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http://www.intel.com/software/products/documentation/vlin/mergedprojects/analyzer_ec/mergedprojects/reference_olh/mergedProjects/instructions/instruct32_hh/vc270.htm this might be helpful Less

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unsigned int a=123; // Given number unsigned int b=0; /// temp variable int i; for(i=0; i>1; } Less

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package array; import java.util.Hashtable; /* * You are given an array of numbers. * You need to print the length of the maximum continuous sequence that you encounter. * For example if input is [3,8,10,1,9,6,5,7,2 ], * the continuous sequences are{1,2,3} and {5,6,7,8,9,10} * the latter is the longest one so the answer becomes 6. * O(n) solution was asked for, * assuming you have a hash map which supports O(1) insertion and fetching operations * src: http://www.glassdoor.com/Interview/Microsoft-India-Interview-Questions-EI_IE1651.0,9_IL.10,15_IN115_IP6.htm */ public class FindLengthOfLongestRandomlyDistContinousSequenceOfNumber { public static void main(String[] str){ int a[] = {3,6,8,1,5,7,0,9,2,4,10,14,13,12,11}; System.out.println(findLength(a)); int b[] = {5,2,3,1,6,7,10,8}; System.out.println(findLength(b)); } private static int findLength(int[] a) { int result = 0; Hashtable h = new Hashtable(); for(int i=0; i result){ result = curr.maxValue; } if(next != null){ next = h.get( curr.max); next.maxValue = curr.maxValue; h.put(a[i]+1, next); next.min = curr.min; } if(prev != null){ prev = h.get(curr.min); prev.maxValue = curr.maxValue; h.put(a[i]-1, prev); prev.max = curr.max; } h.put(a[i], curr); } } return result; } } class Node{ int maxValue; int max; int min; public Node(int v, int m, int n){ maxValue = v; max = m; min = n; } } Less

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static int findlongconsecutivesubseq(int[] arr) { HashSet s= new HashSet(); int ans=0; //put all entries in hashset for(int i=0;i ans) ans=j-arr[i]; } } return ans; } Less

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@Ajit: What're the initial values of the max heap? What happens to the first value that comes in? Less

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Use selection sort for 'max' ( or for 'min') If K > series.length/2 then reverse the criteria- like in stead of looking for 15th highest out of 20 element array - look for (20 -15 =) 5th lowest and so on.... Less

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I can think of two ways: 1) traverse from both heads, get two length: long, short traverse again starting from Array1[long-short] and Array2[0], looking for the same pointer O(4n) time, O(1) space 2) use a hash table holds all seen pointers. traverse from both heads O(2n) time, O(n) space Less

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Step 1: if the linked lists are not empty, find the length of each linked list. Since they have a common end point, we will get length of common area of the linked lists too. Step 2: find the difference between the length of both list. Start traverse the biggest list till the difference in length between them, provided one of them is bigger than the other. Step 3: now they are at the same remaining length. Traverse both of the together till both are pointing to the same next node.That node is the common node which is the merging point. If there is no such point, then they are not merged. Less

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What types of coding questions did you get? Do you feel like the recruiters prepared you for all the technical portions of the interview? Less

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Hi! As an Amazon employee who interviewed and hired a lot of people here, I've created a guide that has all the questions and winning answers from an Amazonian recruiter perspective. Please check it out at interviewjoy.com/services/interview-process-details/amazon-senior-manager-interview-questions/ . Pls also check the positive feedback at the bottom of that page! Thanks. Less

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brute force method: O(n^2) 1st pass, count the number of occurrences for each of the 3 colours, x y and z 2nd pass: modify the original array: index 0 to x will be colour 1, x+1 to x+y is colour 2, and x+y+1 to x+y+z will be colour 3 Method 2, this can be sorted using a modified version of textbook sorting algorithm, if we treat the 3 colours as a primitive data type like an integer, for example: Red = 1, Blue = 2, Green = 3. Typical choices would be mergesort or quicksort for O(nlogn) public int[] colourSort(Colour[] array){ Colour[] helper = new Colour[array.length]; mergeSort(array, helper, 0, array.length); } mergeSort(Colour[] array, Colour[] helper, int start, int end){ int mid = start + (end - start)/2; while(start Less

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brute force method: O(n^2) 1st pass, count the number of occurrences for each of the 3 colours, x y and z 2nd pass: modify the original array: index 0 to x will be colour 1, x+1 to x+y is colour 2, and x+y+1 to x+y+z will be colour 3 Method 2, this can be sorted using a modified version of textbook sorting algorithm, if we treat the 3 colours as a primitive data type like an integer, for example: Red = 1, Blue = 2, Green = 3. Typical choices would be mergesort or quicksort for O(nlogn) public int[] colourSort(Colour[] array){ Colour[] helper = new Colour[array.length]; mergeSort(array, helper, 0, array.length); } mergeSort(Colour[] array, Colour[] helper, int start, int end){ int mid = start + (end - start)/2; while(start Less

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DP can also be used... O(m+n) again...

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import java.util.*; class gAlg{ public static void main(String[] args){ int[] a = new int[5]; int[] b = new int[5]; int[] c = new int[5]; int sum= 0; a[0]=2;a[1]=8;a[2]=9;a[3]=7;a[4]=10; b[0]=3;b[1]=2;b[2]=10;b[3]=9;b[4]=9; System.out.println(Arrays.toString(a)); System.out.println(Arrays.toString(b)); int i; for(i=0;i<5;i++){ c[i]=Math.max(a[i],b[i]); sum = sum+c[i]; } System.out.println(Arrays.toString(c)); System.out.println(sum); } } Less

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public class LinkedListElement { private E _element; private LinkedListElement _next; public LinkedListElement(E element) { _element = element; _next = null; } public LinkedListElement(E element, LinkedListElement next) { _element = element; _next = next; } public E get() { return _element; } public LinkedListElement next() { return _next; } public void setNext(LinkedListElement next) { _next = next; } public static LinkedListElement reverse(LinkedListElement e) { if (e == null) { return null; } LinkedListElement current = e; LinkedListElement next = e.next(); e.setNext(null); while (next != null) { LinkedListElement forward = next.next(); next.setNext(e); e = next; next = forward; } return e; } } Less

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public void reverse(ReverseSinglyLinkedList singlyLinkedList){ if(singlyLinkedList.isEmpty()){ return; } Node currentNode = head; Node nextNode = head.nextNode; Node markNode; head.nextNode = null; while(nextNode!=null){ markNode = nextNode.nextNode; nextNode.nextNode = currentNode; currentNode = nextNode; nextNode = markNode; } head = currentNode; } Less

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package test; import java.util.ArrayList; import java.util.List; public class TreeTraversal { static List preOrder = new ArrayList(); static List inOrder = new ArrayList(); static List postOrder = new ArrayList(); static void traverseTree(Node root) { Node left = root.getLeftNode(); Node right = root.getRightNode(); if (null == left.getLeftNode() && null == left.getRightNode() && null == right.getLeftNode() && null == right.getRightNode()) { preOrder.add(root.getValue()); preOrder.add(left.getValue()); preOrder.add(right.getValue()); inOrder.add(left.getValue()); inOrder.add(root.getValue()); inOrder.add(right.getValue()); postOrder.add(left.getValue()); postOrder.add(right.getValue()); postOrder.add(root.getValue()); } else { preOrder.add(root.getValue()); traverseTree(left); inOrder.add(root.getValue()); traverseTree(right); postOrder.add(root.getValue()); } } public static void main(String[] args) { Node left = new Node(2, new Node(1), new Node(3)); Node right = new Node(6, new Node(5), new Node(7)); Node root = new Node(4, left, right, true); traverseTree(root); System.out.print("Pre Order Traversal ::"); System.out.println(preOrder); System.out.print("In Order Traversal ::"); System.out.println(inOrder); System.out.print("Post Order Traversal ::"); System.out.println(postOrder); } } Less

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package test; import java.util.ArrayList; import java.util.Arrays; import java.util.Iterator; import java.util.List; enum TraversalType { PREORDER, INORDER, POSTORDER; } public class ConstructTreeBack { static Node root = new Node(); static TraversalType traversalType; static void formSubTrees(List treeList) { List leftSubTree = new ArrayList(); List rightSubTree = new ArrayList(); Iterator it = treeList.iterator(); int rootNodeVal = root.getValue(); while (it.hasNext()) { int nodeVal = it.next(); if (rootNodeVal > nodeVal) { leftSubTree.add(nodeVal); } else if (rootNodeVal treeList) { Node node = new Node(); if (traversalType.equals(TraversalType.PREORDER)) { if (null != treeList.get(0)) { node.setValue(treeList.get(0)); } if (null != treeList.get(1)) { node.setLeftNode(new Node(treeList.get(1))); } if (null != treeList.get(2)) { node.setRightNode(new Node(treeList.get(2))); } } else if (traversalType.equals(TraversalType.INORDER)) { if (null != treeList.get(1)) { node.setValue(treeList.get(1)); } if (null != treeList.get(0)) { node.setLeftNode(new Node(treeList.get(0))); } if (null != treeList.get(2)) { node.setRightNode(new Node(treeList.get(2))); } } else if (traversalType.equals(TraversalType.POSTORDER)) { if (null != treeList.get(2)) { node.setValue(treeList.get(2)); } if (null != treeList.get(0)) { node.setLeftNode(new Node(treeList.get(0))); } if (null != treeList.get(1)) { node.setRightNode(new Node(treeList.get(1))); } } return node; } public static void main(String[] args) { int rootNodeVal = 0; List treeList; /*PRE ORDER TRAVERSAL*/ Integer treeArrPreOrder[] = { 4, 2, 1, 3, 6, 5, 7 }; rootNodeVal = treeArrPreOrder[0]; root.setValue(rootNodeVal); root.setRoot(true); treeList = Arrays.asList(treeArrPreOrder); traversalType = TraversalType.PREORDER; formSubTrees(treeList); /*IN ORDER TRAVERSAL*/ Integer treeArrInOrder[] = { 1, 2, 3, 4, 5, 6, 7 }; int rootIndex = 3; root.setValue(treeArrInOrder[rootIndex]); root.setRoot(true); treeList = Arrays.asList(treeArrInOrder); traversalType = TraversalType.INORDER; formSubTrees(treeList); /*POST ORDER TRAVERSAL*/ Integer treeArrPostOrder[] = { 1, 3, 2, 5, 7, 6, 4 }; rootNodeVal = treeArrPostOrder[treeArrPostOrder.length - 1]; root.setValue(rootNodeVal); root.setRoot(true); treeList = Arrays.asList(treeArrPostOrder); traversalType = TraversalType.POSTORDER; formSubTrees(treeList); } } Less