Another 2D dynamic programming problem, this time with strings. You can work in any programming language of your choice - Python, C++, Java, or anything else you want. Let’s get started!

Problem Statement

Given two strings, find a longest subsequence common to both strings. Note that unlike substrings, subsequences are not required to be a consecutive section of the original strings.

Your solution should have run-time O($nm$).

Note: There may be multiple subsequences of the same length. You are required to calculate any subsequence of the maximum possible length.

Examples

Example 1:

String 1: ABBBAABB
String 2: ABBABA

Answer: “ABBBA” (length=5)

Example 2:

String 1: ABACCAA
String 2: BAACABAC

Answer: “BACAA” (length=5)

Example 3:

String 1: DBCCBDDACD
String 2: CBDCB

Answer: “CBDC” (length=4)

Submitting your solution

Once you are confidant of your solution, take the quiz! It will provide you with some inputs to run your code on.

Progressive ‘hint-by-hint’ solution

Only look at the hints if you can’t solve the problem for at-least 20-minutes. Read only one hint that you could not figure out on your own. Then spend at-least 20-minutes trying to solve the problem again.

Hint 1 (dp state):

lcs(i, j) = length of longest common subsequence of string1[0 … i] and string2[0 … j]

Hint 2 (recursion):

There are two cases: string1[i] == string2[j] or string1[i] != string2[j].

Hint 3:

The string1[i] == string2[j] case is left for you. If string1[i] != string2[j], then the LCS either does not include string1[i] or does not include string2[j]. Hence, lcs(i, j) = max(lcs(i-1, j), lcs(i, j-1)).

Hint 4 (constructing the actual subsequence):

You’ll need to define previous(i, j) to remember the path you took from (0, 0) to (i, j). previous(i, j) should be one of (i-1, j), or (i, j-1) or (i-1, j-1).

Full solution

The solution for this assignment is included at the end of the quiz.