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 are implementations of above algorithm

C++

// C++ program to demonstrate working of an algorithm that finds
// an element in an array of infinite size
#include<bits/stdc++.h>
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;
}

Java

// Java program to demonstrate working of
// an algorithm that finds an element in an
// array of infinite size

class Test
{
    // Simple binary search algorithm
    static 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;
    }

    // Method 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
    static 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
            //check that 2*h doesn't exceeds array
            //length to prevent ArrayOutOfBoundException
            if(2*h < arr.length-1)
               h = 2*h;
            else
               h = arr.length-1;

            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 method to test the above function
    public static void main(String[] args)
    {
        int arr[] = new int[]{3, 5, 7, 9, 10, 90,
                            100, 130, 140, 160, 170};
        int ans = findPos(arr,10);

        if (ans==-1)
            System.out.println("Element not found");
        else
            System.out.println("Element found at index " + ans);
    }
}

Python3

# Python Program to demonstrate working of an algorithm that finds
# an element in an array of infinite size

# Binary search algorithm implementation
def binary_search(arr,l,r,x):

    if r >= l:
        mid = l+(r-l)//2

        if arr[mid] == x:
            return mid

        if arr[mid] > x:
            return binary_search(arr,l,mid-1,x)

        return binary_search(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 a[] and we can assume size to be
# infinite in this function.
# NOTE THAT THIS FUNCTION ASSUMES a[] TO BE OF INFINITE SIZE
# THEREFORE, THERE IS NO INDEX OUT OF BOUND CHECKING
def findPos(a, key):

    l, h, val = 0, 1, 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 binary_search(a, l, h, key)

# Driver function
arr = [3, 5, 7, 9, 10, 90, 100, 130, 140, 160, 170]
ans = findPos(arr,10)
if ans == -1:
    print ("Element not found")
else:
    print("Element found at index",ans)

C#

// C# program to demonstrate working of an
// algorithm that finds an element in an
// array of infinite size
using System;

class GFG {

    // Simple binary search algorithm
    static 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;
    }

    // Method 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
    static 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 method to test the above
    // function
    public static void Main()
    {
        int []arr = new int[]{3, 5, 7, 9,
            10, 90, 100, 130, 140, 160, 170};
        int ans = findPos(arr, 10);

        if (ans == -1)
            Console.Write("Element not found");
        else
            Console.Write("Element found at "
                            + "index " + ans);
    }
}

// This code is contributed by nitin mittal.

PHP

<?php
// PHP program to demonstrate working
// of an algorithm that finds an
// element in an array of infinite size

// Simple binary search algorithm
function binarySearch($arr, $l,
                      $r, $x)
{
    if ($r >= $l)
    {
        $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
function findPos( $arr, $key)
{
    $l = 0; $h = 1;
    $val = $arr[0];

    // Find h to do binary search
    while ($val < $key)
    {

        // store previous high
        $l = $h;

         // double high index
        $h = 2 * $h;

        // update new val
        $val = $arr[$h];
    }

    // at this point we have
    // updated low and high
    // indices, Thus use binary
    // search between them
    return binarySearch($arr, $l,
                        $h, $key);
}

    // Driver Code
    $arr = array(3, 5, 7, 9, 10, 90, 100,
                 130, 140, 160, 170);
    $ans = findPos($arr, 10);
    if ($ans==-1)
        echo "Element not found";
    else
        echo "Element found at index " , $ans;

// This code is contributed by anuj_67.
?>

Javascript

<script>
// JavaScript program to demonstrate working of an algorithm that finds
// an element in an array of infinite size

// Simple binary search algorithm
function binarySearch(arr, l, r, x)
{
    if (r>=l)
    {
        let mid = l + Math.floor((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
function findPos(arr, key)
{
    let l = 0, h = 1;
    let 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
    let arr = [3, 5, 7, 9, 10, 90, 100, 130,
                            140, 160, 170];
    let ans = findPos(arr, 10);
    if (ans==-1)
        document.write("Element not found");
    else
        document.write("Element found at index " + ans);



// This code is contributed by Surbhi Tyagi.
</script>

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).
This article is contributed by Gaurav Sharma. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

Approach: The problem can be solved based on the following observation: 

• Since array is sorted we apply binary search but the length of array is infinite so that we take start = 0 and end = 1 .
• After that check value of target is greater than the value at end index,if it is true then change newStart = end + 1  and newEnd = end +(end – start +1)*2 and apply binary search .
• Otherwise , apply binary search in the old index values.

Below are implementations of above algorithm:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Simple binary search algorithm
int binarySearch(int arr[], int target, int start, int end)
{
    while (start <= end) {

        int mid = start + (end - start) / 2;

        if (target < arr[mid]) {
            end = mid - 1;
        }
        else if (target > arr[mid]) {
            start = mid + 1;
        }
        else {
            // ans found
            return mid;
        }
    }
    return -1;
}


// an algorithm that finds an element in an
// array of infinite size

int findPos(int arr[], int target)
{
    // first find the range
    // first start with a box of size 2
    int start = 0;
    int end = 1;

    // condition for the target to lie in the range
    while (target > arr[end]) {
        int temp = end + 1; // this is my new start
        // double the box value
        // end = previous end + sizeofbox*2
        end = end + (end - start + 1) * 2;
        start = temp;
    }
    return binarySearch(arr, target, start, end);
}

// Driver code
int main()
{

    int arr[]
        = { 3, 5, 7, 9, 10, 90, 100, 130, 140, 160, 170 };
    int target = 10;
    // Function call
    int ans = findPos(arr, target);
    if (ans == -1)
        cout << "Element not found";
    else
        cout << "Element found at index " << ans;

    return 0;
}

Java

// Java code to implement the approach

import java.io.*;
import java.util.*;

// Java program to demonstrate working of
// an algorithm that finds an element in an
// array of infinite size

public class GFG {

    static int findPos(int[] arr, int target)
    {
        // first find the range
        // first start with a box of size 2
        int start = 0;
        int end = 1;

        // condition for the target to lie in the range
        while (target > arr[end]) {
            int temp = end + 1; // this is my new start
            // double the box value
            // end = previous end + sizeofbox*2
            end = end + (end - start + 1) * 2;
            start = temp;
        }
        return binarySearch(arr, target, start, end);
    }
    static int binarySearch(int[] arr, int target,
                            int start, int end)
    {
        while (start <= end) {
            // find the middle element
            //            int mid = (start + end) / 2; //
            //            might be possible that (start +
            //            end) exceeds the range of int in
            //            java
            int mid = start + (end - start) / 2;

            if (target < arr[mid]) {
                end = mid - 1;
            }
            else if (target > arr[mid]) {
                start = mid + 1;
            }
            else {
                // ans found
                return mid;
            }
        }
        return -1;
    }

    // Driver code
    public static void main(String[] args)
    {
        int[] arr = { 3,   5,   7,   9,   10, 90,
                      100, 130, 140, 160, 170 };
        int target = 10;
        // Function call
        int ans = findPos(arr, target);
        if (ans == -1)
            System.out.println("Element not found");
        else
            System.out.println("Element found at index "
                               + ans);
    }
}

Python3

# Python program to demonstrate working of
# an algorithm that finds an element in an
# array of infinite size

import sys

def findPos(arr, target):
    # first find the range
    # first start with a box of size 2
    start = 0
    end = 1

    # condition for the target to lie in the range
    while target > arr[end]:
        temp = end + 1  # this is my new start
        # double the box value
        # end = previous end + sizeofbox*2
        end = end + (end - start + 1) * 2
        start = temp
    return binarySearch(arr, target, start, end)

def binarySearch(arr, target, start, end):
    while start <= end:
        # find the middle element
        #            int mid = (start + end) / 2; //
        #            might be possible that (start +
        #            end) exceeds the range of int in
        #            java
        mid = start + (end - start) // 2

        if target < arr[mid]:
            end = mid - 1
        elif target > arr[mid]:
            start = mid + 1
        else:
            # ans found
            return mid
    return -1

# Driver code
if __name__ == '__main__':
    arr = [3, 5, 7, 9, 10, 90, 100, 130, 140, 160, 170]
    target = 10
    # Function call
    ans = findPos(arr, target)
    if ans == -1:
        print("Element not found")
    else:
        print("Element found at index", ans)

Javascript

<script>

 // Javascript  code to implement the approach

// an algorithm that finds an element in an
// array of infinite size
function findPos( arr,  target)
    {
        // first find the range
        // first start with a box of size 2
        let start = 0;
        let end = 1;

        // condition for the target to lie in the range
        while (target > arr[end]) {
            let temp = end + 1; // this is my new start
            // double the box value
            // end = previous end + sizeofbox*2
            end = end + (end - start + 1) * 2;
            start = temp;
        }
        return binarySearch(arr, target, start, end);
    }
// Simple binary search algorithm
function binarySearch(arr, target, start, end)
{
    while (start <= end) {

        let mid = start + Math.floor((end - start) / 2);

        if (target < arr[mid]) {
            end = mid - 1;
        }
        else if (target > arr[mid]) {
            start = mid + 1;
        }
        else {
            // ans found
            return mid;
        }
    }
    return -1;
}
// Driver code
    let arr = [3, 5, 7, 9, 10, 90, 100, 130,
                            140, 160, 170];
// Function call
    let ans = findPos(arr, 10);
    if (ans==-1)
        document.write("Element not found");
    else
        document.write("Element found at index " + ans);


</script>

C#

// C# code to implement the approach

using System;

class GFG
{
static int FindPos(int[] arr, int target)
{
// First find the range
// First start with a box of size 2
int start = 0;
int end = 1;
    // Condition for the target to lie in the range
    while (target > arr[end])
    {
        int temp = end + 1; // This is my new start
        // Double the box value
        // End = previous end + sizeofbox*2
        end = end + (end - start + 1) * 2;
        start = temp;
    }
    return BinarySearch(arr, target, start, end);
}

static int BinarySearch(int[] arr, int target, int start, int end)
{
    while (start <= end)
    {
        // Find the middle element
        //            int mid = (start + end) / 2; //
        //            might be possible that (start +
        //            end) exceeds the range of int in
        //            java
        int mid = start + (end - start) / 2;

        if (target < arr[mid])
        {
            end = mid - 1;
        }
        else if (target > arr[mid])
        {
            start = mid + 1;
        }
        else
        {
            // ans found
            return mid;
        }
    }
    return -1;
}

// Driver code
public static void Main()
{
    int[] arr = { 3,   5,   7,   9,   10, 90,
                  100, 130, 140, 160, 170 };
    int target = 10;
    // Function call
    int ans = FindPos(arr, target);
    if (ans == -1)
        Console.WriteLine("Element not found");
    else
        Console.WriteLine("Element found at index "
                           + ans);
}
}

Output

Element found at index 4

Time Complexity: O(logN) 
Auxiliary Space: O(1)