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 algorithmint 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 CHECKINGint 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 programint 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 sizeclass 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 implementationdef 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 CHECKINGdef 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 functionarr = [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 sizeusing 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 algorithmfunction 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 CHECKINGfunction 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 algorithmfunction 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 CHECKINGfunction 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 algorithmint 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 sizeint 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 codeint 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 approachimport java.io.*;import java.util.*;// Java program to demonstrate working of// an algorithm that finds an element in an// array of infinite sizepublic 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 sizeimport sysdef 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 codeif __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 sizefunction 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 algorithmfunction 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 approachusing System;class GFG{static int FindPos(int[] arr, int target){// First find the range// First start with a box of size 2int 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 codepublic 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)
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