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++ 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 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 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# 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 |
<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)