Check if Decimal representation of given Binary String is divisible by K or not
Last Updated :
03 Aug, 2021
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Given a binary string S, the task is to find that the decimal representation of the given binary string is divisible by integer K or not.
Examples:
Input: S = 1010, k = 5
Output: Yes
Explanation: Decimal representation of 1010 (=10) is divisible by 5Input: S = 1010, k = 6
Output: No
Approach: Since the modulo operator is distributive over addition, the given binary string can be checked bit by bit to see whether the decimal % k is equal to zero or not. Follow the steps below for the approach:
- Initialize an array poweroftwo[], of size of binary string, to store the powers of two.
- Iterate till size and for each i store 2i % K in poweroftwo[].
- Initialize variables, say rem = 0, to store the current remaining number till i.
- Iterate till size and for each i and if S[size - i -1] is 1 then update rem equals rem + poweroftwo[i].
- Finally, return Yes if rem equals zero else return No.
Below is the implementation of the above approach:
// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;
// Function to check the binary number is
// divisible by K
string divisibleByk(string s, int n, int k)
{
// Array poweroftwo will store pow(2, i)%k
int poweroftwo[n];
// Initializing the first element in Array
poweroftwo[0] = 1 % k;
for (int i = 1; i < n; i++) {
// Storing every pow(2, i)%k value in
// the array
poweroftwo[i] = (poweroftwo[i - 1]
* (2 % k))
% k;
}
// To store the remaining
int rem = 0;
// Iterating till N
for (int i = 0; i < n; i++) {
// If current bit is 1
if (s[n - i - 1] == '1') {
// Updating rem
rem += (poweroftwo[i]);
rem %= k;
}
}
// If completely divisible
if (rem == 0) {
return "Yes";
}
// If not Completely divisible
else
return "No";
}
// Driver Code
int main()
{
// Given Input
string s = "1010001";
int k = 9;
// length of string s
int n = s.length();
// Function Call
cout << divisibleByk(s, n, k);
return 0;
}
// Java program for above approach
class GFG
{
// Function to check the binary number is
// divisible by K
public static String divisibleByk(String s, int n, int k) {
// Array poweroftwo will store pow(2, i)%k
int[] poweroftwo = new int[n];
// Initializing the first element in Array
poweroftwo[0] = 1 % k;
for (int i = 1; i < n; i++) {
// Storing every pow(2, i)%k value in
// the array
poweroftwo[i] = (poweroftwo[i - 1] * (2 % k)) % k;
}
// To store the remaining
int rem = 0;
// Iterating till N
for (int i = 0; i < n; i++) {
// If current bit is 1
if (s.charAt(n - i - 1) == '1') {
// Updating rem
rem += (poweroftwo[i]);
rem %= k;
}
}
// If completely divisible
if (rem == 0) {
return "Yes";
}
// If not Completely divisible
else
return "No";
}
// Driver Code
public static void main(String args[])
{
// Given Input
String s = "1010001";
int k = 9;
// length of string s
int n = s.length();
// Function Call
System.out.println(divisibleByk(s, n, k));
}
}
// This code is contributed by _saurabh_jaiswal,
# python 3 program for above approach
# Function to check the binary number is
# divisible by K
def divisibleByk(s, n, k):
# Array poweroftwo will store pow(2, i)%k
poweroftwo = [0 for i in range(n)]
# Initializing the first element in Array
poweroftwo[0] = 1 % k
for i in range(1,n,1):
# Storing every pow(2, i)%k value in
# the array
poweroftwo[i] = (poweroftwo[i - 1] * (2 % k)) % k
# To store the remaining
rem = 0
# Iterating till N
for i in range(n):
# If current bit is 1
if (s[n - i - 1] == '1'):
# Updating rem
rem += (poweroftwo[i])
rem %= k
# If completely divisible
if (rem == 0):
return "Yes"
# If not Completely divisible
else:
return "No"
# Driver Code
if __name__ == '__main__':
# Given Input
s = "1010001"
k = 9
# length of string s
n = len(s)
# Function Call
print(divisibleByk(s, n, k))
# This code is contributed by SURENDRA_GANGWAR.
// C# program for above approach
using System;
class GFG
{
// Function to check the binary number is
// divisible by K
public static String divisibleByk(String s, int n, int k) {
// Array poweroftwo will store pow(2, i)%k
int[] poweroftwo = new int[n];
// Initializing the first element in Array
poweroftwo[0] = 1 % k;
for (int i = 1; i < n; i++) {
// Storing every pow(2, i)%k value in
// the array
poweroftwo[i] = (poweroftwo[i - 1] * (2 % k)) % k;
}
// To store the remaining
int rem = 0;
// Iterating till N
for (int i = 0; i < n; i++) {
// If current bit is 1
if (s[n - i - 1] == '1') {
// Updating rem
rem += (poweroftwo[i]);
rem %= k;
}
}
// If completely divisible
if (rem == 0) {
return "Yes";
}
// If not Completely divisible
else
return "No";
}
// Driver Code
public static void Main(String []args)
{
// Given Input
String s = "1010001";
int k = 9;
// length of string s
int n = s.Length;
// Function Call
Console.Write(divisibleByk(s, n, k));
}
}
// This code is contributed by shivanisinghss2110
<script>
// Javascript program for above approach
// Function to check the binary number is
// divisible by K
function divisibleByk(s, n, k)
{
// Array poweroftwo will store pow(2, i)%k
let poweroftwo = new Array(n);
// Initializing the first element in Array
poweroftwo[0] = 1 % k;
for (let i = 1; i < n; i++) {
// Storing every pow(2, i)%k value in
// the array
poweroftwo[i] = (poweroftwo[i - 1]
* (2 % k))
% k;
}
// To store the remaining
let rem = 0;
// Iterating till N
for (let i = 0; i < n; i++) {
// If current bit is 1
if (s[n - i - 1] == '1') {
// Updating rem
rem += (poweroftwo[i]);
rem %= k;
}
}
// If completely divisible
if (rem == 0) {
return "Yes";
}
// If not Completely divisible
else
return "No";
}
// Driver Code
// Given Input
let s = "1010001";
let k = 9;
// length of string s
let n = s.length;
// Function Call
document.write(divisibleByk(s, n, k));
// This code is contributed by gfgking.
</script>
Output
Yes
Time Complexity: O(N)
Auxiliary Space: O(N)
