Number of subsequences in a given binary string divisible by 2
Given binary string str of length N, the task is to find the count of subsequences of str which are divisible by 2. Leading zeros in a sub-sequence are allowed.
Examples:
Input: str = “101”
Output: 2
“0” and “10” are the only subsequences
which are divisible by 2.
Input: str = “10010”
Output: 22
Naive approach: A naive approach will be to generate all possible sub-sequences and check if they are divisible by 2. The time complexity for this will be O(2N * N).
Efficient approach: It can be observed that any binary number is divisible by 2 only if it ends with a 0. Now, the task is to just count the number of subsequences ending with 0. So, for every index i such that str[i] = ‘0’, find the number of subsequences ending at i. This value is equal to 2i (0-based indexing). Thus, the final answer will be equal to the summation of 2i for all i such that str[i] = ‘0’.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to return the count // of the required subsequences int countSubSeq(string str, int len) { // To store the final answer int ans = 0; // Multiplier int mul = 1; // Loop to find the answer for ( int i = 0; i < len; i++) { // Condition to update the answer if (str[i] == '0' ) ans += mul; // updating multiplier mul *= 2; } return ans; } // Driver code int main() { string str = "10010" ; int len = str.length(); cout << countSubSeq(str, len); return 0; } |
Java
// Java implementation of the approach class GFG { // Function to return the count // of the required subsequences static int countSubSeq(String str, int len) { // To store the final answer int ans = 0 ; // Multiplier int mul = 1 ; // Loop to find the answer for ( int i = 0 ; i < len; i++) { // Condition to update the answer if (str.charAt(i) == '0' ) ans += mul; // updating multiplier mul *= 2 ; } return ans; } // Driver code public static void main(String[] args) { String str = "10010" ; int len = str.length(); System.out.print(countSubSeq(str, len)); } } // This code is contributed by 29AjayKumar |
Python3
# Python3 implementation of the approach # Function to return the count # of the required subsequences def countSubSeq(strr, lenn): # To store the final answer ans = 0 # Multiplier mul = 1 # Loop to find the answer for i in range (lenn): # Condition to update the answer if (strr[i] = = '0' ): ans + = mul # updating multiplier mul * = 2 return ans # Driver code strr = "10010" lenn = len (strr) print (countSubSeq(strr, lenn)) # This code is contributed by Mohit Kumar |
C#
// C# implementation of the approach using System; class GFG { // Function to return the count // of the required subsequences static int countSubSeq( string str, int len) { // To store the final answer int ans = 0; // Multiplier int mul = 1; // Loop to find the answer for ( int i = 0; i < len; i++) { // Condition to update the answer if (str[i] == '0' ) ans += mul; // updating multiplier mul *= 2; } return ans; } // Driver code static public void Main () { string str = "10010" ; int len = str.Length; Console.WriteLine(countSubSeq(str, len)); } } // This code is contributed by AnkitRai01 |
Javascript
<script> // Javascript implementation of the approach // Function to return the count // of the required subsequences function countSubSeq(str, len) { // To store the final answer var ans = 0; // Multiplier var mul = 1; // Loop to find the answer for ( var i = 0; i < len; i++) { // Condition to update the answer if (str[i] == '0' ) ans += mul; // updating multiplier mul *= 2; } return ans; } // Driver code var str = "10010" ; var len = str.length; document.write( countSubSeq(str, len)); </script> |
Output:
22
Time Complexity: O(len), where len is the size of the given string
Auxiliary Space: O(1)