Longest common sequence - Dynamic programming


LCS Problem Statement: Given two sequences, find the length of longest subsequence present in both of them. A subsequence is a sequence that appears in the same relative order, but not necessarily contiguous. For example, “abc”, “abg”, “bdf”, “aeg”, ‘”acefg”, .. etc are subsequences of “abcdefg”. So a string of length n has 2^n different possible subsequences.
It is a classic computer science problem, the basis of diff (a file comparison program that outputs the differences between two files), and has applications in bioinformatics.

Examples:
LCS for input Sequences “ABCDGH” and “AEDFHR” is “ADH” of length 3.
LCS for input Sequences “AGGTAB” and “GXTXAYB” is “GTAB” of length 4.

/* Dynamic Programming C/C++ implementation of LCS problem */

 #include<bits/stdc++.h>
  
int max(int a, int b);
  
/* 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];
}
  
/* Utility function to get max of 2 integers */
int max(int a, int b)
{
    return (a > b)? a : b;
}
  
/* Driver program to test above function */
int main()
{
  char X[] = "AGGTAB";
  char Y[] = "GXTXAYB";
  
  int m = strlen(X);
  int n = strlen(Y);
  
  printf("Length of LCS is %d\n", lcs( X, Y, m, n ) );
  return 0;
}

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