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

05 Jul, 2021
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 infinite, 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++
Java
Python3
C#
PHP
Javascript
// 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 appliedint 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'sint 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 aboveint main(){    int arr[] = { 0, 0, 1, 1, 1, 1 };    cout << "Index = "         << posOfFirstOne(arr);    return 0;}
// JAVA Code For Find the index of first 1// in an infinite sorted array of 0s and 1simport 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.
# 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 applieddef 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'sdef 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 programarr = [ 0, 0, 1, 1, 1, 1 ]print( "Index = ", posOfFirstOne(arr)) # This code is contributed by Nikita Tiwari.
// C# Code For Find the index of first 1// in an infinite sorted array of 0's and 1'susing 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// PHP 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 appliedfunction indexOfFirstOne($arr, $low, $high){    $mid;    while ($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'        else if ($arr[$mid] == 1)            $high = $mid - 1;         // first '1' lies to        // the right of 'mid'        else            $low = $mid + 1;    }     // required index    return ceil($mid);} // function to find the index of// first 1 in an infinite sorted// array of 0's and 1'sfunction 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, $h);} // Driver Code$arr = array(0, 0, 1, 1, 1, 1);echo "Index = " ,      posOfFirstOne($arr); // This code is contributed by anuj_67.?>
<script> // Javascript 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 appliedfunction indexOfFirstOne(arr, low, high){    var mid;    while (low <= high) {        mid = parseInt((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'sfunction posOfFirstOne(arr){    // find the upper and lower bounds between    // which the first '1' would be present    var 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 abovevar arr = [0, 0, 1, 1, 1, 1];document.write( "Index = "        + posOfFirstOne(arr)); // This code is contributed by rrrtnx.</script>
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 write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
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