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Repeatedly search an element by doubling it after every successful search
• Difficulty Level : Easy
• Last Updated : 15 Mar, 2019

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
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

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

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 ``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 code``int` `main()``{``    ``int` `a[] = { 1, 2, 3 };``    ``int` `n = ``sizeof``(a) / ``sizeof``(a);``    ``int` `b = 1;``    ``cout << findElement(a, n, b);``    ``return` `0;``}`

## Java

 `// Java program to repeatedly search an element by``// doubling it after every successful search``import` `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 code``a ``=` `[ ``1``, ``2``, ``3` `]``n ``=` `len``(a)``b ``=` `1``print` `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 search``using` `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

 ` ``\$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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