计算替换“？”的方法在二进制字符串中使 0 和 1 的计数与另一个字符串的计数相同

给定两个大小分别为N和M的二进制字符串S1和S2 ，使得字符串S2也包含字符'?' ，任务是找到替换“？”的方法的数量。在字符串S2中，使得字符串S2中0和1的计数与字符串S1中的计数相同。

例子：

Input: S1 = “1010”, S2 = “10??”
Output: 2
Explanation:
Following are the ways to replace ‘?’ in the string S2:

1. Replace “??” in the string S2 with “01” modifies the string S2 = “1001”. Now the counts of 0s and 1s in both the strings S1 and S2 are the same.
2. Replace “??” in the string S2 with “10” modifies the string S2 = “1010”. Now the counts of 0s and 1s in both the strings S1 and S2 are the same.

Therefore, the total number of ways is 2.

Input: S1 = “0000”, S2 = “??10?”
Output: 0

方法：给定的问题可以通过使用组合的概念来解决。请按照以下步骤解决问题：

• 初始化变量，例如sum1为0 ， sum2为0 ，它们存储给定字符串S1和S2中0和1的数量。
• 初始化一个变量，比如说 作为0存储替换“？”的方式总数在满足给定条件的字符串S2中。
• 遍历字符串S1 ，如果当前字符为1 ，则将sum1的值递增 1. 否则，将sum1的值减1 。
• 遍历字符串S2 ，如果当前字符为1则将sum2的值加1或者如果当前字符为0则将sum2的值减1 。
• 遍历字符串t ，如果当前字符是'+'则将 sum2的值增加1 。否则，将K的值增加1 。
• 初始化一个变量，比如P存储sum1和sum2的绝对差。
• 如果P的值至少为 K或(K – P)的值是奇数，则没有可能的方法来替换“？”因此打印0 。否则，打印^K C _(P+K)/2的值作为路数的结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to find the factorial of
// the given number N
int fact(int n)
{
    // Stores the factorial
    int res = 1;
 
    // Iterate over the range [2, N]
    for (int i = 2; i <= n; i++) {
        res = res * i;
    }
 
    // Return the resultant result
    return res;
}
 
// Function to find the number of ways
// of choosing r objects from n
// distinct objects
int nCr(int n, int r)
{
    return fact(n) / (fact(r) * fact(n - r));
}
 
// Function to find the number of ways
// to replace '?' in string t to get
// the same count of 0s and 1s in the
// string S and T
void countWays(string s, string t)
{
    int n = s.length();
    int sum1 = 0, sum2 = 0, K = 0;
 
    // Traverse the string s
    for (int i = 0; i < n; i++) {
 
        // If the current character
        // is 1
        if (s[i] == '1') {
 
            // Update the value of
            // the sum1
            sum1++;
        }
 
        // Otherwise
        else
            sum1--;
    }
 
    int m = t.length();
 
    // Traverse the string t
    for (int i = 0; i < m; i++) {
 
        // If the current character
        // is 1, then update the
        // value of sum2
        if (t[i] == '1') {
            sum2++;
        }
 
        // If the current character
        // is 0
        else if (t[i] == '0') {
            sum2--;
        }
 
        // Otherwise, update the
        // value of K
        else
            K++;
    }
 
    int P = abs(sum1 - sum2);
 
    // Check if P is greater than K
    // or if K-P is odd
    if (P > K or (K - P) % 2) {
        cout << 0;
        return;
    }
 
    // Print the count of ways
    cout << nCr(K, (P + K) / 2);
}
 
// Driver Code
int main()
{
    string S1 = "1010";
    string S2 = "10??";
    countWays(S1, S2);
 
    return 0;
}

Java

// Java program for above approach
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;
 
class GFG{
 
    // Function to find the factorial of
    // the given number N
    static int fact(int n)
    {
        // Stores the factorial
        int res = 1;
 
        // Iterate over the range [2, N]
        for (int i = 2; i <= n; i++) {
            res = res * i;
        }
 
        // Return the resultant result
        return res;
    }
 
    // Function to find the number of ways
    // of choosing r objects from n
    // distinct objects
    static int nCr(int n, int r)
    {
        return fact(n) / (fact(r) * fact(n - r));
    }
 
    // Function to find the number of ways
    // to replace '?' in string t to get
    // the same count of 0s and 1s in the
    // string S and T
    static void countWays(String s, String t)
    {
        int n = s.length();
        int sum1 = 0, sum2 = 0, K = 0;
 
        // Traverse the string s
        for (int i = 0; i < n; i++) {
 
            // If the current character
            // is 1
            if (s.charAt(i) == '1') {
 
                // Update the value of
                // the sum1
                sum1++;
            }
 
            // Otherwise
            else
                sum1--;
        }
 
        int m = t.length();
 
        // Traverse the string t
        for (int i = 0; i < m; i++) {
 
            // If the current character
            // is 1, then update the
            // value of sum2
            if (t.charAt(i) == '1') {
                sum2++;
            }
 
            // If the current character
            // is 0
            else if (t.charAt(i) == '0') {
                sum2--;
            }
 
            // Otherwise, update the
            // value of K
            else
                K++;
        }
 
        int P = Math.abs(sum1 - sum2);
 
        // Check if P is greater than K
        // or if K-P is odd
        if ((P > K) || (K - P) % 2==1) {
            System.out.println(0);
        return;
    }
 
        // Print the count of ways
        System.out.println(nCr(K, (P + K) / 2));
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        String S1 = "1010";
        String S2 = "10??";
        countWays(S1, S2);
    }
}
 
// This code is contributed by hritikrommie.

Python3

# Python 3 program for the above approach
 
# Function to find the factorial of
# the given number N
def fact(n):
    # Stores the factorial
    res = 1
 
    # Iterate over the range [2, N]
    for i in range(2,n+1,1):
        res = res * i
 
    # Return the resultant result
    return res
 
# Function to find the number of ways
# of choosing r objects from n
# distinct objects
def nCr(n, r):
    return fact(n) // (fact(r) * fact(n - r))
 
# Function to find the number of ways
# to replace '?' in string t to get
# the same count of 0s and 1s in the
# string S and T
def countWays(s, t):
    n = len(s);
    sum1 = 0
    sum2 = 0
    K = 0
 
    # Traverse the string s
    for i in range(n):
        # If the current character
        # is 1
        if (s[i] == '1'):
 
            # Update the value of
            # the sum1
            sum1 += 1
 
        # Otherwise
        else:
            sum1 -= 1
 
    m = len(t)
 
    # Traverse the string t
    for i in range(m):
        # If the current character
        # is 1, then update the
        # value of sum2
        if (t[i] == '1'):
            sum2 += 1
     
        # If the current character
        # is 0
        elif (t[i] == '0'):
            sum2 -= 1
 
        # Otherwise, update the
        # value of K
        else:
            K += 1
 
    P = abs(sum1 - sum2)
 
    # Check if P is greater than K
    # or if K-P is odd
    if (P > K or (K - P) % 2):
        print(0)
        return
 
    # Print the count of ways
    print(nCr(K, (P + K) // 2))
 
# Driver Code
if __name__ == '__main__':
    S1 = "1010"
    S2 = "10??"
    countWays(S1, S2)
     
    # This code is contributed by ipg2016107.

C#

// C# program for the above approach
using System;
 
class GFG{
     
// Function to find the factorial of
// the given number N
static int fact(int n)
{
    // Stores the factorial
    int res = 1;
 
    // Iterate over the range [2, N]
    for (int i = 2; i <= n; i++) {
        res = res * i;
    }
 
    // Return the resultant result
    return res;
}
 
// Function to find the number of ways
// of choosing r objects from n
// distinct objects
static int nCr(int n, int r)
{
    return fact(n) / (fact(r) * fact(n - r));
}
 
// Function to find the number of ways
// to replace '?' in string t to get
// the same count of 0s and 1s in the
// string S and T
static void countWays(string s, string t)
{
    int n = s.Length;
    int sum1 = 0, sum2 = 0, K = 0;
 
    // Traverse the string s
    for (int i = 0; i < n; i++) {
 
        // If the current character
        // is 1
        if (s[i] == '1') {
 
            // Update the value of
            // the sum1
            sum1++;
        }
 
        // Otherwise
        else
            sum1--;
    }
 
    int m = t.Length;
 
    // Traverse the string t
    for (int i = 0; i < m; i++) {
 
        // If the current character
        // is 1, then update the
        // value of sum2
        if (t[i] == '1') {
            sum2++;
        }
 
        // If the current character
        // is 0
        else if (t[i] == '0') {
            sum2--;
        }
 
        // Otherwise, update the
        // value of K
        else
            K++;
    }
 
    int P = Math.Abs(sum1 - sum2);
 
    // Check if P is greater than K
    // or if K-P is odd
    if ((P > K) || ((K - P) % 2) != 0) {
        Console.WriteLine(0);
        return;
    }
 
    // Print the count of ways
    Console.WriteLine( nCr(K, (P + K) / 2));
}
 
 
// Driver code
static public void Main()
{
    string S1 = "1010";
    string S2 = "10??";
    countWays(S1, S2);
}
}
 
// This code is contributed by target_2.

Javascript

 
// JavaScript program for the above approach
 
// Function to find the factorial of
// the given number N
function fact(n)
{
     
    // Stores the factorial
    let res = 1;
 
    // Iterate over the range [2, N]
    for(let i = 2; i <= n; i++)
    {
        res = res * i;
    }
 
    // Return the resultant result
    return res;
}
 
// Function to find the number of ways
// of choosing r objects from n
// distinct objects
function nCr(n, r)
{
    return fact(n) / (fact(r) * fact(n - r));
}
 
// Function to find the number of ways
// to replace '?' in string t to get
// the same count of 0s and 1s in the
// string S and T
function countWays(s, t)
{
    let n = s.length;
    let sum1 = 0, sum2 = 0, K = 0;
 
    // Traverse the string s
    for(let i = 0; i < n; i++)
    {
         
        // If the current character
        // is 1
        if (s[i] == '1')
        {
             
            // Update the value of
            // the sum1
            sum1++;
        }
 
        // Otherwise
        else
            sum1--;
    }
 
    let m = t.length;
 
    // Traverse the string t
    for(let i = 0; i < m; i++)
    {
         
        // If the current character
        // is 1, then update the
        // value of sum2
        if (t[i] == '1')
        {
            sum2++;
        }
 
        // If the current character
        // is 0
        else if (t[i] == '0')
        {
            sum2--;
        }
 
        // Otherwise, update the
        // value of K
        else
            K++;
    }
 
    let P = Math.abs(sum1 - sum2);
 
    // Check if P is greater than K
    // or if K-P is odd
    if (P > K || (K - P) % 2)
    {
        document.write(0);
        return;
    }
 
    // Print the count of ways
    document.write(nCr(K, (P + K) / 2));
}
 
// Driver Code
let S1 = "1010";
let S2 = "10??";
 
countWays(S1, S2);
 
// This code is contributed by Potta Lokesh
 

输出：

2

时间复杂度： O(max(N, M))
辅助空间： O(1)