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: 2
There are two ways to split the string:
{“10”, “0”} and {“100”}Input: S = “110”
Output: 1
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 as 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 |
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