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.

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If you are going to use it in your books, please contact the author via
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