# Longest Common Subsequence

$LCS\left(X_{i},Y_{j}\right) = \begin{cases} \empty & \mbox{ if }\ i = 0 \mbox{ or } j = 0 \\ \textrm{ } LCS\left(X_{i-1},Y_{j-1}\right) \frown x_{i} & \mbox{ if } x_i = y_j \\ \mbox{longest}\left(LCS\left(X_{i},Y_{j-1}\right),LCS\left(X_{i-1},Y_{j}\right)\right) & \mbox{ if } x_i \ne y_j \\ \end{cases}$

/* Returns length of LCS for X[0..m-1], Y[0..n-1] */
int lcs( char *X, char *Y, int m, int n )
{
   if (m == 0 || n == 0)
     return 0;
   if (X[m-1] == Y[n-1])
     return 1 + lcs(X, Y, m-1, n-1);
   else
     return max(lcs(X, Y, m, n-1), lcs(X, Y, m-1, n));
}

/* Returns length of LCS for X[0..m-1], Y[0..n-1] */
int lcs( char *X, char *Y, int m, int n )
{
   int L[m+1][n+1];
   int i, j;

   /* Following steps build L[m+1][n+1] in bottom up fashion. Note
      that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1] */
   for (i=0; i<=m; i++)
   {
     for (j=0; j<=n; j++)
     {
       if (i == 0 || j == 0)
         L[i][j] = 0;

       else if (X[i-1] == Y[j-1])
         L[i][j] = L[i-1][j-1] + 1;

       else
         L[i][j] = max(L[i-1][j], L[i][j-1]);
     }
   }

   /* L[m][n] contains length of LCS for X[0..n-1] and Y[0..m-1] */
   return L[m][n];
}

// Following code is used to print LCS
   int index = L[m][n];
   // Create a character array to store the lcs string
   char lcs[index+1];
   lcs[index] = '\0'; // Set the terminating character
   // Start from the right-most-bottom-most corner and
   // one by one store characters in lcs[]
   int i = m, j = n;
   while (i > 0 && j > 0)
   {
      // If current character in X[] and Y are same, then
      // current character is part of LCS
      if (X[i-1] == Y[j-1])
      {
          lcs[index-1] = X[i-1]; // Put current character in result
          i--; j--; index--;     // reduce values of i, j and index
      }
      // If not same, then find the larger of two and
      // go in the direction of larger value
      else if (L[i-1][j] > L[i][j-1])
         i--;
      else
         j--;
   }
   // Print the lcs
   cout << "LCS of " << X << " and " << Y << " is " << lcs;

### C Questions

C Questions
C Questions

Note : All the programs are tested under Turbo C/C++ compilers.
It is assumed that,
Programs run under DOS environment, The underlying machine is an x86 system, Program is compiled using Turbo C/C++ compiler.
The program output may depend on the information based on this assumptions (for example sizeof(int) == 2 may be assumed).
Predict the output or error(s) for the following:

void main()
{
int const * p=5; printf("%d",++(*p));
}
Compiler error: Cannot modify a constant value.
Explanation:
p is a pointer to a "constant integer". But we tried to change the value of the "constant integer".
main()
{
char s[ ]="man"; int i;
for(i=0;s[ i ];i++)
printf("\n%c%c%c%c",s[ i ],*(s+i),*(i+s),i[s]);
}
aaaa nnnn
Explanation

### Zoho Interview | Set 1 (Advanced Programming Round)

Third Round: (Advanced Programming Round) Here they asked us to create a “Railway reservation system” and gave us 4 modules. The modules were:
1. Booking
2. Availability checking
3. Cancellation
4. Prepare chart
We were asked to create the modules for representing each data first and to continue with the implementation phase.

My Solution :