Skip to content
Related Articles

Related Articles

Improve Article

Number of ways to split a binary number such that every part is divisible by 2

  • Last Updated : 04 Aug, 2021
Geek Week

Given a binary string S, the task is to find the number of ways to split it into parts such that every part is divisible by 2.

Examples: 

Input: S = “100” 
Output:
There are two ways to split the string: 
{“10”, “0”} and {“100”}
Input: S = “110” 
Output:

Approach: One observation is that the string can only be split after a 0. Thus, count the number of zeros in the string. Let’s call this count c_zero. Assuming the case when the string is even, for every 0 except for the rightmost one, there are two choices i.e. either cut the string after that zero or don’t. Thus, the final answer becomes 2(c_zero – 1) for even strings. 
The case, where the string can’t be split is the case when it ends at a 1. Thus, for odd strings answer will always be zero as the last split part will always be odd.

Below is the implementation of the above approach: 



C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
#define maxN 20
#define maxM 64
 
// Function to return the required count
int cntSplits(string s)
{
    // If the splitting is not possible
    if (s[s.size() - 1] == '1')
        return 0;
 
    // To store the count of zeroes
    int c_zero = 0;
 
    // Counting the number of zeroes
    for (int i = 0; i < s.size(); i++)
        c_zero += (s[i] == '0');
 
    // Return the final answer
    return (int)pow(2, c_zero - 1);
}
 
// Driver code
int main()
{
    string s = "10010";
 
    cout << cntSplits(s);
 
    return 0;
}

Java




// Java implementation of the approach
class GFG
{
 
static int maxN = 20;
static int maxM = 64;
 
// Function to return the required count
static int cntSplits(String s)
{
    // If the splitting is not possible
    if (s.charAt(s.length() - 1) == '1')
        return 0;
 
    // To store the count of zeroes
    int c_zero = 0;
 
    // Counting the number of zeroes
    for (int i = 0; i < s.length(); i++)
        c_zero += (s.charAt(i) == '0') ? 1 : 0;
 
    // Return the final answer
    return (int)Math.pow(2, c_zero - 1);
}
 
// Driver code
public static void main(String []args)
{
    String s = "10010";
 
    System.out.println(cntSplits(s));
}
}
 
// This code is contributed by PrinciRaj1992

Python3




# Python3 implementation of the approach
 
# Function to return the required count
def cntSplits(s) :
 
    # If the splitting is not possible
    if (s[len(s) - 1] == '1') :
        return 0;
 
    # To store the count of zeroes
    c_zero = 0;
 
    # Counting the number of zeroes
    for i in range(len(s)) :
        c_zero += (s[i] == '0');
 
    # Return the final answer
    return int(pow(2, c_zero - 1));
 
# Driver code
if __name__ == "__main__" :
 
    s = "10010";
 
    print(cntSplits(s));
     
# This code is contributed by AnkitRai01

C#




// C# implementation of the approach
using System;
                     
class GFG
{
 
static int maxN = 20;
static int maxM = 64;
 
// Function to return the required count
static int cntSplits(String s)
{
    // If the splitting is not possible
    if (s[s.Length - 1] == '1')
        return 0;
 
    // To store the count of zeroes
    int c_zero = 0;
 
    // Counting the number of zeroes
    for (int i = 0; i < s.Length; i++)
        c_zero += (s[i] == '0') ? 1 : 0;
 
    // Return the final answer
    return (int)Math.Pow(2, c_zero - 1);
}
 
// Driver code
public static void Main(String []args)
{
    String s = "10010";
 
    Console.WriteLine(cntSplits(s));
}
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
// Javascript implementation of the approach
 
var maxN = 20;
var maxM = 64;
 
// Function to return the required count
function cntSplits(s)
{
    // If the splitting is not possible
    if (s[s.length - 1] == '1')
        return 0;
 
    // To store the count of zeroes
    var c_zero = 0;
 
    // Counting the number of zeroes
    for (var i = 0; i < s.length; i++)
        c_zero += (s[i] == '0');
 
    // Return the final answer
    return Math.pow(2, c_zero - 1);
}
 
// Driver code
var s = "10010";
document.write( cntSplits(s));
 
</script>
Output: 
4

 

Time Complexity: O(N)
Auxiliary Space: O(1)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :