通过替换“？”来计算可能的排列在二进制字符串中的字符

给定一个由字符0 、 1和‘?’组成的字符串S , 任务是计算通过替换‘?’形成的二进制字符串的所有可能组合。由0或1 。

例子：

Input: S = “0100?110”
Output: 2
Explanation: Replacing each ‘?’s with ‘1’ and ‘0’, the count of such strings formed will be equal to “01001110” and “01000110”. Therefore, the total count of such strings formed is 2.

Input: S = “00?0?111”
Output: 4

方法：根据以下观察可以解决给定的问题：

• 由于每个‘?’可以替换为‘0’或‘1’ ，每个‘?’都有两种可能的选择。字符。
• 现在，任务减少到找到“?”的计数。 s，说计数。
• 因此，可以形成的不同字符串的总数为2 ^count 。

请按照以下步骤解决问题：

• 计算“?”出现的次数，说数
• 打印2 ^count的值作为形成的字符串的结果组合。

下面是上述方法的实现：

C++14

#include 
 
using namespace std;
 
//Function to count all possible
//permutations of the string s
void countPermutations(string s){
 
    //Stores frequency of the
    //character '?'
    int count = 0;
 
    //Traverse the string
    for (char i:s){
        if (i == '?')
            count += 1;
      }
 
    //Print the answer
    cout<

---

Java

// Java program for the above approach
class GFG{
 
// Function to count all possible
// permutations of the string s
static void countPermutations(String s)
{
 
    // Stores frequency of the
    // character '?'
    int count = 0;
 
    // Traverse the string
    for (char i : s.toCharArray())
    {
        if (i == '?')
            count += 1;
      }
 
    // Print the answer
    System.out.print((int)Math.pow(2,count));
}
 
 
// Driver Code
public static void main(String[] args)
{
     
    // Given string S
  String s = "0100?110";
 
  // Function call to count
  // the number of permutations
  countPermutations(s);
}
}
 
// This code is contributed by code_hunt.

Python3

# Python3 program for the above approach
 
# Function to count all possible
# permutations of the string s
def countPermutations(s):
 
    # Stores frequency of the
    # character '?'
    count = 0
 
    # Traverse the string
    for i in s:
        if i == '?':
 
            # Increment count
            count += 1
 
    # Print the answer
    print(2**count)
 
# Driver Code
 
# Given string S
s = "0100?110"
 
# Function call to count
# the number of permutations
countPermutations(s)
 
# This code is contribute by mohit kumar 29.

C#

// C# program for above approach
using System;
 
public class GFG
{
   
// Function to count all possible
// permutations of the string s
static void countPermutations(string s)
{
 
    // Stores frequency of the
    // character '?'
    int count = 0;
 
    // Traverse the string
    foreach (char i in s)
    {
        if (i == '?')
            count += 1;
      }
 
    // Print the answer
    Console.WriteLine(Math.Pow(2,count));
}
 
// Driver code
public static void Main(String[] args)
{
   
    // Given string S
  string s = "0100?110";
 
  // Function call to count
  // the number of permutations
  countPermutations(s);
}
}
 
// This code is contributed by splevel62.

Javascript

 
 
//Function to count all possible
//permutations of the string s
function countPermutations(s){
 
    //Stores frequency of the
    //character '?'
    var count = 0;
 
    //Traverse the string
    for(var i =0; i< s.length; i++)
    {
        if (s[i] == '?')
            count += 1;
    }
 
    //Print the answer
    document.write( Math.pow(2,count));
}
 
//Driver Code
 
//Given string S
var s = "0100?110";
 
//Function call to count
//the number of permutations
countPermutations(s);
 
 

输出：

2

时间复杂度： O(N)
辅助空间： O(1)

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