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

• Last Updated : 06 Oct, 2021

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:

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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 ` `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`

## Javascript

 ``

Output:

`10100101`

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

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