What is the time complexity and pseudocode of the above algorithm?
public class Soru32
{
// Returns length of LCS for X[0..m-1], Y[0..n-1]
static void lcs(String X, String Y, int m, int n)
{
int[][] L = new int[m+1][n+1];
// 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 (int i=0; i<=m; i++)
{
for (int j=0; j<=n; j++)
{
if (i == 0 || j == 0)
L[i][j] = 0;
else if (X.charAt(i-1) == Y.charAt(j-1))
L[i][j] = L[i-1][j-1] + 1;
else
L[i][j] = Math.max(L[i-1][j], L[i][j-1]);
}
}
// Following code is used to print LCS
int index = L[m][n];
int temp = index;
// Create a character array to store the lcs string
char[] lcs = new char[index+1];
lcs[index] = '\u0000'; // Set the terminating character
// Start from the right-most-bottom-most corner and
// one by one store characters in lcs[]
int i = m;
int 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.charAt(i-1) == Y.charAt(j-1))
{
// Put current character in result
lcs[index-1] = X.charAt(i-1);
// reduce values of i, j and index
i--;
j--;
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
System.out.print("LCS of "+X+" and "+Y+" is ");
for(int k=0;k<=temp;k++)
System.out.print(lcs[k]);
}
// driver program
public static void main (String[] args)
{
String S1 = "BCDAACD";
String S2 = "ACDBAC";
int m = S1.length();
int n = S2.length();
lcs(S1, S2, m, n);
}
}
What is the time complexity and pseudocode of the above
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