Given a _sorted_ array A of distinct natural numbers, return the index of the fixed point if one exists, or otherwise, return `-1` to signal that no fixed point exists. Your algorithm must be as efficient as possible.

def find_fixed_point_natural(a): # YOUR CODE HERE

THE ALGORITHM'S TIME COMPLEXITY SHOULD BE FASTER THAN O(loglogn). If you think to answer like this below:

#include using namespace std; int binarySearch(int arr[], int low, int high) { if(high >= low) { int mid = (low + high)/2; /*low + (high - low)/2;*/ if(mid == arr[mid]) return mid; if(mid > arr[mid]) return binarySearch(arr, (mid + 1), high); else return binarySearch(arr, low, (mid -1)); } /* Return -1 if there is no Fixed Point */ return -1; } /* Driver code */ int main() { int n ; cin>>n; int arr[1000]; for(int i = 0; i>arr[i]; cout<<"Fixed Point is "<< binarySearch(arr, 0, n-1); return 0; } 

IF YOU SOLVE LIKE ABOVE PLEASE DO NOT ANSWER THIS QUESTION. THIS IS NOT A ANSWER FOR MY QUESTION. PLEASE FIND ANOTHER ALGORITHM TO ANSWER CORRECTLY..... YOU CAN NOT CHANGE THE INPUT AND NAME OF THE FUNCTION, PLEASE DO NOT ADD MORE ARGUMENTS. The ONLY ARGUMENT SHOULD BE SORTED ARRAY "A".

I ALSO NEED DIVIDE AND CONQUER STRATEGY AND PLEASE THINK ABOUT WHAT IS THE DIFFERENCE BTW NATURAL NUMBERS AND INTEGERS.