Decimal representation of given binary string is divisible by 10 or not
Last Updated :
05 Apr, 2023
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++
// 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;
}
// 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.
# 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
else if (posFromRight % 4 == 2) :
sum = sum + 4
else if (posFromRight % 4 == 3) :
sum = sum + 8
else if (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.
// 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
<?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.
?>
<script>
// Javascript 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)
{
let 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
let sum = 0;
// traverse from the 2nd last up to
// 1st digit in 'bin'
for (let i = n - 2; i >= 0; i--)
{
// if digit in '1'
if (bin[i] == '1')
{
// calculate digit's position
// from the right
let 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 function
let bin = "11000111001110";
if (isDivisibleBy10(bin))
document.write("Yes");
else
document.write("No");
</script>
Output:
Yes
Time Complexity: O(N)
Auxiliary Space: O(1)
Method: Convert the given binary string in to decimal using int function then check if it is divisible by 10 or not using modulo division.
C++
#include <iostream>
using namespace std;
int main()
{
char s[] = "1010";
// converting binary string in to
// decimal number using stoi function
int n = stoi(s, 0, 2);
if (n % 10 == 0) {
cout << "Yes";
}
else {
cout << "No";
}
return 0;
}
// this code is contributed by Gangarajula Laxmi
# Python code to check
# decimal representation of
# a given binary string is
# divisible by 10 or not
str1 = "101000"
# converting binary string in to
# decimal number using int function
decnum = int(str1, 2)
# checking if number is divisible by 10
# or not if divisible print yes else no
if decnum % 10 == 0:
print("Yes")
else:
print("No")
# this code is contributed by gangarajula laxmi
// java code to check
// decimal representation of
// a given binary string is
// divisible by 10 or not
import java.io.*;
class GFG {
public static void main (String[] args) {
String s="1010";
//converting binary string in to
//decimal number using Convert.ToInt function
int n=Integer.parseInt(s,2);
if (n%10==0)
{
System.out.println("Yes");
}
else
{
System.out.println("No");
}
}
}
<script>
// JavaScript code for the above approach
str1 = "101000"
// converting binary string in to
// decimal number using int function
decnum = Number.parseInt(str1, 2)
// checking if number is divisible by 10
// or not if divisible print yes else no
if (decnum % 10 == 0)
document.write("Yes")
else
document.write("No")
// This code is contributed by Potta Lokesh
</script>
// C# code to check
// decimal representation of
// a given binary string is
// divisible by 10 or not
using System;
public class GFG{
static public void Main (){
string s="1010";
//converting binary string in to
//decimal number using Convert.ToInt function
int n=Convert.ToInt32(s,2);
if (n%10==0)
{
Console.Write("Yes");
}
else
{
Console.Write("No");
}
}
}
Time Complexity: O(n), where n is the number of digits in the binary number.
Auxiliary Space: O(1)
Method 3: Using the concept of modular arithmetic
We can also solve this problem without converting the binary string to decimal by using the concept of modular arithmetic. The idea is to iterate through the binary string from right to left, and keep track of the remainder after dividing the current number by 10. If the final remainder is 0, then the binary string is divisible by 10.
Python3
def is_divisible_by_10(binary_str):
remainder = 0
for digit in binary_str:
remainder = (remainder * 2 + int(digit)) % 10
return remainder == 0
# Example usage:
binary_str = '101010'
print(is_divisible_by_10(binary_str))
// THIS CODE IS CONTRIBUTED BY CHANDAN AGARWAL
function is_divisible_by_10(binary_str) {
let remainder = 0;
for (let i = 0; i < binary_str.length; i++) {
remainder = (remainder * 2 + parseInt(binary_str[i])) % 10;
}
return remainder == 0;
}
// Example usage:
let binary_str = '101010';
console.log(is_divisible_by_10(binary_str));
// Define a public class named GFG
public class GFG {
public static boolean
is_divisible_by_10(String binary_str)
{
int remainder = 0;
// Iterate over the binary digits of the input
// string
for (int i = 0; i < binary_str.length(); i++) {
// Convert the current binary digit to an
// integer
char digit = binary_str.charAt(i);
int num = Character.getNumericValue(digit);
// Update the remainder by shifting the bits to
// the left by 1 position and adding the current
// binary digit
remainder = (remainder * 2 + num) % 10;
}
// Return true if the remainder is 0 (i.e., the
// binary number is divisible by 10), otherwise
// return false
return remainder == 0;
}
// Define the main method
// The main method calls the is_divisible_by_10 method
// with an example binary string and prints the result
// to the console
public static void main(String[] args)
{
String binary_str = "101010";
System.out.println(is_divisible_by_10(binary_str));
}
}
#include <iostream>
#include <string>
using namespace std;
bool is_divisible_by_10(string binary_str) {
int remainder = 0;
for (int i = 0; i < binary_str.length(); i++) {
char digit = binary_str[i];
int num = digit - '0';
remainder = (remainder * 2 + num) % 10;
}
return remainder == 0;
}
int main() {
string binary_str = "101010";
cout << is_divisible_by_10(binary_str) << endl;
return 0;
}
using System;
class Program
{
static bool IsDivisibleBy10(string binaryStr)
{
int remainder = 0;
foreach (char digit in binaryStr)
{
remainder = (remainder * 2 + int.Parse(digit.ToString())) % 10;
}
return remainder == 0;
}
static void Main(string[] args)
{
string binaryStr = "101010";
Console.WriteLine(IsDivisibleBy10(binaryStr));
}
}
Time complexity: O(n)
Auxiliary Space: O(1)
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