# Check if ceil of number divided by power of two exist in sorted array

Given a sorted array arr[] and an integer K, the task is to check if there exists a ceil of number K divided by some power of 2 in the array.
Note: If there is no such element print -1.
Examples:

Input: arr[] = {3, 5, 7, 8, 10}, K = 4
Output: -1
Explanation:
There is no such element.

Input: arr[] = {1, 2, 3, 5, 7, 8}, K = 4
Output:
Explanation:
When 4 is divided by 2 to the power 1, there exist an element in the array.

Approach: The idea is to try for every power of 2 and check that there exists a ceil of that number starting from power as 0. Checking an element exists in the array or not can be done using Binary Search.

Below is the implementation of the above approach:

## C++14

 `// C++14 implementation to check ` `// if a number divided by power ` `// of two exist in the sorted array ` `#include   ` `using` `namespace` `std;  ` ` `  `// Function to find there exist a ` `// number or not in the array ` `int` `findNumberDivByPowerofTwo(``int` `ar[],  ` `                              ``int` `k, ``int` `n) ` `{ ` `    ``int` `found = -1, m = k; ` ` `  `    ``// Loop to check if there exist ` `    ``// a number by divided by power of 2 ` `    ``while` `(m > 0) ` `    ``{ ` `        ``int` `l = 0; ` `        ``int` `r = n - 1; ` ` `  `        ``// Binary Search ` `        ``while` `(l <= r) ` `        ``{ ` `            ``int` `mid = (l + r) / 2; ` ` `  `            ``if` `(ar[mid] == m) ` `            ``{ ` `                ``found = m; ` `                ``break``; ` `            ``} ` `            ``else` `if` `(ar[mid] > m) ` `            ``{ ` `                ``r = mid - 1; ` `            ``} ` `            ``else` `if` `(ar[mid] < m) ` `            ``{ ` `                ``l = mid + 1; ` `            ``} ` `        ``} ` ` `  `        ``// Condition to check the number ` `        ``// is found in the array or not ` `        ``if` `(found != -1) ` `        ``{ ` `            ``break``; ` `        ``} ` ` `  `        ``// Otherwise divide the number ` `        ``// by increasing the one more ` `        ``// power of 2 ` `        ``m = m / 2; ` `    ``} ` `    ``return` `found; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 3, 5, 7, 8, 10 }; ` `    ``int` `k = 4, n = 5; ` `     `  `    ``cout << findNumberDivByPowerofTwo(arr, k, n); ` `} ` ` `  `// This code is contributed by code_hunt `

## Java

 `// Java implementation to check ` `// if a number divided by power ` `// of two exist in the sorted array ` ` `  `import` `java.util.Scanner; ` ` `  `public` `class` `GreeksForGreeksQuestions { ` ` `  `    ``// Function to find there exist a ` `    ``// number or not in the array ` `    ``static` `int` `findNumberDivByPowerofTwo( ` `        ``int``[] ar, ``int` `k, ``int` `n) ` `    ``{ ` `        ``int` `found = -``1``, m = k; ` ` `  `        ``// Loop to check if there exist ` `        ``// a number by divided by power of 2 ` `        ``while` `(m > ``0``) { ` `            ``int` `l = ``0``; ` `            ``int` `r = n - ``1``; ` ` `  `            ``// Binary Search ` `            ``while` `(l <= r) { ` `                ``int` `mid = (l + r) / ``2``; ` ` `  `                ``if` `(ar[mid] == m) { ` `                    ``found = m; ` `                    ``break``; ` `                ``} ` `                ``else` `if` `(ar[mid] > m) { ` `                    ``r = mid - ``1``; ` `                ``} ` `                ``else` `if` `(ar[mid] < m) { ` `                    ``l = mid + ``1``; ` `                ``} ` `            ``} ` ` `  `            ``// Condition to check the number ` `            ``// is found in the array or not ` `            ``if` `(found != -``1``) { ` `                ``break``; ` `            ``} ` ` `  `            ``// Otherwise divide the number ` `            ``// by increasing the one more ` `            ``// power of 2 ` `            ``m = m / ``2``; ` `        ``} ` ` `  `        ``return` `found; ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `arr[] = { ``3``, ``5``, ``7``, ``8``, ``10` `}; ` `        ``int` `k = ``4``, n = ``5``; ` ` `  `        ``System.out.println( ` `            ``findNumberDivByPowerofTwo( ` `                ``arr, k, n)); ` `    ``} ` `} `

## Python3

 `# Python3 implementation to check ` `# if a number divided by power ` `# of two exist in the sorted array ` ` `  `# Function to find there exist a ` `# number or not in the array ` `def` `findNumberDivByPowerofTwo(ar, k, n): ` `     `  `    ``found ``=` `-``1` `    ``m ``=` `k ` ` `  `    ``# Loop to check if there exist ` `    ``# a number by divided by power of 2 ` `    ``while` `(m > ``0``): ` `        ``l ``=` `0` `        ``r ``=` `n ``-` `1` ` `  `        ``# Binary Search ` `        ``while` `(l <``=` `r): ` `            ``mid ``=` `(l ``+` `r) ``/``/` `2` ` `  `            ``if` `(ar[mid] ``=``=` `m): ` `                ``found ``=` `m ` `                ``break` `             `  `            ``elif` `(ar[mid] > m): ` `                ``r ``=` `mid ``-` `1` `             `  `            ``elif` `(ar[mid] < m): ` `                ``l ``=` `mid ``+` `1` `             `  `        ``# Condition to check the number ` `        ``# is found in the array or not ` `        ``if` `(found !``=` `-``1``): ` `            ``break` `         `  `        ``# Otherwise divide the number ` `        ``# by increasing the one more ` `        ``# power of 2 ` `        ``m ``=` `m ``/``/` `2` `     `  `    ``return` `found ` ` `  `# Driver Code ` `arr ``=` `[ ``3``, ``5``, ``7``, ``8``, ``10` `] ` `k ``=` `4` `n ``=` `5` ` `  `print``(findNumberDivByPowerofTwo(arr, k, n)) ` ` `  `# This code is contributed by code_hunt `

## C#

 `// C# implementation to check ` `// if a number divided by power ` `// of two exist in the sorted array ` `using` `System; ` ` `  `class` `GFG{ ` `     `  `// Function to find there exist a ` `// number or not in the array ` `static` `int` `findNumberDivByPowerofTwo(``int``[] ar, ``int` `k, ` `                                     ``int` `n) ` `{ ` `    ``int` `found = -1, m = k; ` ` `  `    ``// Loop to check if there exist ` `    ``// a number by divided by power of 2 ` `    ``while` `(m > 0)  ` `    ``{ ` `        ``int` `l = 0; ` `        ``int` `r = n - 1; ` ` `  `        ``// Binary Search ` `        ``while` `(l <= r) ` `        ``{ ` `            ``int` `mid = (l + r) / 2; ` ` `  `            ``if` `(ar[mid] == m) ` `            ``{ ` `                ``found = m; ` `                ``break``; ` `            ``} ` `            ``else` `if` `(ar[mid] > m) ` `            ``{ ` `                ``r = mid - 1; ` `            ``} ` `            ``else` `if` `(ar[mid] < m) ` `            ``{ ` `                ``l = mid + 1; ` `            ``} ` `        ``} ` ` `  `        ``// Condition to check the number ` `        ``// is found in the array or not ` `        ``if` `(found != -1) ` `        ``{ ` `            ``break``; ` `        ``} ` ` `  `        ``// Otherwise divide the number ` `        ``// by increasing the one more ` `        ``// power of 2 ` `        ``m = m / 2; ` `    ``} ` `    ``return` `found; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `[]arr = { 3, 5, 7, 8, 10 }; ` `    ``int` `k = 4, n = 5; ` ` `  `    ``Console.WriteLine(findNumberDivByPowerofTwo( ` `                      ``arr, k, n)); ` `} ` `} ` ` `  `// This code is contributed by princi singh `

Output:

```-1
```

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Improved By : princi singh, code_hunt