Розбір

Let's consider the input string $s$ — the resulting string after Huseyn has performed all operations. Initialize two pointers: $i=0$ at the start of the string and $j=n-1$ at the end.

If $s_i\ne s_j$, then the characters at the ends of the string are different: either $0$ on the left and $1$ on the right, or $1$ on the left and $0$ on the right. This means the previous string could have been extended by applying one of the specified operations. In this case, move both pointers $i$ and $j$ one position towards each other and check again whether the substring $s[i...j]$ could have been obtained through Huseyn's operations.

As soon as the current substring $s[i...j]$ satisfies $s_i=s_j$, it can no longer be produced by the given operations. Print its length — this is the original string that Huseyn had.

Example

Let's perform Huseyn's operations in reverse order.

The string $s="1"$ was originally the one Huseyn had.

Algorithm realization

Read the input string and calculate its length, storing it in the variable $res$.

cin >> s;
res = s.size();

Initialize two pointers: $i=0$ at the start and $j=n-1$ at the end of the string.

i = 0; j = s.size() - 1;

If $s_i\ne s_j$, then the characters at the ends of the string are different. Huseyn could have obtained the substring $s[i...j]$ from the substring $s[i+\mathrm{1...}j-1]$.

while ((i < j) && (s[i] != s[j]))
{

Move both pointers $i$ and $j$ one position towards each other and decrease the current size $res$ of the substring $s[i...j]$.

  i++; j--;
  res -= 2;
}

Print the answer.

cout << res << endl;