Decimal representation of given binary string is divisible by 20 or not
Last Updated :
19 Apr, 2023
The problem is to check whether the decimal representation of the given binary number is divisible by 20 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 the minimum numbers of multiplication and division operations. No leadings 0’s are there in the input.
Examples:
Input : 101000
Output : Yes
(10100)2 = (40)10
and 40 is divisible by 20.
Input : 110001110011100
Output : Yes
Approach:
- Initialize a variable 'dec' to 0, which will store the decimal value of the binary number.
- Traverse the binary number from left to right and for each digit, multiply the current value of 'dec' by 2 and add the current binary digit (0 or 1) to it.
- After the traversal is complete, the final value of 'dec' will be the decimal representation of the binary number.
Once we have the decimal representation of the binary number, we can check whether it is divisible by 20 by using the modulo operator (%). If the decimal representation is divisible by 20, then the binary number is also divisible by 20.
Implementation:
C++
// C++ implementation to check whether
// decimal representation of given binary
// number is divisible by 20 or not
#include <bits/stdc++.h>
using namespace std;
// function to check whether decimal
// representation of given binary number
// is divisible by 20 or not
bool isDivisibleBy20(char bin[], int n)
{
// convert binary number to decimal
int dec = 0;
for (int i = 0; i < n; i++) {
dec = dec * 2 + (bin[i] - '0');
}
// check if decimal representation is divisible by 20
if (dec % 20 == 0) {
return true;
}
else {
return false;
}
}
// Driver program to test above
int main()
{
char bin[] = "101000";
int n = sizeof(bin) / sizeof(bin[0]);
if (isDivisibleBy20(bin, n - 1))
cout << "Yes";
else
cout << "No";
return 0;
}
// Java implementation to check whether
// decimal representation of given binary
// number is divisible by 20 or not
import java.util.*;
class Main {
// function to check whether decimal
// representation of given binary number
// is divisible by 20 or not
static boolean isDivisibleBy20(char[] bin, int n)
{
// convert binary number to decimal
int dec = 0;
for (int i = 0; i < n; i++) {
dec = dec * 2 + (bin[i] - '0');
}
// check if decimal representation is divisible by
// 20
if (dec % 20 == 0) {
return true;
}
else {
return false;
}
}
// Driver program to test above
public static void main(String[] args)
{
char[] bin = { '1', '0', '1', '0', '0', '0' };
int n = bin.length;
if (isDivisibleBy20(bin, n)) {
System.out.println("Yes");
}
else {
System.out.println("No");
}
}
}
// This code is contributed by Prajwal Kandekar
# function to check whether decimal
# representation of given binary number
# is divisible by 20 or not
def is_divisible_by_20(bin, n):
# convert binary number to decimal
dec = 0
for i in range(n):
dec = dec * 2 + int(bin[i])
# check if decimal representation is divisible by 20
if dec % 20 == 0:
return True
else:
return False
# Driver program to test above
if __name__ == '__main__':
bin = "101000"
n = len(bin)
if is_divisible_by_20(bin, n):
print("Yes")
else:
print("No")
using System;
class MainClass {
// function to check whether decimal
// representation of given binary number
// is divisible by 20 or not
static bool IsDivisibleBy20(char[] bin, int n)
{
// convert binary number to decimal
int dec = 0;
for (int i = 0; i < n; i++) {
dec = dec * 2 + (bin[i] - '0');
}
// check if decimal representation is divisible by
// 20
if (dec % 20 == 0) {
return true;
}
else {
return false;
}
}
// Driver program to test above
public static void Main()
{
char[] bin = "101000".ToCharArray();
int n = bin.Length;
if (IsDivisibleBy20(bin, n - 1)) {
Console.WriteLine("Yes");
}
else {
Console.WriteLine("No");
}
}
}
function isDivisibleBy20(bin) {
// Convert binary number to decimal
let dec = 0;
for (let i = 0; i < bin.length; i++) {
dec = dec * 2 + parseInt(bin[i]);
}
// Check if decimal representation is divisible by 20
if (dec % 20 === 0) {
return true;
} else {
return false;
}
}
// Driver program to test above
let bin = "101000";
if (isDivisibleBy20(bin)) {
console.log("Yes");
} else {
console.log("No");
}
Output: Yes
Time Complexity: O(N)
Space Complexity: O(1)
Approach: Following are the steps:
- Let the binary string be bin[].
- Let the length of bin[] be n.
- If bin[n-1] == '1', then it is an odd number and thus not divisible by 20.
- Else check if bin[0..n-2] is divisible by 10. Refer to this post.
C++
// C++ implementation to check whether
// decimal representation of given binary
// number is divisible by 20 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(char bin[], int n)
{
// 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;
}
// function to check whether decimal
// representation of given binary number
// is divisible by 20 or not
bool isDivisibleBy20(char bin[], int n)
{
// if 'bin' is an odd number
if (bin[n - 1] == '1')
return false;
// check if bin(0..n-2) is divisible
// by 10 or not
return isDivisibleBy10(bin, n - 1);
}
// Driver program to test above
int main()
{
char bin[] = "101000";
int n = sizeof(bin) / sizeof(bin[0]);
if (isDivisibleBy20(bin, n - 1))
cout << "Yes";
else
cout << "No";
return 0;
}
// Java implementation to check whether
// decimal representation of given binary
// number is divisible by 20 or not
import java.io.*;
class GFG {
// function to check whether decimal
// representation of given binary number
// is divisible by 10 or not
static boolean isDivisibleBy10(char bin[], int n)
{
// 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;
}
// function to check whether decimal
// representation of given binary number
// is divisible by 20 or not
static boolean isDivisibleBy20(char bin[], int n)
{
// if 'bin' is an odd number
if (bin[n - 1] == '1')
return false;
// check if bin(0..n-2) is divisible
// by 10 or not
return isDivisibleBy10(bin, n - 1);
}
// Driver program to test above
public static void main(String args[])
{
char bin[] = "101000".toCharArray();
int n = bin.length;
if (isDivisibleBy20(bin, n - 1))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed
// by Nikita Tiwari.
# Python 3 implementation to check whether
# decimal representation of given binary
# number is divisible by 20 or not
# function to check whether decimal
# representation of given binary number
# is divisible by 10 or not
def isDivisibleBy10(bin, n):
# 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 in range(n - 2, -1, -1):
# 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
# if last digit is 0, then
# divisible by 10
if (sum % 10 == 0):
return True
# not divisible by 10
return False
# function to check whether decimal
# representation of given binary number
# is divisible by 20 or not
def isDivisibleBy20(bin, n):
# if 'bin' is an odd number
if (bin[n - 1] == '1'):
return false
# check if bin(0..n-2) is divisible
# by 10 or not
return isDivisibleBy10(bin, n - 1)
# Driver program to test above
bin = ['1','0','1','0','0','0']
n = len(bin)
if (isDivisibleBy20(bin, n - 1)):
print("Yes")
else:
print("No")
# This code is contributed by Smitha Dinesh Semwal
// C# implementation to check whether
// decimal representation of given binary
// number is divisible by 20 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)
{
// 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;
}
// function to check whether decimal
// representation of given binary number
// is divisible by 20 or not
static bool isDivisibleBy20(string bin, int n)
{
// if 'bin' is an odd number
if (bin[n - 1] == '1')
return false;
// check if bin(0..n-2) is divisible
// by 10 or not
return isDivisibleBy10(bin, n - 1);
}
// Driver program to test above
public static void Main()
{
string bin = "101000";
int n = bin.Length;
if (isDivisibleBy20(bin, n - 1))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed
// by vt_m.
<?php
// PHP implementation to check whether
// decimal representation of given binary
// number is divisible by 20 or not
// function to check whether decimal
// representation of given binary number
// is divisible by 10 or not
function isDivisibleBy10($bin, $n)
{
// if last digit is '1', then
// number is not divisible by 10
if ($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;
}
// function to check whether decimal
// representation of given binary number
// is divisible by 20 or not
function isDivisibleBy20($bin, $n)
{
// if 'bin' is an odd number
if ($bin[$n - 1] == '1')
return false;
// check if bin(0..n-2) is divisible
// by 10 or not
return isDivisibleBy10($bin, $n - 1);
}
// Driver code
$bin= "101000";
$n = strlen($bin);
if (isDivisibleBy20($bin, $n - 1))
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 20 or not
// function to check whether decimal
// representation of given binary number
// is divisible by 10 or not
function isDivisibleBy10(bin, n)
{
// 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
var sum = 0;
// traverse from the 2nd last up
// to 1st digit in 'bin'
for (var i = n - 2; i >= 0; i--) {
// if digit in '1'
if (bin[i] == '1') {
// calculate digit's position from
// the right
var 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;
}
// function to check whether decimal
// representation of given binary number
// is divisible by 20 or not
function isDivisibleBy20(bin, n)
{
// if 'bin' is an odd number
if (bin[n - 1] == '1')
return false;
// check if bin(0..n-2) is divisible
// by 10 or not
return isDivisibleBy10(bin, n - 1);
}
// Driver program to test above
var bin = "101000";
var n = bin.length;
if (isDivisibleBy20(bin, n - 1))
document.write( "Yes");
else
document.write( "No");
</script>
Time Complexity: O(n), where n is the number of digits in the binary number.
Auxiliary Space: O(1)
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