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Make Palindrome binary string with exactly a 0s and b 1s by replacing wild card ?

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  • Last Updated : 30 May, 2022
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Given a string S of N characters consisting of ‘?’, ‘0‘, and ‘1‘ and two integers a and b, the task is to find a palindromic binary string with exactly a 0s and b 1s by replacing the ‘?‘ with either ‘0‘ or ‘1‘.

Examples:

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.

 

Approach: The given problem can be solved by using the following observations:

  • For the string S of N characters to be a palindrome, S[i] = S[N-i-1] must hold true for all values of i in the range [0, N).
  • If only one of the S[i] and S[N-i-1] is equal to ‘?‘, then it must be replaced with the corresponding character of the other index.
  • If the value of both S[i] and S[N-i-1] is ‘?‘, the most optimal choice is to replace both of them with the character whose required count is more.

Using the above-stated observations, follow the below steps to solve the problem:

  • Traverse through the string in the range [0, N/2), and for the cases where only one of the S[i] and S[N – i – 1] is equal to ‘?‘, replace it with the corresponding character.
  • Update the values of a and b by subtracting the count of ‘0‘ and ‘1‘ after the above operation. The count can be easily calculated using the std::count function.
  • Now traverse the given string in the range [0, N/2), and if both S[i] =?‘ and S[N-i-1] = ‘?‘, update their value with the character whose required count is more( i.e, with ‘0‘ if a>b otherwise with ‘1‘).
  • In the case of odd string length with the middle character as ‘?‘, update it with character with the remaining count.
  • If the value of a = 0 and b = 0, the string stored in S is the required string. Otherwise, the required string does not exist, hence return -1.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
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;
}

Java




// Java program for the above approach
 
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";
        }
    }
 
    public static void main(String[] args)
    {
        String S = "10?????1";
        int a = 4, b = 4;
 
        System.out.println(convertString(S, a, b));
    }
}
 
// this code is contributed by phasing17

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




<script>
    // JavaScript program for the above approach
 
    // Function to convert the given string
    // to a palindrome with a 0's and b 1's
    const convertString = (S, a, b) => {
        // Stores the size of the string
        let N = S.length;
 
        // Loop to iterate over the string
        for (let i = 0; i < parseInt(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
        let cnt_0 = 0;
        for (let i = 0; i < N; ++i) {
            if (S[i] == '0') cnt_0++;
        }
        a = a - cnt_0;
 
        // Subtract the count of 1 from
        // required number of ones
        let cnt_1 = 0;
        for (let i = 0; i < N; ++i) {
            if (S[i] == '0') cnt_1++;
        }
        b = b - cnt_1;
 
        // Traverse the string
        for (let i = 0; i < parseInt(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[parseInt(N / 2)] == '?') {
 
            // If a is greater than b
            if (a > b) {
 
                // Update middle character
                // with '0'
                S[parseInt(N / 2)] = '0';
                a--;
            }
            else {
 
                // Update middle character
                // by '1'
                S[parseInt(N / 2)] = '1';
                b--;
            }
        }
 
        // Return Answer
        if (a == 0 && b == 0) {
            return S;
        }
        else {
            return "-1";
        }
    }
 
    // Driver Code
    let S = "10?????1";
    S = S.split('');
    let a = 4, b = 4;
 
    document.write(convertString(S, a, b).join(''));
 
// This code is contributed by rakeshsahni
 
</script>

 
 

Output: 

10100101

 

 

Time Complexity: O(N)
Auxiliary Space: O(1)

 


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