# Check divisibility of binary string by 2^k

Last Updated : 29 Sep, 2023

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.

Step-by-step approach:

• Convert the binary string to decimal by iterating over each digit from left to right.
• Calculate the value of 2^k using bitwise left shift operation (1 << k).
• Check if the decimal value obtained in step 1 is divisible by the value obtained in step 2 using the modulus operator.
• If the modulus is zero, return true, otherwise return false.

Below is the implementation of the above approach:

## C++

 #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) {     // convert binary string to decimal     int decimal_num = 0;     int base = 1;     int n = strlen(str);     for (int i = n - 1; i >= 0; i--) {         if (str[i] == '1') {             decimal_num += base;         }         base *= 2;     }       // check if decimal_num is divisible by 2^k     return (decimal_num % (1 << k)) == 0; }   // 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

 import java.util.Arrays;   public class Main {       // function to check whether     // given binary number is     // evenly divisible by 2^k or not     public static boolean isDivisible(String str, int k) {         // convert binary string to decimal         int decimal_num = 0;         int base = 1;         int n = str.length();         for (int i = n - 1; i >= 0; i--) {             if (str.charAt(i) == '1') {                 decimal_num += base;             }             base *= 2;         }           // check if decimal_num is divisible by 2^k         return (decimal_num % (1 << k)) == 0;     }       // Driver program to test above     public static void main(String[] args) {         // first example         String str1 = "10101100";         int k = 2;         if (isDivisible(str1, k))             System.out.println("Yes");         else             System.out.println("No");           // Second example         String str2 = "111010100";         k = 2;         if (isDivisible(str2, k))             System.out.println("Yes");         else             System.out.println("No");     } }

## Python3

 # function to check whether # given binary number is # evenly divisible by 2^k or not def is_divisible(binary_str, k):     # convert binary string to decimal     decimal_num = 0     base = 1     n = len(binary_str)     for i in range(n - 1, -1, -1):         if binary_str[i] == '1':             decimal_num += base         base *= 2       # check if decimal_num is divisible by 2^k     return decimal_num % (1 << k) == 0   # Driver program to test above if __name__ == "__main__":     # first example     str1 = "10101100"     k = 2     if is_divisible(str1, k):         print("Yes")     else:         print("No")       # Second example     str2 = "111010100"     k = 2     if is_divisible(str2, k):         print("Yes")     else:         print("No")   # This code is contributed by shivamgupta0987654321

## C#

 using System;   class GFG {     // function to check whether     // given binary number is     // evenly divisible by 2^k or not     static bool IsDivisible(string binaryStr, int k)     {         // convert binary string to decimal         int decimalNum = 0;         int baseValue = 1;         int n = binaryStr.Length;         for (int i = n - 1; i >= 0; i--)         {             if (binaryStr[i] == '1')             {                 decimalNum += baseValue;             }             baseValue *= 2;         }           // check if decimalNum is divisible by 2^k         return (decimalNum % (1 << k)) == 0;     }       // Driver program to test above     static void Main()     {         // first example         string str1 = "10101100";         int k = 2;         if (IsDivisible(str1, k))             Console.WriteLine("Yes");         else             Console.WriteLine("No");           // Second example         string str2 = "111010100";         k = 2;         if (IsDivisible(str2, k))             Console.WriteLine("Yes");         else             Console.WriteLine("No");     } }

## Javascript

 // Javascript code   // Function to check whether the given binary number is evenly divisible by 2^k or not function isDivisible(binaryStr, k) {     // Convert binary string to decimal     let decimalNum = 0;     let base = 1;     for (let i = binaryStr.length - 1; i >= 0; i--) {         if (binaryStr[i] === '1') {             decimalNum += base;         }         base *= 2;     }       // Check if decimalNum is divisible by 2^k     return decimalNum % (1 << k) === 0; }   // Driver program to test above const str1 = "10101100"; let k = 2; if (isDivisible(str1, k)) {     console.log("Yes"); } else {     console.log("No"); }   const str2 = "111010100"; k = 2; if (isDivisible(str2, k)) {     console.log("Yes"); } else {     console.log("No"); }   // This code is contributed by guptapratik

Output

Yes
Yes

Time Complexity:  O(n), where n is the length of the binary string.
Auxiliary Space: O(1), as we are not using any extra space.

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



Output

Yes
Yes

Complexity Analysis:

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