Find position of an element in a sorted array of infinite numbers

Suppose you have a sorted array of infinite numbers, how would you search an element in the array?

Source: Amazon Interview Experience.

Since array is sorted, the first thing clicks into mind is binary search, but the problem here is that we don’t know size of array.
If the array is infinite, that means we don’t have proper bounds to apply binary search. So in order to find position of key, first we find bounds and then apply binary search algorithm.

Let low be pointing to 1st element and high pointing to 2nd element of array, Now compare key with high index element,
->if it is greater than high index element then copy high index in low index and double the high index.
->if it is smaller, then apply binary search on high and low indices found.
Below is the C++ implementation of above algorithm

// C++ program to demonstrate working of an algorithm that finds // an element in an array of infinite size #include<iostream> using namespace std;   // Simple binary search algorithm int binarySearch(int arr[], int l, int r, int x) {     if (r>=l)     {         int mid = l + (r - l)/2;         if (arr[mid] == x)             return mid;         if (arr[mid] > x)             return binarySearch(arr, l, mid-1, x);         return binarySearch(arr, mid+1, r, x);     }     return -1; }   // function takes an infinite size array and a key to be //  searched and returns its position if found else -1. // We don't know size of arr[] and we can assume size to be // infinite in this function. // NOTE THAT THIS FUNCTION ASSUMES arr[] TO BE OF INFINITE SIZE // THEREFORE, THERE IS NO INDEX OUT OF BOUND CHECKING int findPos(int arr[], int key) {     int l = 0, h = 1;     int val = arr[0];       // Find h to do binary search     while (val < key)     {         l = h;        // store previous high         h = 2*h;      // double high index         val = arr[h]; // update new val     }       // at this point we have updated low and high indices,     // thus use binary search between them     return binarySearch(arr, l, h, key); }   // Driver program int main() {     int arr[] = {3, 5, 7, 9, 10, 90, 100, 130, 140, 160, 170};     int ans = findPos(arr, 10);     if (ans==-1)         cout << "Element not found";     else         cout << "Element found at index " << ans;     return 0; } // C++ program to demonstrate working of an algorithm that finds // an element in an array of infinite size #include<iostream> using namespace std;   // Simple binary search algorithm int binarySearch(int arr[], int l, int r, int x) {     if (r>=l)     {         int mid = l + (r - l)/2;         if (arr[mid] == x)             return mid;         if (arr[mid] > x)             return binarySearch(arr, l, mid-1, x);         return binarySearch(arr, mid+1, r, x);     }     return -1; }   // function takes an infinite size array and a key to be //  searched and returns its position if found else -1. // We don't know size of arr[] and we can assume size to be // infinite in this function. // NOTE THAT THIS FUNCTION ASSUMES arr[] TO BE OF INFINITE SIZE int findPos(int arr[], int key) {     int l = 0, h = 1;     int val = arr[0];       // Find h to do binary search     while (val < key)     {         l = h;        // store previous high         h = 2*h;      // double high index         val = arr[h]; // update new val     }       // at this point we have updated low and high indices,     // thus use binary search between them     return binarySearch(arr, l, h, key); }   // Driver program int main() {     int arr[] = {3, 5, 7, 9, 10, 90, 100, 130, 140, 160, 170};     int ans = findPos(arr, 10);     if (ans==-1)         cout << "Element not found";     else         cout << "Element found at index " << ans;     return 0; }

Output:

Element found at index 4

Let p be the position of element to be searched. Number of steps for finding high index ‘h’ is O(Log p). The value of ‘h’ must be less than 2*p. The number of elements between h/2 and h must be O(p). Therefore, time complexity of Binary Search step is also O(Log p) and overall time complexity is 2*O(Log p) which is O(Log p).