Cracking the Coding Interview, 4 Edition - 150 Programming Interview Questions and Solutions

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Solutions to Chapter 11 | System Design and Memory Limits 11.4 You have an array with all the numbers from 1 to N, where N is at most 32,000. The array may have duplicate entries and you do not know what N is. With only 4KB of memory available, how would you print all duplicate elements in the array? SOLUTION pg 72 We have 4KB of memory which means we can address up to 8 * 4 * (2^10) bits. Note that 32* (2^10) bits is greater than 32000. We can create a bit vector with 32000 bits, where each bit represents one integer. NOTE: While this isn’t an especially difficult problem, it’s important to implement this cleanly. We will define our own bit vector class to hold a large bit vector. 1 public static void checkDuplicates(int[] array) { 2 BitSet bs = new BitSet(32000); 3 for (int i = 0; i < array.length; i++) { 4 int num = array[i]; 5 int num0 = num - 1; // bitset starts at 0, numbers start at 1 6 if (bs.get(num0)) { 7 System.out.println(num); 8 } else { 9 bs.set(num0); 10 } 11 } 12 } 13 14 class BitSet { 15 int[] bitset; 16 17 public BitSet(int size) { 18 bitset = new int[size >> 5]; // divide by 32 19 } 20 21 boolean get(int pos) { 22 int wordNumber = (pos >> 5); // divide by 32 23 int bitNumber = (pos & 0x1F); // mod 32 24 return (bitset[wordNumber] & (1 > 5); // divide by 32 29 int bitNumber = (pos & 0x1F); // mod 32 30 bitset[wordNumber] |= 1
Solutions to Chapter 11 | System Design and Memory Limits 11.5 If you were designing a web crawler, how would you avoid getting into infinite loops? SOLUTION pg 72 First, how does the crawler get into a loop? The answer is very simple: when we re-parse an already parsed page. This would mean that we revisit all the links found in that page, and this would continue in a circular fashion. Be careful about what the interviewer considers the “same” page. Is it URL or content? One could easily get redirected to a previously crawled page. So how do we stop visiting an already visited page? The web is a graph-based structure, and we commonly use DFS (depth first search) and BFS (breadth first search) for traversing graphs. We can mark already visited pages the same way that we would in a BFS/DFS. We can easily prove that this algorithm will terminate in any case. We know that each step of the algorithm will parse only new pages, not already visited pages. So, if we assume that we have N number of unvisited pages, then at every step we are reducing N (N-1) by 1. That proves that our algorithm will continue until they are only N steps. SUGGESTIONS AND OBSERVATIONS »» This question has a lot of ambiguity. Ask clarifying questions! »» Be prepared to answer questions about coverage. »» What kind of pages will you hit with a DFS versus a BFS? » » What will you do when your crawler runs into a honey pot that generates an infinite subgraph for you to wander about? CareerCup.com 2 0 6
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CRACKING THE FOURTH EDITION C O D I
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Table of Contents Chapter 7 | Objec
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Introduction Something’s Wrong We
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Behind the Scenes For many candidat
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Behind the Scenes | The Amazon Inte
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Behind the Scenes | The Apple Inter
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Interview War Stories The View from
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Interview War Stories | Pop Divas P
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Interview War Stories | You Can (Ma
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Interview War Stories | Spider Sens
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Before the Interview | Resume Advic
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Before the Interview | Behavioral P
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Before the Interview | Technical Pr
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The Interview and Beyond
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At the Interview | Handling Behavio
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At the Interview | Handling Technic
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At the Interview | Five Algorithm A
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At the Interview | Five Algorithm A
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At the Interview | The Offer and Be
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At the Interview | Top Ten Mistakes
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At the Interview | Frequently Asked
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Interview Questions
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Chapter 1 | Arrays and Strings Hash
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Chapter 2 | Linked Lists How to App
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Chapter 3 | Stacks and Queues How t
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Chapter 4 | Trees and Graphs How to
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Part 2 Concepts and Algorithms
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Chapter 5 | Bit Manipulation 5.1 Yo
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Chapter 6 | Brain Teasers 6.1 Add a
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Chapter 7 | Object Oriented Design
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Chapter 8 | Recursion 8.1 Write a m
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Chapter 9 | Sorting and Searching 9
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Chapter 10 | Mathematical 10.1 You
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Chapter 11 | Testing 11.1 Find the
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Chapter 12 | System Design and Memo
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Chapter 13 | C++ How To Approach: A
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Chapter 14 | Java How to Approach:
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Chapter 15 | Databases How to Appro
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Chapter 16 | Low Level How to Appro
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Chapter 17 | Networking How to Appr
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Chapter 18 | Threads and Locks How
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Part 4 Additional Review Problems
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Chapter 19 | Moderate _____________
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Chapter 20 | Hard Input: DAMP, LIKE
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Solutions
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Solutions to Chapter 1 | Arrays and
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Solutions to Chapter 2 | Linked Lis
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Solutions to Chapter 3 | Stacks and
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Solutions to Chapter 4 | Trees and
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Solutions to Chapter 6 | Brain Teas
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Solutions to Chapter 7 | Object Ori
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Solutions to Chapter 18 | Threads a
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Solutions to Chapter 19 | Moderate
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Solutions to Chapter 20 | Hard 20.3
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Solutions to Chapter 20 | Hard 10 1
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Solutions to Chapter 20 | Hard Val_
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Index A arithmetic 108, 131, 143, 1
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Mock Interviews Mock Interviews Stu
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