寻找最长的公共子串[DP]时如何获得空间复杂度O（n）？

Question

¡Hello!

I'm trying to find the longest common substring between two strings with a good time and space complexity, following using dynamic programming. I could find a solution with O(n^2) time and space complexity:

public static String LCS(String s1, String s2){
    int maxlen = 0;            // stores the max length of LCS      
    int m = s1.length();
    int n = s2.length();
    int endingIndex = m;  // stores the ending index of LCS in X

    // lookup[i][j] stores the length of LCS of substring
    // X[0..i-1], Y[0..j-1]
    int[][] lookup = new int[m + 1][n + 1];

    // fill the lookup table in bottom-up manner
    for (int i = 1; i <= m; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            // if current character of X and Y matches
            if (s1.charAt(i - 1) == s2.charAt(j - 1))
            {
                lookup[i][j] = lookup[i - 1][j - 1] + 1;

                // update the maximum length and ending index
                if (lookup[i][j] > maxlen)
                {
                    maxlen = lookup[i][j];
                    endingIndex = i;
                }
            }
        }
    }

    // return Longest common substring having length maxlen
    return s1.substring(endingIndex - maxlen, endingIndex);

}

My question is: How can I get better space complexity?

Thanks in advance!

Answer 1

The best time complexity you can get for finding the LCS of two Strings is O(n^2) using dynamic programming. I tried to find another algorithm for the problem since it was one of my University projects. But the best thing I could find was an algorithm with O(n^3) complexity. The main solution of this problem uses "Recurrence relation" which uses less space but far more process. But like "Fibonacci series" computer scientists used dynamic programming to reduce the time complexity. The Recurrence relation code:

void calculateLCS(string &lcs , char frstInp[] , char secInp[] , int lengthFrstInp ,  int lengthSecInp) {

if (lengthFrstInp == -1 || lengthSecInp == -1)
    return;

if (frstInp[lengthFrstInp] == secInp[lengthSecInp]) {
    lcs += frstInp[lengthFrstInp];
    lengthFrstInp--;
    lengthSecInp--;
    calculateLCS(lcs, frstInp, secInp, lengthFrstInp, lengthSecInp);

}


else {

    string lcs1 ="";
    string lcs2 ="";    
    lcs1 = lcs;
    lcs2 = lcs;
    calculateLCS(lcs1, frstInp, secInp, lengthFrstInp, lengthSecInp - 1);
    calculateLCS(lcs2, frstInp, secInp, lengthFrstInp - 1, lengthSecInp);

    if (lcs1.size() >= lcs2.size())
        lcs = lcs1;
    else
        lcs = lcs2;

}