# Find the index of first 1 in an infinite sorted array of 0s and 1s

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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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)
Declare l = 0, h = 1
while arr[h] == 0
l = h
h = 2*h;
return indexOfFirstOne(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 ` `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. `

## Python3

 `# Python 3 implementation to find the ` `# index of first 1 in an infinite  ` `# sorted array of 0's and 1's ` ` `  `# function to find the index of first  ` `# '1' binary search technique is applied ` `def` `indexOfFirstOne(arr, low, high) : ` `     `  `    ``while` `(low <``=` `high) : ` `        ``mid ``=` `(low ``+` `high) ``/``/` `2` `  `  `        ``# if true, then 'mid' is the index ` `        ``# of first '1' ` `        ``if` `(arr[mid] ``=``=` `1` `and` `(mid ``=``=` `0` `or`  `                       ``arr[mid ``-` `1``] ``=``=` `0``)) : ` `            ``break` `  `  `        ``# first '1' lies to the left of 'mid' ` `        ``elif` `(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 ` `def` `posOfFirstOne(arr) : ` `   `  `    ``# find the upper and lower bounds between ` `    ``# which the first '1' would be present ` `    ``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 ` `arr ``=` `[ ``0``, ``0``, ``1``, ``1``, ``1``, ``1` `] ` `print``( ``"Index = "``, posOfFirstOne(arr)) ` ` `  `# This code is contributed by Nikita Tiwari. `

## C#

 `// C# Code For Find the index of first 1 ` `// in an infinite sorted array of 0's and 1's ` `using` `System; ` ` `  `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() ` `    ``{ ` `        ``int``[] arr = { 0, 0, 1, 1, 1, 1 }; ` `        ``Console.Write(``"Index = "` `+ posOfFirstOne(arr)); ` `    ``} ` `} ` ` `  `// This code is contributed by Sam007 `

## PHP

 ` `

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).

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