Given an infinite sorted array consisting 0s and 1s. The problem is to find the index of first ‘1’ in that array. As the array is infinite, therefore it is guaranteed that number ‘1’ will be present in the array.

Examples:

Input : arr[] = {0, 0, 1, 1, 1, 1} Output : 2 Input : arr[] = {1, 1, 1, 1,, 1, 1} Output : 0

**Approach:** The problem is closely related to the problem of finding position of an element in a sorted array of infinite numbers. As the array is infinte, therefore we do not know the upper and lower bounds between which we have to find the occurrence of first ‘1’. Below is an algorithm to find the upper and lower bounds.

**Algorithm:**

posOfFirstOne(arr)Declarel= 0,h= 1 whilearr[h]== 0 l = h h = 2*h; returnindexOfFirstOne(arr, l, h)}

Here **h** and **l** are the required upper and lower bounds. **indexOfFirstOne(arr, l, h)** is used to find the index of occurrence of first ‘1’ between these two bounds. Refer this post.

## C++

// C++ implementation to find the index of first 1 // in an infinite sorted array of 0's and 1's #include <bits/stdc++.h> using namespace std; // function to find the index of first '1' // binary search technique is applied int indexOfFirstOne(int arr[], int low, int high) { int mid; while (low <= high) { mid = (low + high) / 2; // if true, then 'mid' is the index of first '1' if (arr[mid] == 1 && (mid == 0 || arr[mid - 1] == 0)) break; // first '1' lies to the left of 'mid' else if (arr[mid] == 1) high = mid - 1; // first '1' lies to the right of 'mid' else low = mid + 1; } // required index return mid; } // function to find the index of first 1 in // an infinite sorted array of 0's and 1's int posOfFirstOne(int arr[]) { // find the upper and lower bounds between // which the first '1' would be present int l = 0, h = 1; // as the array is being considered infinite // therefore 'h' index will always exist in // the array while (arr[h] == 0) { // lower bound l = h; // upper bound h = 2 * h; } // required index of first '1' return indexOfFirstOne(arr, l, h); } // Driver program to test above int main() { int arr[] = { 0, 0, 1, 1, 1, 1 }; cout << "Index = " << posOfFirstOne(arr); return 0; }

## Java

// JAVA Code For Find the index of first 1 // in an infinite sorted array of 0s and 1s import java.util.*; class GFG { // function to find the index of first '1' // binary search technique is applied public static int indexOfFirstOne(int arr[], int low, int high) { int mid=0; while (low <= high) { mid = (low + high) / 2; // if true, then 'mid' is the index of // first '1' if (arr[mid] == 1 && (mid == 0 || arr[mid - 1] == 0)) break; // first '1' lies to the left of 'mid' else if (arr[mid] == 1) high = mid - 1; // first '1' lies to the right of 'mid' else low = mid + 1; } // required index return mid; } // function to find the index of first 1 in // an infinite sorted array of 0's and 1's public static int posOfFirstOne(int arr[]) { // find the upper and lower bounds // between which the first '1' would // be present int l = 0, h = 1; // as the array is being considered // infinite therefore 'h' index will // always exist in the array while (arr[h] == 0) { // lower bound l = h; // upper bound h = 2 * h; } // required index of first '1' return indexOfFirstOne(arr, l, h); } /* Driver program to test above function */ public static void main(String[] args) { int arr[] = { 0, 0, 1, 1, 1, 1 }; System.out.println("Index = " + posOfFirstOne(arr)); } } // This code is contributed by Arnav Kr. Mandal.

Output:

Index = 2

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 **Ayush Jauhari**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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