Understand the time complexity for this LCS (longest common subsequence) solution

Question

I would appreciate an intuitive way to find the time complexity of dynamic programming problems. Can anyone explain me “#subproblems * time/subproblem”? I am not able to grok it.

Code for LCS -

public static String findLCS(String str1, String str2 ) {
    // If either string is empty, return the empty string
    if(null == str1 || null == str2)
        return "";
    if("".equals(str1) || "".equals(str2)) {
        return "";
    }
    // are the last characters identical?
    if(str1.charAt(str1.length()-1) == str2.charAt(str2.length()-1)) {
        // yes, so strip off the last character and recurse
        return findLCS(str1.substring(0, str1.length() -1), str2.substring(0, str2.length()-1)) + str1.substring(str1.length()-1, str1.length());
    } else {
       // no, so recurse independently on (str1_without_last_character, str2)
       // and (str1, str2_without_last_character)
       String opt1 = findLCS(str1.substring(0, str1.length() -1), str2); 
       String opt2 = findLCS(str1, str2.substring(0, str2.length()-1));
       // return the longest LCS found
       if(opt1.length() >= opt2.length())
           return opt1;
       else
           return opt2;
    }
}

I am just providing the actual code instead of pseudo code (i hope pseudo code or the algo is pretty self explanatory from above)

See here for general techniques on runtime analysis. – Raphael Mar 04 '15 at 07:22 — Raphael, Mar 04 '15 at 07:22

score 1 · Accepted Answer · answered Aug 11 '13 at 14:45

I assume you are interested in worst-case complexity.

A dynamic programming approach (recursively) breaks a problem into "smaller" problems, and records the solution of these subproblems to make sure that subproblems are not computed several times. So when the time required to combine subproblems is constant (it is the case for LCS), you only have to find an upper bound for the number of possible subproblems that will be solved along the program.

Here the subproblems are of the form findLCS(s1,s2) where s1 is a prefix of str1, and s2 a prefix of str2. There are (1+str1.length()) possible prefixes for str1, and (1+str2.length()) for str2.
The number of possible subproblems is therefore (1+str1.length()) * (1+str2.length()). Furthermore, a solution is obtained in constant time from its subproblems. Hence complexity O( str1.length()*str2.length() ).

N.B. on some sequences all subproblems will be indeed computed, e.g., when letters in str1 and str2 are distinct. Consequently, the quadratic bound is "tight".

N.B.2 you may find (short) explanations on the wikipedia page for LCS

"So when the time required to combine subproblems is constant (it is the case for LCS)".. so this is because i m not doing anything once i get solution to the just smaller subproblem?? — abipc, Aug 11 '13 at 16:04
Yes. Well hardly anything (you only compare the length of the two answers opt1 and opt2.).
N.B.: I assumed that list.length() has constant complexity. It is true in Python, and would be negligible in practice for most languages unless you consider extremely large strings. Anyway you could always store the length of the string together with the string if this became an issue. — Joseph Stack, Aug 12 '13 at 11:47

Christian Vielma · Answer 2 · 2015-03-05T01:37:45.680

-1

The code posted doesn't implement Dynamic Programming, so the time complexity is in fact O(2^n).

See this Wikipedia article and this GeeksforGeeks post for pseudocode and specific implementations.This post also shows how to get the LCS in a recursive/iteratively way using DP.

To implement Dynamic Programming is necessary to implement a cache mechanism to prevent calculating the same sub-results multiple times.

edited Mar 05 '15 at 01:37

answered Mar 04 '15 at 00:36

Christian Vielma

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4

1

Please get rid of the source code and replace it with ideas, pseudo code and arguments of correctness. See here and here for related meta discussions. – Raphael Mar 04 '15 at 07:21

Understand the time complexity for this LCS (longest common subsequence) solution

2 Answers2