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# Check divisibility of binary string by 2^k

• Difficulty Level : Easy
• Last Updated : 17 Aug, 2022

Given a binary string and a number k, the task is to check whether the binary string is evenly divisible by 2k or not.

Examples :

```Input : 11000  k = 2
Output : Yes
Explanation :
(11000)2 = (24)10
24 is evenly divisible by 2k i.e. 4.

Input : 10101  k = 3
Output : No
Explanation :
(10101)2 = (21)10
21 is not evenly divisible by 2k i.e. 8.```

Naive Approach: Compute the binary string into decimal and then simply check whether it is divisible by 2k. But, this approach is not feasible for the long binary strings as time complexity will be high for computing decimal number from binary string and then dividing it by 2k.

Efficient Approach: In the binary string, check for the last k bits. If all the last k bits are 0, then the binary number is evenly divisible by 2k else it is not evenly divisible. Time complexity using this approach is O(k).

Below is the implementation of the approach.

## C++

 `// C++ implementation to check whether` `// given binary number is evenly` `// divisible by 2^k or not` `#include ` `using` `namespace` `std;`   `// function to check whether` `// given binary number is ` `// evenly divisible by 2^k or not` `bool` `isDivisible(``char` `str[], ``int` `k)` `{` `    ``int` `n = ``strlen``(str);` `    ``int` `c = 0;` `    `  `    ``// count of number of 0 from last` `    ``for` `(``int` `i = 0; i < k; i++)    ` `        ``if` `(str[n - i - 1] == ``'0'``)         ` `            ``c++;` `    `  `    ``// if count = k, number is evenly ` `    ``// divisible, so returns true else ` `    ``// false` `    ``return` `(c == k);` `}`   `// Driver program to test above` `int` `main()` `{` `    ``// first example` `    ``char` `str1[] = ``"10101100"``;` `    ``int` `k = 2;` `    ``if` `(isDivisible(str1, k))` `        ``cout << ``"Yes"` `<< endl;` `    ``else` `        ``cout << ``"No"` `             ``<< ``"\n"``;`   `    ``// Second example` `    ``char` `str2[] = ``"111010100"``;` `    ``k = 2;` `    ``if` `(isDivisible(str2, k))` `        ``cout << ``"Yes"` `<< endl;` `    ``else` `        ``cout << ``"No"` `<< endl;`   `    ``return` `0;` `}`

## Java

 `// Java implementation to check whether` `// given binary number is evenly` `// divisible by 2^k or not` `class` `GFG {` `    `  `    ``// function to check whether` `    ``// given binary number is ` `    ``// evenly divisible by 2^k or not` `    ``static` `boolean` `isDivisible(String str, ``int` `k)` `    ``{` `        ``int` `n = str.length();` `        ``int` `c = ``0``;` `    `  `        ``// count of number of 0 from last` `        ``for` `(``int` `i = ``0``; i < k; i++) ` `            ``if` `(str.charAt(n - i - ``1``) == ``'0'``)         ` `                ``c++;` `    `  `        ``// if count = k, number is evenly ` `        ``// divisible, so returns true else ` `        ``// false` `        ``return` `(c == k);` `    ``}`   `    ``// Driver program to test above` `    ``public` `static` `void` `main(String args[])` `    ``{ ` `        `  `        ``// first example` `        ``String str1 = ``"10101100"``;` `        ``int` `k = ``2``;` `        ``if` `(isDivisible(str1, k) == ``true``)` `            ``System.out.println(``"Yes"``);` `        ``else` `            ``System.out.println(``"No"``);`   `    ``// Second example` `        ``String str2 = ``"111010100"``;` `        ``k = ``2``;` `        ``if` `(isDivisible(str2, k) == ``true``)` `            ``System.out.println(``"Yes"``);` `        ``else` `            ``System.out.println(``"No"``);` `    ``}` `}`   `// This code is contributed by JaideepPyne.`

## Python3

 `# Python 3 implementation to check` `# whether given binary number is` `# evenly divisible by 2^k or not`   `# function to check whether` `# given binary number is ` `# evenly divisible by 2^k or not` `def` `isDivisible(``str``, k):` `    ``n ``=` `len``(``str``)` `    ``c ``=` `0` `    `  `    ``# count of number of 0 from last` `    ``for` `i ``in` `range``(``0``, k):` `        ``if` `(``str``[n ``-` `i ``-` `1``] ``=``=` `'0'``):     ` `            ``c ``+``=` `1` `    `  `    ``# if count = k, number is evenly ` `    ``# divisible, so returns true else ` `    ``# false` `    ``return` `(c ``=``=` `k)`   `# Driver program to test above` `# first example` `str1 ``=` `"10101100"` `k ``=` `2` `if` `(isDivisible(str1, k)):` `    ``print``(``"Yes"``)` `else``:` `    ``print``(``"No"``)`   `# Second example` `str2 ``=` `"111010100"` `k ``=` `2` `if` `(isDivisible(str2, k)):` `    ``print``(``"Yes"``)` `else``:` `    ``print``(``"No"``)`   `# This code is contributed by Smitha`

## C#

 `// C# implementation to check whether` `// given binary number is evenly` `// divisible by 2^k or not` `using` `System;`   `class` `GFG {` `    `  `    ``// function to check whether` `    ``// given binary number is ` `    ``// evenly divisible by 2^k or not` `    ``static` `bool` `isDivisible(String str, ``int` `k)` `    ``{` `        ``int` `n = str.Length;` `        ``int` `c = 0;` `    `  `        ``// count of number of 0 from last` `        ``for` `(``int` `i = 0; i < k; i++) ` `            ``if` `(str[n - i - 1] == ``'0'``)     ` `                ``c++;` `    `  `        ``// if count = k, number is evenly ` `        ``// divisible, so returns true else ` `        ``// false` `        ``return` `(c == k);` `    ``}`   `    ``// Driver program to test above` `    ``public` `static` `void` `Main()` `    ``{ ` `        `  `        ``// first example` `        ``String str1 = ``"10101100"``;` `        ``int` `k = 2;` `        `  `        ``if` `(isDivisible(str1, k) == ``true``)` `            ``Console.Write(``"Yes\n"``);` `        ``else` `            ``Console.Write(``"No"``);`   `        ``// Second example` `        ``String str2 = ``"111010100"``;` `        ``k = 2;` `        `  `        ``if` `(isDivisible(str2, k) == ``true``)` `            ``Console.Write(``"Yes"``);` `        ``else` `            ``Console.Write(``"No"``);` `    ``}` `}`   `// This code is contributed by Smitha.`

## PHP

 ``

## Javascript

 ``

Output

```Yes
Yes```

Complexity Analysis:

• Time Complexity: O(k)
• Auxiliary Space: O(1)

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