We used dynamic programming to solve the longest common subsequence (LCS) proble
ID: 3694481 • Letter: W
Question
We used dynamic programming to solve the longest common subsequence (LCS) problem in class. This algorithm needs to use a 2 dimensional array with size |x| * |y| given that x and y are two input sequences. Now, modify the dynamic programming algorithm so that we only need to use a linear table with size either k|x| or k|y|, where k is a small constant, such as 2. Describe your idea (or modification) and then write your pseudocode. Note: you don’t need to find the longest common subsequence between x and y. You only need to find the size of an LCS
Explanation / Answer
Answer for Question:
This below psedocode or c code will fine the length of Longest common subsequnce of given strings using dynamic programming algorithm.
See the below code..
/* 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 ", lcs( X, Y, m, n ) );
return 0;
}
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