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Divisibility by 64 with removal of bits allowed

  • Difficulty Level : Medium
  • Last Updated : 27 Apr, 2021

Given a binary string, we need to check whether that number is divisible by 64 or not after removing of some bits. If yes then print “possible” else “not possible”. We cannot make number 0 to make it divisible.
Example : 
 

Input: 100010001 
Output: Possible
Explanation: We can get string 1 000 000 
after removing two ones which is a 
representation of number 64 in the binary 
numerical system.

Input: 100
Output: Not possible
Explanation : The number is 4 which is not 
divisible by 64 or cannot be made possible 
my removing some digits.

 

If we have 6 zeros after any one, then we can remove other bits represent it as a multiple of 64. So we just need to check if there is a 1 before six zeros. 
 

C++




// CPP program to find if given binary string can
// become divisible by 64 after removing some bits.
#include <iostream>
using namespace std;
 
// function to check if it is possible
// to make it a multiple of 64.
bool checking(string s)
{
    int c = 0; // counter to count 0's
    int n = s.length(); // length of the string
 
    // loop which traverses right to left and
    // calculates the number of zeros before 1.
    for (int i = n - 1; i >= 0; i--) {
        if (s[i] == '0')
            c++;
 
        if (c >= 6 and s[i] == '1')
            return true;
    }
 
    return false;
}
 
// driver code
int main()
{
    string s = "100010001";
    if (checking(s))
        cout << "Possible";
    else
        cout << "Not possible";
    return 0;
}

Java




// Java program to find if
// given binary string can
// become divisible by
// 64 after removing some bits
import java.io.*;
 
class GFG
{
    // function to check if it is possible
    // to make it a multiple of 64.
    static boolean checking(String s)
    {
        // counter to count 0's
        int c = 0;
         
        // length of the string
        int n = s.length();
     
        // loop which traverses right to left and
        // calculates the number of zeros before 1.
        for (int i = n - 1; i >= 0; i--)
        {
            if (s.charAt(i) == '0')
                c++;
     
            if (c >= 6 && s.charAt(i) == '1')
                return true;
        }
     
        return false;
    }
 
    // Driver code
    public static void main (String[] args)
    {
        String s = "100010001";
        if (checking(s))
            System.out.println ( "Possible");
        else
            System.out.println ( "Not possible");
     
    }
}
 
// This code  is contributed by vt_m

Python3




# Python 3 program to find if given binary
# string can become divisible by 64 after
# removing some bits.
 
# function to check if it is possible
# to make it a multiple of 64.
def checking(s):
    c = 0
     
    # counter to count 0's
    n = len(s)
     
    # length of the string
 
    # loop which traverses right to left and
    # calculates the number of zeros before 1.
    i = n - 1
    while(i >= 0):
        if (s[i] == '0'):
            c += 1
 
        if (c >= 6 and s[i] == '1'):
            return True
         
        i -= 1
     
    return False
 
# Driver code
if __name__ == '__main__':
    s = "100010001"
    if (checking(s)):
        print("Possible")
    else:
        print("Not possible")
     
# This code is contributed by
# Surendra_Gangwar

C#




// C# program to find if given binary
// string can become divisible by 64
// after removing some bits
using System;
 
class GFG {
     
    // function to check if it is possible
    // to make it a multiple of 64.
    static bool checking(string s)
    {
         
        // counter to count 0's
        int c = 0;
         
        // length of the string
        int n = s.Length;
     
        // loop which traverses right to
        // left and calculates the number
        // of zeros before 1.
        for (int i = n - 1; i >= 0; i--)
        {
            if (s[i] == '0')
                c++;
     
            if (c >= 6 && s[i] == '1')
                return true;
        }
     
        return false;
    }
 
    // Driver code
    public static void Main ()
    {
        String s = "100010001";
         
        if (checking(s))
            Console.WriteLine (
                          "Possible");
        else
            Console.WriteLine(
                     "Not possible");
    }
}
 
// This code is contributed by vt_m.

PHP




<?php
// PHP program to find if
// given binary string can
// become divisible by 64
// after removing some bits.
 
// function to check if
// it is possible
// to make it a multiple of 64.
 
function checking($s)
{
    // counter to count 0's
    $c = 0;
     
    // length of the string
    $n = strlen($s);
 
    // loop which traverses right
    // to left and calculates the
    // number of zeros before 1.
    for ($i = $n - 1; $i >= 0; $i--)
    {
        if ($s[$i] == '0')
            $c++;
 
        if ($c >= 6 and $s[$i] == '1')
            return true;
    }
 
    return false;
}
 
// Driver Code
$s = "100010001";
if (checking($s))
    echo "Possible";
else
    echo "Not possible";
 
// This code is contributed by ajit
?>

Javascript




<script>
 
// Javascript program to find if given binary
// string can become divisible by 64
// after removing some bits
 
    // function to check if it is possible
    // to make it a multiple of 64.
    function checking(s)
    {
           
        // counter to count 0's
        let c = 0;
           
        // length of the string
        let n = s.length;
       
        // loop which traverses right to
        // left and calculates the number
        // of zeros before 1.
        for (let i = n - 1; i >= 0; i--)
        {
            if (s[i] == '0')
                c++;
       
            if (c >= 6 && s[i] == '1')
                return true;
        }
       
        return false;
    }
     
// Driver code
        let s = "100010001";
           
        if (checking(s))
            document.write(
                          "Possible");
        else
            document.write(
                     "Not possible");
    
   // This code is contributed by code_hunt.
</script>

Output : 
 

Possible

Time Complexity: O(length of string)
 



This article is contributed by Twinkle Bajaj. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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