asked at 'Bind Software Innovations'.This question is still unanswered. Can you please provide an answer. | ||||
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Ans. First Find the number of nodes in the linked list and take it as n.then find the data at n-3 element. | ||||
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Ans. We can merge two sorted array by using quick sort algorithm | ||||
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