Question: I was confronted not so long ago to an algorithmic problem.

I needed to find if a value stored in an array was at it "place".
An example will be easier to understand.

Let's take an Array A = {-10, -3, 3, 5, 7}. The algorithm would return 3, because the number 3 is at A[2] (3rd place).
On the contrary, if we take an Array B = {5, 7, 9, 10}, the algorithm will return 0 or false or whatever.

**The array is always sorted !**

I wasn't able to find a solution with a good complexity. (Looking at each value individualy is not good !) Maybe it is possible to resolve that problem by using an approach similar to merge sorting, by cuting in half and verifying on those halves ?

Can somebody help me on this one ? 
*Java algorithm would be the best, but pseudocode would also help me a lot !*

Answer: 
Since there can be no duplicates, you can use the fact that the function `f(x): A[x] - x` is monotonous and apply binary search to solve the problem in `O(log n)` worst-case complexity.

You want to find a point where that function `A[x] - x` takes value zero. This code should work:

```
boolean binarySearch(int[] data, int size)
{
    int low = 0;
    int high = size - 1;

    while(high >= low) {
        int middle = (low + high) / 2;
        if(data[middle] - 1 == middle) {
            return true;
        }
        if(data[middle] - 1 < middle) {
            low = middle + 1;
        }
        if(data[middle] - 1 > middle) {
            high = middle - 1;
        }
    }
    return false;
}

```

Watch out for the fact that arrays in Java are 0-indexed - that is the reason why I subtract `-1` from the array.
Using binary search (or a similar algorithm) you could get better than O(n). Since the array is sorted, we can make the following assumptions:

* if the value at index `x` is smaller than `x-1` (`a[x] <= x`), you know that all previous values also must be smaller than their index (because no duplicates are allowed)
* if `a[x] > x + 1` all following values must be greater than their index (again no duplicates allowed).

Using that you can use a binary approach and pick the center value, check for its index and discard the left/right part if it matches one of the conditions above. Of course you stop when `a[x] = x + 1`.