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Decimal representation of given binary string is divisible by 5 or not

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  • Difficulty Level : Hard
  • Last Updated : 12 Jul, 2022
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The problem is to check whether the decimal representation of the given binary number is divisible by 5 or not. Take care, the number could be very large and may not fit even in long long int. The approach should be such that there are zero or minimum number of multiplication and division operations. No leading 0’s are there in the input. 

Examples:

Input : 1010
Output : YES
(1010)2 = (10)10,
and 10 is divisible by 5.

Input : 10000101001
Output : YES
Recommended Practice

Approach: 

The following steps are:

  1. Convert the binary number to base 4.
  2. Numbers in base 4 contains only 0, 1, 2, 3 as their digits.
  3. 5 in base 4 is equivalent to 11.
  4. Now apply the rule of divisibility by 11 where you add all the digits at odd places and add all the digits at even places and then subtract one from the other. If the result is divisible by 11(which remember is 5), then the binary number is divisible by 5.

How to convert binary number to base 4 representation?

  1. Check whether the length of binary string is even or odd.
  2. If odd, the add ‘0’ in the beginning of the string.
  3. Now, traverse the string from left to right.
  4. One by extract substrings of size 2 and add their equivalent decimal to the resultant string.

Implementation:

C++




// C++ implementation to check whether decimal representation
// of given binary number is divisible by 5 or not
#include <bits/stdc++.h>
 
using namespace std;
 
// function to return equivalent base 4 number
// of the given binary number
int equivalentBase4(string bin)
{
    if (bin.compare("00") == 0)
        return 0;
    if (bin.compare("01") == 0)
        return 1;
    if (bin.compare("10") == 0)
        return 2;
    return 3;
}
 
// function to check whether the given binary
// number is divisible by 5 or not
string isDivisibleBy5(string bin)
{
    int l = bin.size();
     
    if (l % 2 != 0)
    // add '0' in the beginning to make
    // length an even number
        bin = '0' + bin;
     
    // to store sum of digits at odd and
    // even places respectively
    int odd_sum, even_sum = 0;
     
    // variable check for odd place and
    // even place digit
    int isOddDigit = 1;
    for (int i = 0; i<bin.size(); i+= 2)
    {
        // if digit of base 4 is at odd place, then
        // add it to odd_sum
        if (isOddDigit)
            odd_sum += equivalentBase4(bin.substr(i, 2));
        // else digit of base 4 is at even place,
        // add it to even_sum
        else
            even_sum += equivalentBase4(bin.substr(i, 2));
         
        isOddDigit ^= 1;
    }
     
    // if this diff is divisible by 11(which is 5 in decimal)
    // then, the binary number is divisible by 5
    if (abs(odd_sum - even_sum) % 5 == 0)
        return "Yes";
     
    // else not divisible by 5
    return "No";
             
}
 
// Driver program to test above
int main()
{
    string bin = "10000101001";
    cout << isDivisibleBy5(bin);
    return 0;
}

Java




//Java implementation to check whether decimal representation
//of given binary number is divisible by 5 or not
 
class GFG
{
    // Method to return equivalent base 4 number
    // of the given binary number
    static int equivalentBase4(String bin)
    {
        if (bin.compareTo("00") == 0)
            return 0;
        if (bin.compareTo("01") == 0)
            return 1;
        if (bin.compareTo("10") == 0)
            return 2;
        return 3;
    }
     
    // Method to check whether the given binary
    // number is divisible by 5 or not
    static String isDivisibleBy5(String bin)
    {
        int l = bin.length();
         
        if (l % 2 != 0)
        // add '0' in the beginning to make
        // length an even number
            bin = '0' + bin;
         
        // to store sum of digits at odd and
        // even places respectively
        int odd_sum=0, even_sum = 0;
         
        // variable check for odd place and
        // even place digit
        int isOddDigit = 1;
        for (int i = 0; i<bin.length(); i+= 2)
        {
            // if digit of base 4 is at odd place, then
            // add it to odd_sum
            if (isOddDigit != 0)
                odd_sum += equivalentBase4(bin.substring(i, i+2));
            // else digit of base 4 is at even place,
            // add it to even_sum
            else
                even_sum += equivalentBase4(bin.substring(i, i+2));
             
            isOddDigit ^= 1;
        }
         
        // if this diff is divisible by 11(which is 5 in decimal)
        // then, the binary number is divisible by 5
        if (Math.abs(odd_sum - even_sum) % 5 == 0)
            return "Yes";
         
        // else not divisible by 5
        return "No";
                 
    }
     
    public static void main (String[] args)
    {
        String bin = "10000101001";
        System.out.println(isDivisibleBy5(bin));
    }
}

Python 3




# Python3 implementation to check whether
# decimal representation of given binary
# number is divisible by 5 or not
 
# function to return equivalent base 4
# number of the given binary number
def equivalentBase4(bin):
    if(bin == "00"):
        return 0
    if(bin == "01"):
        return 1
    if(bin == "10"):
        return 2
    if(bin == "11"):
        return 3
     
# function to check whether the given
# binary number is divisible by 5 or not    
def isDivisibleBy5(bin):
    l = len(bin)
    if((l % 2) == 1):
         
    # add '0' in the beginning to
    # make length an even number    
        bin = '0' + bin
         
    # to store sum of digits at odd
    # and even places respectively
    odd_sum = 0
    even_sum = 0
    isOddDigit = 1
    for i in range(0, len(bin), 2):
         
        # if digit of base 4 is at odd place,
        # then add it to odd_sum
        if(isOddDigit):
            odd_sum += equivalentBase4(bin[i:i + 2])
             
        # else digit of base 4 is at
        # even place, add it to even_sum    
        else:
            even_sum += equivalentBase4(bin[i:i + 2])
             
        isOddDigit = isOddDigit ^ 1
 
    # if this diff is divisible by 11(which is
    # 5 in decimal) then, the binary number is
    # divisible by 5
    if(abs(odd_sum - even_sum) % 5 == 0):
        return "Yes"
    else:
        return "No"
 
# Driver Code
if __name__=="__main__":
    bin = "10000101001"
    print(isDivisibleBy5(bin))
 
# This code is contributed
# by Sairahul Jella

C#




// C# implementation to check whether
// decimal representation of given
// binary number is divisible by 5 or not
using System;
 
class GFG
{
// Method to return equivalent base
// 4 number of the given binary number
public static int equivalentBase4(string bin)
{
    if (bin.CompareTo("00") == 0)
    {
        return 0;
    }
    if (bin.CompareTo("01") == 0)
    {
        return 1;
    }
    if (bin.CompareTo("10") == 0)
    {
        return 2;
    }
    return 3;
}
 
// Method to check whether the
// given binary number is divisible
// by 5 or not
public static string isDivisibleBy5(string bin)
{
    int l = bin.Length;
 
    if (l % 2 != 0)
    {
        // add '0' in the beginning to
        // make length an even number
        bin = '0' + bin;
    }
 
    // to store sum of digits at odd
    // and even places respectively
    int odd_sum = 0, even_sum = 0;
 
    // variable check for odd place
    // and even place digit
    int isOddDigit = 1;
    for (int i = 0; i < bin.Length; i += 2)
    {
        // if digit of base 4 is at odd
        // place, then add it to odd_sum
        if (isOddDigit != 0)
        {
            odd_sum += equivalentBase4(
                             bin.Substring(i, 2));
        }
         
        // else digit of base 4 is at even 
        // place, add it to even_sum
        else
        {
            even_sum += equivalentBase4(
                              bin.Substring(i, 2));
        }
 
        isOddDigit ^= 1;
    }
 
    // if this diff is divisible by
    // 11(which is 5 in decimal) then,
    // the binary number is divisible by 5
    if (Math.Abs(odd_sum - even_sum) % 5 == 0)
    {
        return "YES";
    }
 
    // else not divisible by 5
    return "NO";
 
}
 
// Driver Code
public static void Main(string[] args)
{
    string bin = "10000101001";
    Console.WriteLine(isDivisibleBy5(bin));
}
}
 
// This code is contributed by Shrikant13

Javascript




<script>
 
// Javascript implementation to check whether
// decimal representation of given binary
// number is divisible by 5 or not
 
// function to return equivalent base 4
// number of the given binary number
function equivalentBase4(bin){
    if(bin == "00")
        return 0
    if(bin == "01")
        return 1
    if(bin == "10")
        return 2
    if(bin == "11")
        return 3
}
     
// function to check whether the given
// binary number is divisible by 5 or not   
function isDivisibleBy5(bin){
    let l = bin.length
    if((l % 2) == 1){
         
    // add '0' in the beginning to
    // make length an even number   
        bin = '0' + bin
    }
         
    // to store sum of digits at odd
    // and even places respectively
    let odd_sum = 0
    let even_sum = 0
    let isOddDigit = 1
    for(let i=0;i<bin.length;i+=2){
         
        // if digit of base 4 is at odd place,
        // then add it to odd_sum
        if(isOddDigit)
            odd_sum += equivalentBase4(bin.substring(i,i + 2))
             
        // else digit of base 4 is at
        // even place, add it to even_sum   
        else
            even_sum += equivalentBase4(bin.substring(i,i + 2))
             
        isOddDigit = isOddDigit ^ 1
    }
 
    // if this diff is divisible by 11(which is
    // 5 in decimal) then, the binary number is
    // divisible by 5
    if(Math.abs(odd_sum - even_sum) % 5 == 0)
        return "Yes"
    else
        return "No"
}
 
// Driver Code
 
let    bin = "10000101001"
document.writeln(isDivisibleBy5(bin))
 
// This code is contributed by shinjanpatra
 
</script>

Output

No

Time Complexity: O(n), where n is the number of digits in the binary number. 
Auxiliary Space: O(1)

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


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