Friday, October 10, 2014

No. 57 - Integer Identical to Index


Problem: Integers in an array are unique and increasingly sorted. Please write a function/method to find an integer from the array who equals to its index. For example, in the array {-3, -1, 1, 3, 5}, the number 3 equals its index 3.
Analysis: If we scan all integers in the array from beginning to end, we may check whether every element equals its index. Obviously, this solution costs O(n) time.
Since numbers are sorted in the array, let's try to utilize the binary search algorithm to optimize. Supposing we reach the ith element in the array at some step. If the value of element is also i, it is a target integer and let's return it.
What would happen when the value m is greater than the index i? For any k (k>0), the value of element with index i+k should be greater than or equal to m+k, because integers are unique and increasingly sorted in the array. Additionally because m>i, m+k>i+k. Therefore, every element on the right side of index i should be greater than its index in such a case.
Similarly, when the value of element with index i is less than i, every integer on the left side should be less than its index. Please prove it if you are interested.
Therefore, we could reduce the search scope to half for the next step, and it is a typical process for binary search. The solution can be implemented with the following Java code:
public static int getNumberSameAsIndex(int[] numbers) {
    if(numbers == null || numbers.length == 0) {
        return -1;
    }
       
    int left = 0;
    int right = numbers.length - 1;
    while(left <= right) {
        int middle = left + ((right - left) >>> 1);
        if(numbers[middle] == middle) {
            return middle;
        }
           
        if(numbers[middle] > middle) {
            right = middle - 1;
        }
        else {
            left = middle + 1;
        }
    }
       
    return -1;
}

The source code with unit test cases is shared at http://ideone.com/ZSd9kG.
More coding interview questions are discussed in my book< Coding Interviews: Questions, Analysis & Solutions>. You may find the details of this book on Amazon.com, or Apress.

The author Harry He owns all the rights of this post. If you are going to use part of or the whole of this article in your blog or webpages, please add a reference to http://codercareer.blogspot.com/. If you are going to use it in your books, please contact the author via zhedahht@gmail.com . Thanks.

36 comments:

  1. May "because m>k, m+k>i+k" should be "because m>i, m+k>i+k" ?

    ReplyDelete
    Replies
    1. Thanks for your findings. It's a typo, and been fixed.

      Delete
  2. Very good explanation about this program
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    ReplyDelete
  3. It looks like there are better solutions available in the book
    Cracking Programming Interviews: 500 Questions With Solutions by Sergei Nakariakov
    http://www.amazon.com/Cracking-Programming-Interviews-Questions-Solutions/dp/1495459802/

    ReplyDelete
  4. what if the list contains more than 1 element whose value equals index? e.g.
    [-2, 1, 2, 3]

    ReplyDelete
    Replies
    1. The function which returns anyone of 1, 2, or 3 should be Ok.

      Delete
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  11. Hi There,

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  15. //Hello this solution is also acceptable right? Response will be highy appreciable Here according to my solution all the conncection between the element and the index is if we do i-a[i] on the left side is positive and on the right side is negative

    //Based on that

    public static int getIndexIdentical(int[] a,int low,int high)
    {
    if(a==null || a.length==0)
    return -1;
    int mid;
    if(low<=high)
    {
    mid=(low+high)/2;
    if(mid-a[mid]==0)
    return mid;
    else if(mid-a[mid]>0)
    return getIndexIdentical(a,mid+1,high);
    else
    return getIndexIdentical(a,low,mid+1);
    }
    return -1;
    }

    ReplyDelete
  16. In the Else part its is
    return getIndexIdentical(a,low,mid-1);

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