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Maximum number of splits of a binary number
  • Last Updated : 11 Nov, 2019

Given a binary string S, the task is to find the maximum number of parts that you can split it into such that every part is divisible by 2. If the string can’t be split satisfying the given conditions then print -1.

Examples:

Input: S = “100”
Output: 2
The splits are as follows:
“10” ans “0”.

Input: S = “110”
Output: 1

Approach: This problem can be solved greedily, start from the left end and put a cut at an index j such that j is the smallest index for which sub-string upto j is divisible by 2. Now, continue this step with the rest of the left-over string. It is also known that any binary number ending with a 0 is divisible by 2. Thus, put a cut after each and every zero and the answer will be equal to the number of zeros in the string. The only case where the answer is not possible is when the given string is odd i.e. no matter how cuts are made on the string, the last split part will always be odd.



Below is the implementation of the above approach:

C++

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

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Java

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// Java implementation of the approach
class GFG
{
  
// 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 -1;
  
    // To store the final ans
    int ans = 0;
  
    // Counting the number of zeros
    for (int i = 0; i < s.length(); i++)
        ans += (s.charAt(i) == '0') ? 1 : 0;
  
    // Return the final answer
    return ans;
}
  
// Driver code
public static void main(String []args) 
{
    String s = "10010";
  
    System.out.println(cntSplits(s));
}
}
  
// This code is contributed by 29AjayKumar

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Python3

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# 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 -1
  
    # To store the count of zeroes 
    ans = 0
  
    # Counting the number of zeroes 
    for i in range(len(s)) :
        ans += (s[i] == '0'); 
  
    # Return the final answer 
    return ans ; 
  
# Driver code 
if __name__ == "__main__"
  
    s = "10010"
  
    print(cntSplits(s));
      
# This code is contributed by AnkitRai01

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C#

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

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Output:

3

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