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Top 100 Interview Questions & Answers in a convenient and easy to read book!

“I bought this guide a few days ago to prepare for my interview with Oracle. Many of the questions they asked me were from this guide. I found this book absolutely great!”

– Ravi, California

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  1. Submitted By: J1gs4w — June 2, 2008
    -6 votes
      + -

    This forum sucks to the core. What’s the point in setting up a forum and having a fictitious Ravi’s statement when the forum is not upto the standard.

  2. Submitted By: kk — June 3, 2008
    -8 votes
      + -

    Answer for 2:

    merge the two linked list directly
    and sort the combined linked list

  3. Submitted By: Shailesh Mittal — June 21, 2008
    +25 votes
      + -

    This code is written in C#. Again performance and error handling is not given a thought while writting this code. You can optimize it to level you want.

    class CoinGames
    {
    static int _NoofWays = 0;

    static void Main(string[] args)
    {
    int[] Coins = new int[] { 50, 25, 75, 15, 10 };

    // iterate through increasing set of numbers
    for (int i = 1; i

  4. Submitted By: Anirudh Aggarwal — July 15, 2008
    +26 votes
      + -

    int a[] = {50,25,75,15,10};
    int nSize = 5 // Array size
    int nWays= 0; //no of ways
    int nWordCount = 1;

    while(nWordCount

  5. Submitted By: Smitha Nataraj — July 30, 2008
    +27 votes
      + -

    Hi,

    I wrote a program for this..

    Not sure if someone can come up with a more efficient version!..

    public string permCoins()
    {
    string s = “”;
    int[] coins = new int[] {50, 25, 75, 15, 10};

    for (int combCount = 2; combCount

  6. Submitted By: Ankush Bindlish — October 23, 2008
    +4 votes
      + -

    2.Write a method to combine two two sorted linked list into one in sorted form with out using temporary Node.

    NODE MergedSortedList(NODE node1, NODE node2)
    {
    if(NULL == node1 ) return node2;
    if(NULL == node2) return node1;

    if(node1->info info)
    node1->next = MergedSortedList(node1->next,node2);
    return node1;
    }
    else if(node1->info > node2->info)
    node1->next = MergedSortedList(node1,node2->next);
    return node2;
    }
    else if(node1->info == node2->info)
    node1->next= node2;
    node2->next= MergedSortedList(node1->next,node2->next);
    return node1;
    }
    }

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