通过替换通配符使回文二进制字符串恰好具有 0 和 b 1 吗？

给定一个包含N个字符的字符串S ，其中包含'?' , ' 0 ', 和' 1 ' 以及两个字符串a和b ，任务是通过替换 ' ? ' 带有 ' 0 ' 或 ' 1 '。

例子：

Input: S = “10?????1”, a = 4, b = 4
Output: 10100101
Explanation: The output string is a palindromic binary string with exactly 4 0’s and 4 1’s.

Input: S = “??????”, a = 5, b = 1
Output: -1
Explanation: There does not exist any palindromic string with exactly a 0’s and b 1’s.

编程需要懂一点英语

方法：给定的问题可以通过使用以下观察来解决：

对于由N个字符组成的字符串S是回文， S[i] = S[Ni-1]必须对[0, N)范围内的i的所有值都成立。
如果S[i]和S[Ni-1]中只有一个等于 ' ？ '，则必须替换为其他索引的对应字符。
如果S[i]和S[Ni-1]的值都是 ' ? '，最佳的选择是用所需计数更多的字符替换它们。

使用上述观察结果，请按照以下步骤解决问题：

遍历[0, N/2)范围内的字符串，对于只有S[i]和S[N – i – 1]之一等于 ' ? '，将其替换为对应的字符。
在上述操作之后，通过减去“ 0 ”和“ 1 ”的计数来更新a和b的值。使用 std::count函数可以轻松计算计数。
现在遍历[0, N/2)范围内的给定字符串，如果两者S[i] = ' ？ ' 和S[Ni-1] = ' ? '，用所需计数更多的字符更新它们的值（即，如果a>b ，则使用' 0 '，否则使用' 1 '）。
在奇数字符串长度的情况下，中间字符为 ' ? '，用剩余计数的字符更新它。
如果a = 0且b = 0的值，则存储在S中的字符串是所需的字符串。否则，所需的字符串不存在，因此返回-1 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to convert the given string
// to a palindrome with a 0's and b 1's
string convertString(string S, int a, int b)
{
    // Stores the size of the string
    int N = S.size();
 
    // Loop to iterate over the string
    for (int i = 0; i < N / 2; ++i) {
 
        // If one of S[i] or S[N-i-1] is
        // equal to '?', replace it with
        // corresponding character
        if (S[i] == '?'
            && S[N - i - 1] != '?') {
            S[i] = S[N - i - 1];
        }
        else if (S[i] != '?'
                 && S[N - i - 1] == '?') {
            S[N - i - 1] = S[i];
        }
    }
 
    // Subtract the count of 0 from the
    // required number of zeroes
    a = a - count(S.begin(), S.end(), '0');
 
    // Subtract the count of 1 from
    // required number of ones
    b = b - count(S.begin(), S.end(), '1');
 
    // Traverse the string
    for (int i = 0; i < N / 2; ++i) {
 
        // If both S[i] and S[N-i-1] are '?'
        if (S[i] == '?' && S[N - i - 1] == '?') {
 
            // If a is greater than b
            if (a > b) {
 
                // Update S[i] and S[N-i-1] to '0'
                S[i] = S[N - i - 1] = '0';
 
                // Update the value of a
                a -= 2;
            }
            else {
 
                // Update S[i] and S[N-i-1] to '1'
                S[i] = S[N - i - 1] = '1';
 
                // Update the value of b
                b -= 2;
            }
        }
    }
 
    // Case with middle character '?'
    // in case of odd length string
    if (S[N / 2] == '?') {
 
        // If a is greater than b
        if (a > b) {
 
            // Update middle character
            // with '0'
            S[N / 2] = '0';
            a--;
        }
        else {
 
            // Update middle character
            // by '1'
            S[N / 2] = '1';
            b--;
        }
    }
 
    // Return Answer
    if (a == 0 && b == 0) {
        return S;
    }
    else {
        return "-1";
    }
}
 
// Driver Code
int main()
{
    string S = "10?????1";
    int a = 4, b = 4;
 
    cout << convertString(S, a, b);
 
    return 0;
}

Python3

# Python program for the above approach
 
# Function to convert the given string
# to a palindrome with a 0's and b 1's
def convertString(S, a, b):
    print(S)
     
    # Stores the size of the string
    N = len(S)
 
    # Loop to iterate over the string
    for i in range(0, N // 2):
 
        # If one of S[i] or S[N-i-1] is
        # equal to '?', replace it with
        # corresponding character
        if (S[i] == '?' and S[N - i - 1] != '?'):
            S[i] = S[N - i - 1]
        elif (S[i] != '?' and S[N - i - 1] == '?'):
            S[N - i - 1] = S[i]
         
    # Subtract the count of 0 from the
    # required number of zeroes
    cnt_0 = 0
    for i in range(0, N):
        if (S[i] == '0'):
            cnt_0 += 1
     
    a = a - cnt_0
 
    # Subtract the count of 1 from
    # required number of ones
    cnt_1 = 0
 
    for i in range(0, N):
        if (S[i] == '0'):
             cnt_1 += 1
 
 
    b = b - cnt_1
 
        # Traverse the string
    for i in range(0, N // 2):
 
        # If both S[i] and S[N-i-1] are '?'
        if (S[i] == '?' and S[N - i - 1] == '?'):
 
            # If a is greater than b
            if (a > b):
 
                # Update S[i] and S[N-i-1] to '0'
                S[i] = S[N - i - 1] = '0'
 
                # Update the value of a
                a -= 2
            else:
 
                # Update S[i] and S[N-i-1] to '1'
                S[i] = S[N - i - 1] = '1'
 
                # Update the value of b
                b -= 2
         
 
    # Case with middle character '?'
    # in case of odd length string
    if (S[N // 2] == '?'):
 
            # If a is greater than b
            if (a > b):
 
                # Update middle character
                # with '0'
                S[N // 2] = '0'
                a -= 1
            else:
 
                # Update middle character
                # by '1'
                S[N // 2] = '1'
                b -= 1
             
    # Return Answer
    if (a == 0 and b == 0):
            return S
    else:
            return "-1"
 
# Driver Code
S = "10?????1"
S = list(S)
a = 4
b = 4
 
print("".join(convertString(S, a, b)))
 
# This code is contributed by gfgking

C#

// C# program for the above approach
using System;
class GFG {
 
  // Function to convert the given string
  // to a palindrome with a 0's and b 1's
  static string convertString(string S, int a, int b)
  {
     
    // Stores the size of the string
    int N = S.Length;
    char[] str = S.ToCharArray();
 
    // Loop to iterate over the string
    for (int i = 0; i < N / 2; ++i) {
 
      // If one of S[i] or S[N-i-1] is
      // equal to '?', replace it with
      // corresponding character
      if (str[i] == '?' && str[N - i - 1] != '?') {
        str[i] = str[N - i - 1];
      }
      else if (str[i] != '?'
               && str[N - i - 1] == '?') {
        str[N - i - 1] = str[i];
      }
    }
 
    // Subtract the count of 0 from the
    // required number of zeroes
    int countA = 0, countB = 0;
    for (int i = 0; i < str.Length; i++) {
      if (str[i] == '0')
        countA++;
      else if (str[i] == '1')
        countB++;
    }
    a = a - countA;
 
    // Subtract the count of 1 from
    // required number of ones
    b = b - countB;
 
    // Traverse the string
    for (int i = 0; i < N / 2; ++i) {
 
      // If both S[i] and S[N-i-1] are '?'
      if (str[i] == '?' && str[N - i - 1] == '?') {
 
        // If a is greater than b
        if (a > b) {
 
          // Update S[i] and S[N-i-1] to '0'
          str[i] = '0';
          str[N - i - 1] = '0';
 
          // Update the value of a
          a -= 2;
        }
        else {
 
          // Update S[i] and S[N-i-1] to '1'
          str[i] = '1';
          str[N - i - 1] = '1';
 
          // Update the value of b
          b -= 2;
        }
      }
    }
 
    // Case with middle character '?'
    // in case of odd length string
    if (str[N / 2] == '?') {
 
      // If a is greater than b
      if (a > b) {
 
        // Update middle character
        // with '0'
        str[N / 2] = '0';
        a--;
      }
      else {
 
        // Update middle character
        // by '1'
        str[N / 2] = '1';
        b--;
      }
    }
 
    // Return Answer
    if (a == 0 && b == 0) {
      return new String(str);
    }
    else {
      return "-1";
    }
  }
 
  // Driver Code
  public static void Main()
  {
    string S = "10?????1";
    int a = 4, b = 4;
 
    Console.Write(convertString(S, a, b));
  }
}
 
// This code is contributed by ukasp.

Javascript

输出：

10100101

时间复杂度： O(N)
辅助空间： O(1)