Repeatedly search an element by doubling it after every successful search
Given an array “a[]” and integer “b”. Find whether b is present in a[] or not. If present, then double the value of b and search again. We repeat these steps until b is not found. Finally we return value of b.
Examples:
Input : a[] = {1, 2, 3}
b = 1
Output :4
Initially we start with b = 1. Since
it is present in array, it becomes 2.
Now 2 is also present in array b becomes 4 .
Since 4 is not present, we return 4.
Input : a[] = {1 3 5 2 12}
b = 3
Output :6
Question Source : Asked in Yatra.com Online Test
1) Sort the input array.
2) Keep doing binary search and doubling until the element is not present.
The below code using binary_search() in STL
C++
// C++ program to repeatedly search an element by// doubling it after every successful search#include <bits/stdc++.h>using namespace std; int findElement(int a[], int n, int b){ // Sort the given array so that binary search // can be applied on it sort(a, a + n); int max = a[n - 1]; // Maximum array element while (b < max) { // search for the element b present or // not in array if (binary_search(a, a + n, b)) b *= 2; else return b; } return b;} // Driver codeint main(){ int a[] = { 1, 2, 3 }; int n = sizeof(a) / sizeof(a[0]); int b = 1; cout << findElement(a, n, b); return 0;} |
Java
// Java program to repeatedly search an element by// doubling it after every successful searchimport java.util.Arrays;public class Test4 { static int findElement(int a[], int n, int b) { // Sort the given array so that binary search // can be applied on it Arrays.sort(a); int max = a[n - 1]; // Maximum array element while (b < max) { // search for the element b present or // not in array if (Arrays.binarySearch(a, b) > -1) b *= 2; else return b; } return b; } // Driver code public static void main(String args[]) { int a[] = { 1, 2, 3 }; int n = a.length; int b = 1; System.out.println(findElement(a, n, b)); }}// This article is contributed by Sumit Ghosh |
Python
# Python program to repeatedly search an element by# doubling it after every successful search def binary_search(a, x, lo = 0, hi = None): if hi is None: hi = len(a) while lo < hi: mid = (lo + hi)//2 midval = a[mid] if midval < x: lo = mid + 1 elif midval > x: hi = mid else: return mid return -1 def findElement(a, n, b): # Sort the given array so that binary search # can be applied on it a.sort() mx = a[n - 1] # Maximum array element while (b < max): # search for the element b present or # not in array if (binary_search(a, b, 0, n) != -1): b *= 2 else: return b return b # Driver codea = [ 1, 2, 3 ]n = len(a)b = 1print findElement(a, n, b) # This code is contributed by Sachin Bisht |
C#
// C# program to repeatedly search an// element by doubling it after every// successful searchusing System; public class GFG { static int findElement(int[] a, int n, int b) { // Sort the given array so that // binary search can be applied // on it Array.Sort(a); // Maximum array element int max = a[n - 1]; while (b < max) { // search for the element b // present or not in array if (Array.BinarySearch(a, b) > -1) b *= 2; else return b; } return b; } // Driver code public static void Main() { int[] a = { 1, 2, 3 }; int n = a.Length; int b = 1; Console.WriteLine(findElement(a, n, b)); }} // This code is contributed by vt_m. |
PHP
<?php// Php program to repeatedly search an element by// doubling it after every successful search function binary_search($a, $x, $lo=0, $hi=NULL){ if ($hi == NULL) $hi = count($a); while ($lo < $hi) { $mid = ($lo + $hi) / 2; $midval = $a[$mid]; if ($midval < $x) $lo = $mid + 1; else if ($midval > $x) $hi = $mid; else return $mid; } return -1;} function findElement($a, $n, $b){// Sort the given array so that binary search// can be applied on it sort($a); $mx = $a[$n - 1]; // Maximum array element while ($b < max($a)) { // search for the element b present or// not in array if (binary_search($a, $b, 0, $n) != -1) $b *= 2; else return $b;}return $b;} // Driver code$a = array(1, 2, 3 );$n = count($a);$b = 1;echo findElement($a, $n, $b); // This code is contributed by Srathore?> |
Output:
4
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