description: A binary string is a string that consists of characters 0 and 1. A bi-table is a table that has exactly two rows of equal length, each being a binary string.
  
  Let \operatorname{MEX} of a bi-table be the smallest digit among 0, 1, or 2 that does not occur in the bi-table. For example, \operatorname{MEX} for \begin{bmatrix} 0011\\\ 1010 \end{bmatrix} is 2, because 0 and 1 occur in the bi-table at least once. \operatorname{MEX} for \begin{bmatrix} 111\\\ 111 \end{bmatrix} is 0, because 0 and 2 do not occur in the bi-table, and 0 < 2.
  
  You are given a bi-table with n columns. You should cut it into any number of bi-tables (each consisting of consecutive columns) so that each column is in exactly one bi-table. It is possible to cut the bi-table into a single bi-table — the whole bi-table.
  
  What is the maximal sum of \operatorname{MEX} of all resulting bi-tables can be?
  
  Input
  
  The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.
  
  The first line of the description of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of columns in the bi-table.
  
  Each of the next two lines contains a binary string of length n — the rows of the bi-table.
  
  It's guaranteed that the sum of n over all test cases does not exceed 10^5.
  
  Output
  
  For each test case print a single integer — the maximal sum of \operatorname{MEX} of all bi-tables that it is possible to get by cutting the given bi-table optimally.
  
  Example
  
  Input
  
  
  4
  7
  0101000
  1101100
  5
  01100
  10101
  2
  01
  01
  6
  000000
  111111
  
  
  Output
  
  
  8
  8
  2
  12
  
  Note
  
  In the first test case you can cut the bi-table as follows:
  
    * \begin{bmatrix} 0\\\ 1 \end{bmatrix}, its \operatorname{MEX} is 2.
    * \begin{bmatrix} 10\\\ 10 \end{bmatrix}, its \operatorname{MEX} is 2.
    * \begin{bmatrix} 1\\\ 1 \end{bmatrix}, its \operatorname{MEX} is 0.
    * \begin{bmatrix} 0\\\ 1 \end{bmatrix}, its \operatorname{MEX} is 2.
    * \begin{bmatrix} 0\\\ 0 \end{bmatrix}, its \operatorname{MEX} is 1.
    * \begin{bmatrix} 0\\\ 0 \end{bmatrix}, its \operatorname{MEX} is 1.
  
  
  
  The sum of \operatorname{MEX} is 8.
solution: import java.util.Scanner;
  
  public class MaxMexCut {
      public static void main(String[] args) {
          Scanner sc = new Scanner(System.in);
          int t = Integer.parseInt(sc.nextLine());
          for(int i=0; i<t; i++) {
              int n = Integer.parseInt(sc.nextLine());
              char[][] biTable = new char[2][n];
              String s1 = sc.nextLine();
              String s2 = sc.nextLine();
  
              for(int j=0; j<n; j++) {
                  biTable[0][j] = s1.charAt(j);
                  biTable[1][j] = s2.charAt(j);
              }
              System.out.println(maxMex(biTable));
          }
      }
  
      public static int maxMex(char[][] biTable) {
          int maxMex = 0;
          boolean prevZero = false;
          boolean prevOne = false;
  
          for(int i=0; i< biTable[0].length; i++) {
              if(biTable[0][i] != biTable[1][i]) {
                  maxMex += 2;
                  prevZero = prevOne = false;
              } else {
                  boolean currZero = biTable[0][i] == '0';
                  boolean currOne = biTable[0][i] == '1';
  
                  // Take the current column into account
                  if(currZero) {
                      maxMex += 1;
                  }
                  if((prevZero && currOne) || (prevOne && currZero)) {
                      maxMex += 1;
                      prevZero = false; prevOne = false;
                      continue;
                  }
                  // Set prev zero or prev one
                  prevOne = currOne;
                  prevZero = currZero;
              }
  
          }
  
          return maxMex;
      }
  }
  