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Given a sorted array of distinct interger A[1...N] , you want to find out whether there is an index "i" for which A[i] =i . if this problem is solved using divide and conquer method,then find time complexity.
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We can use modification of binary search here :

Basically we have to find index i such that a[i] = i in other a[i] - i = 0.

Now as the array is sorted , so all the indices j which is less than i will satisfy the property : a[j] - j <= 0 as the array is sorted so a[j] will be at least 1 less than a[i] and left index will also be at least 1 less than i hence the conclusion.

Similarly to the right of the ith index element , we have a[j] - j >= 0 .

So the key idea is just we find :

mid = left + right / 2 where left and right are the boundary indices and then check whether mid is the ith index we are looking for using the above mentioned property.For this we compare only its immediate left and right neighbours.

If it satisfies , well and good and we will report our final answer which is index i as mid only else we apply binary search recursively like normal one.

So time complexity remains the same i.e. O(logn)

Reference : http://stackoverflow.com/questions/4101141/algorithm-to-find-if-there-is-any-i-so-that-arrayi-equals-i

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