Check divisibility of binary string by 2^k

Given a binary string and a number k, 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 long binary string 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 last k bits. If the 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++

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// C++ implementation to check whether
// given binary number is evenly
// divisible by 2^k or not
#include <bits/stdc++.h>
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;
}

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Java














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// 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. 



















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Python 3














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# 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 



















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C#














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// 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. 



















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PHP














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<?php 


// PHP implementation to check whether 


// given binary number is evenly 


  


// function to check whether 


// given binary number is  


// evenly divisible by 2^k or not 


function isDivisible($str, $k) 


{ 


    $n = strlen($str); 


    $c = 0; 


      


    // count of number  


    // of 0 from last 


    for ($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 Code 


    // first example 


    $str1 = "10101100"; 


    $k = 2; 


    if (isDivisible($str1, $k)) 


        echo "Yes", "\n"; 


    else


        echo "No", "\n"; 


  


    // Second example 


    $str2= "111010100"; 


    $k = 2; 


    if (isDivisible($str2, $k)) 


        echo "Yes", "\n"; 


    else


        echo "No", "\n"; 


  


// This code is contributed by Ajit 


?> 



















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Output:


Yes
Yes





Time Complexity: O(k)





    

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