# 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
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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Improved By : vt_m, princi singh

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