给定二进制字符串可被 2 整除的子序列数

给定长度为N 的二进制字符串str ，任务是找到str的可被2整除的子序列的计数。子序列中允许有前导零。

例子：

Input: str = “101” 
Output: 2 
“0” and “10” are the only subsequences 
which are divisible by 2.
Input: str = “10010” 
Output: 22  

朴素的方法：朴素的方法是生成所有可能的子序列并检查它们是否可以被 2 整除。这样做的时间复杂度为 O(2 ^N * N)。

有效的方法：可以观察到，任何二进制数只有以0结尾才能被2整除。现在，任务只是计算以0结尾的子序列的数量。因此，对于每个索引i使得str[i] = ‘0’ ，找到以i结尾的子序列的数量。该值等于2 ⁱ (基于 0 的索引)。因此，最终答案将等于所有i的2 ⁱ的总和，使得str[i] = ‘0’ 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
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

 
// 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));
 

输出：

22

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。