Decimal representation of given binary string is divisible by 10 or not

The problem is to check whether the decimal representation of the given binary number is divisible by 10 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 : 101000
Output : Yes
(101000)2 = (40)10
and 40 is divisible by 10.

Input : 11000111001110
Output : Yes

Approach: First of all we need to know that last digit of pow(2, i) = 2, 4, 8, 6 if i % 4 is equal to 1, 2, 3, 0 respectively, where i is greater than equal to 1. So, in the binary representation we need to know the position of digit ‘1’ from the right, so as to know the perfect power of 2 with which it is going to be multiplied. This will help us to obtain the last digit of the required perfect power’s of 2. We can add these digits and then check whether the last digit of the sum is 0 or not which implies that the number is divisible by 10 or not. Note that if the last digit in the binary representation is ‘1’ then it represents an odd number, and thus not divisible by 10.

C++

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// C++ implementation to check whether decimal
// representation of given binary number is 
// divisible by 10 or not
#include <bits/stdc++.h>
using namespace std;
  
// function to check whether decimal representation 
// of given binary number is divisible by 10 or not
bool isDivisibleBy10(string bin)
{
    int n = bin.size();
      
    // if last digit is '1', then
    // number is not divisible by 10
    if (bin[n-1] == '1')
        return false;
      
    // to accumulate the sum of last digits
    // in perfect powers of 2
    int sum = 0;
      
    // traverse from the 2nd last up to 1st digit
    // in 'bin'
    for (int i=n-2; i>=0; i--)    
    {
        // if digit in '1'
        if (bin[i] == '1')
        {
            // calculate digit's position from
            // the right
            int posFromRight = n - i - 1;
              
            // according to the digit's position, 
            // obtain the last digit of the applicable 
            // perfect power of 2
            if (posFromRight % 4 == 1)
                sum = sum + 2;
            else if (posFromRight % 4 == 2)
                sum = sum + 4;
            else if (posFromRight % 4 == 3)
                sum = sum + 8;
            else if (posFromRight % 4 == 0)
                sum = sum + 6;            
        }
    }
      
    // if last digit is 0, then
    // divisible by 10
    if (sum % 10 == 0)
        return true;
      
    // not divisible by 10    
    return false;    
}
  
// Driver program to test above
int main()
{
    string bin = "11000111001110";
      
    if (isDivisibleBy10(bin))
        cout << "Yes";
    else
        cout << "No";    
          
    return 0;

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Java

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// Java implementation to check whether decimal
// representation of given binary number is 
// divisible by 10 or not
import java.util.*;
  
class GFG {
      
    // function to check whether decimal 
    // representation of given binary number
    // is divisible by 10 or not
    static boolean isDivisibleBy10(String bin)
    {
        int n = bin.length();
           
        // if last digit is '1', then
        // number is not divisible by 10
        if (bin.charAt(n - 1) == '1')
            return false;
           
        // to accumulate the sum of last 
        // digits in perfect powers of 2
        int sum = 0;
           
        // traverse from the 2nd last up to
        // 1st digit in 'bin'
        for (int i = n - 2; i >= 0; i--)    
        {
            // if digit in '1'
            if (bin.charAt(i) == '1')
            {
                // calculate digit's position
                // from the right
                int posFromRight = n - i - 1;
                   
                // according to the digit's 
                // position, obtain the last 
                // digit of the applicable 
                // perfect power of 2
                if (posFromRight % 4 == 1)
                    sum = sum + 2;
                else if (posFromRight % 4 == 2)
                    sum = sum + 4;
                else if (posFromRight % 4 == 3)
                    sum = sum + 8;
                else if (posFromRight % 4 == 0)
                    sum = sum + 6;            
            }
        }
           
        // if last digit is 0, then
        // divisible by 10
        if (sum % 10 == 0)
            return true;
           
        // not divisible by 10    
        return false;    
    }
      
    /* Driver program to test above function */
    public static void main(String[] args) 
    {
        String bin = "11000111001110";
           
        if (isDivisibleBy10(bin))
            System.out.print("Yes");
        else
            System.out.print("No");   
          
        }
    }
  
// This code is contributed by Arnav Kr. Mandal.    

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Python

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# Python implementation to check whether
# decimal representation of given binary
# number is divisible by 10 or not
  
  
# function to check whether decimal 
# representation of given binary number
# is divisible by 10 or not
def isDivisibleBy10(bin) :
    n = len(bin)
      
    #if last digit is '1', then
    # number is not divisible by 10
    if (bin[n - 1] == '1') :
        return False
          
    # to accumulate the sum of last 
    # digits in perfect powers of 2
    sum = 0
      
    #traverse from the 2nd last up to
    # 1st digit in 'bin'
      
    i = n - 2
    while i >= 0 :
          
        # if digit in '1'
        if (bin[i] == '1') :
            # calculate digit's position
            # from the right
            posFromRight = n - i - 1
              
            #according to the digit's 
            # position, obtain the last 
            # digit of the applicable 
            # perfect power of 2
            if (posFromRight % 4 == 1) :
                sum = sum + 2
            elif (posFromRight % 4 == 2) :
                sum = sum + 4
            elif (posFromRight % 4 == 3) :
                sum = sum + 8
            elif (posFromRight % 4 == 0) :
                sum = sum + 6
              
        i = i - 1
          
    # if last digit is 0, then
    # divisible by 10
    if (sum % 10 == 0) :
        return True
          
    # not divisible by 10 
    return False
          
  
# Driver program to test above function
  
bin = "11000111001110"
if (isDivisibleBy10(bin)== True) :
    print("Yes")
else :
    print("No")
  
# This code is contributed by Nikita Tiwari. 

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C#

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// C# implementation to check whether decimal
// representation of given binary number is
// divisible by 10 or not
using System;
  
class GFG {
  
    // function to check whether decimal
    // representation of given binary number
    // is divisible by 10 or not
    static bool isDivisibleBy10(String bin)
    {
        int n = bin.Length;
  
        // if last digit is '1', then
        // number is not divisible by 10
        if (bin[n - 1] == '1')
            return false;
  
        // to accumulate the sum of last
        // digits in perfect powers of 2
        int sum = 0;
  
        // traverse from the 2nd last up to
        // 1st digit in 'bin'
        for (int i = n - 2; i >= 0; i--) {
              
            // if digit in '1'
            if (bin[i] == '1') {
                  
                // calculate digit's position
                // from the right
                int posFromRight = n - i - 1;
  
                // according to the digit's
                // position, obtain the last
                // digit of the applicable
                // perfect power of 2
                if (posFromRight % 4 == 1)
                    sum = sum + 2;
                else if (posFromRight % 4 == 2)
                    sum = sum + 4;
                else if (posFromRight % 4 == 3)
                    sum = sum + 8;
                else if (posFromRight % 4 == 0)
                    sum = sum + 6;
            }
        }
  
        // if last digit is 0, then
        // divisible by 10
        if (sum % 10 == 0)
            return true;
  
        // not divisible by 10
        return false;
    }
  
    /* Driver program to test above function */
    public static void Main()
    {
        String bin = "11000111001110";
  
        if (isDivisibleBy10(bin))
            Console.Write("Yes");
        else
            Console.Write("No");
    }
}
  
// This code is contributed by Sam007

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PHP

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<?php
// PHP implementation to 
// check whether decimal
// representation of given 
// binary number is divisible
// by 10 or not
  
  
// function to check whether 
// decimal representation of 
// given binary number is 
// divisible by 10 or not
function isDivisibleBy10($bin)
{
    $n = strlen($bin);
      
    // if last digit is '1',
    // then number is not 
    // divisible by 10
    if ($bin[$n - 1] == '1')
        return false;
      
    // to accumulate the sum 
    // of last digits in 
    // perfect powers of 2
    $sum = 0;
      
    // traverse from the 2nd 
    // last up to 1st digit 
    // in 'bin'
    for ($i = $n - 2; $i >= 0; $i--) 
    {
        // if digit in '1'
        if ($bin[$i] == '1')
        {
            // calculate digit's 
            // position from the right
            $posFromRight = $n - $i - 1;
              
            // according to the digit's 
            // position, obtain the last 
            // digit of the applicable 
            // perfect power of 2
            if ($posFromRight % 4 == 1)
                $sum = $sum + 2;
            else if ($posFromRight % 4 == 2)
                $sum = $sum + 4;
            else if ($posFromRight % 4 == 3)
                $sum = $sum + 8;
            else if ($posFromRight % 4 == 0)
                $sum = $sum + 6;         
        }
    }
      
    // if last digit is 0, then
    // divisible by 10
    if ($sum % 10 == 0)
        return true;
      
    // not divisible by 10 
    return false; 
}
  
// Driver Code
$bin = "11000111001110";
if(isDivisibleBy10($bin))
    echo "Yes";
else
    echo "No";
  
// This code is contributed by mits. 
?>

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Output:

Yes

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

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