This article is a part of series

- Introduction
**Longest Common Subsequence**

In this article, we will look at using the steps mentioned in the introduction article to arrive at a Dynamic Programming solution to the Longest Common Subsequence problem.

**Longest Common Subsequence (LCS)**

**Problem Statement:** Given two sequences, find the length of longest subsequence present in both of them. A subsequence is a sequence that appears in the same relative order, but not necessarily contiguous.

**Examples:**

- LCS for input sequences “ABCDGH” and “AEDFHR” is “ADH”
- LCS for input sequences “HUMAN” and “CHIMPANZEE” is “HMAN”

The longest common subsequence is used to solve problems such as computing how similar two DNA sequences are; and comparing two different versions of the same file.

You can read more about the Longest Common Subsequence here.

**Profiling**

As mentioned in the introduction article, we will be using google benchmark to help profile our solutions. If you want to know more about how we will be using google benchmark, you can read the introduction article.

For the LCS problem, the worst case occurs when there is no match between the sequences. For the purposes of this article, we will profile against the worst case by creating 2 strings of length * n* and fill one of them with

*and the other with*

**‘a’****‘b’**. We will then attempt to see how long it takes for the function to return.

*will be part of the Benchmark name.*

**n**If you want to see the benchmarks in action, you can find the code here.

**Steps**

In the introduction article, we came up with the following steps to find a dynamic programming approach to our problem

- Find the overlapping subproblem.
- Start with a recursive solution
- Modify the recursive solution to use a top-down memoized version.
- Remove the recursion by making it an iterative solution.
- If you don’t need to keep all the previous results, keep only the ones that are required.

**Overlapping Subproblem**

Let the 2 strings be **X** of length * m* and

**Y**of length

**n**Let **LCS(i,j)** be the length of the LCS. Then we can formalise the problem as follows

LCS(i,j) = 0 if X[i] = '\0' || Y[j] = '\0' LCS(i,j) = 1 + LCS(i+1, j+1) if X[i] = Y[j] LCS(i,j) = max(LCS(i+1, j), LCS(i, j+1) if X[i] != Y[j]

The length of the longest common subsequence will be **LCS(0,0)**.

**Naive Recursive Approach**

The above formalisation can easily be translated into the following recursive method

int cLCS::Recursive(const char* const first, const char* const second) { if (*first == '\0' || *second == '\0') { return 0; } if (*first == *second) { return 1 + Recursive(first + 1, second + 1); } return max(Recursive(first + 1, second), Recursive(first, second + 1)); }

In the recursive approach, it is extremely hard to get the actual LCS string so we are just going to return the length of the LCS.

**Top-Down Recursive approach with Memoization**

LCS subproblems consist of a pair of suffixes of the 2 input strings. To store and look up the subproblem solutions we can use a 2d array. We will use a * -1* to tell the algorithm that nothing has been stored as yet.

string cLCS::Memonized(const string& first, const string& second) { int length1 = first.length(); int length2 = second.length(); if (length1 == 0 || length2 == 0) { return ""; } Memonized(first.data(), second.data(), 0, 0); return GetText(first, second); } int cLCS::Memonized(const char* const first, const char* const second, int i, int j) { if (m_results[GetIndex(i, j)] == -1) { if (first[i] == '\0' || second[j] == '\0') { m_results[GetIndex(i, j)] = 0; } else if (first[i] == second[j]) { m_results[GetIndex(i, j)] = 1 + Memonized(first, second, i + 1, j + 1); } else { int val1 = Memonized(first, second, i + 1, j); int val2 = Memonized(first, second, i, j + 1); m_results[GetIndex(i, j)] = max(val1, val2); } } return m_results[GetIndex(i, j)]; }

In the above algorithm, * m_results(0)*, gives the length of the LCS. We will look at

*in the next section to see how we can get the subsequence after computing the*

**GetText()***array.*

**m_results****Getting the subsequence**

Once we have filled in the * m_results* array, we can find the sequence by traversing forwards through the array.

string cLCS::GetText(const string& first, const string& second) { if (m_results[0] == 0) { return ""; } int i = 0; int j = 0; stringstream ss(""); while (i < first.length() && j < second.length()) { if (first[i] == second[j]) { ss << first[i]; i++; j++; } else if (m_results[GetIndex(i + 1, j)] >= m_results[GetIndex(i, j + 1)]) { i++; } else { j++; } } return ss.str(); }

To find the longest common subsequence, we traverse through both the strings – * first* and

**using indexes**

*second**and*

**i***respectively. If*

**j***then we add this character to the result string and increment both*

**first[i] = second[j]**,*and*

**i***. If there is no match, that means that the subsequence was formed either by deleting*

**j***or*

**first[i]***If*

**second[j]**.*>=*

**m_results[i+1][j]***, this means that the subsequence was formed by deleting*

**m_results[i][j + 1]***. Otherwise, it was formed by deleting*

**first[i]***. Continuing in this way, we can get the common subsequence.*

**second[j]****Bottom Up Approach with Dynamic Programming**

To come up with a DP approach, we just flip the way we are storing the results by traversing the array backwards.

string cLCS::DP(const std::string& first, const std::string& second) { int length1 = first.length(); int length2 = second.length(); if (length1 == 0 || length2 == 0) { return ""; } for (int i = length1; i >= 0; i--) { for (int j = length2; j >= 0; j--) { if (first[i] == '\0' || second[j] == '\0') { m_results[GetIndex(i, j)] = 0; } else if (first[i] == second[j]) { m_results[GetIndex(i, j)] = 1 + m_results[GetIndex(i + 1, j + 1)]; } else { m_results[GetIndex(i, j)] = max(m_results[GetIndex(i + 1, j)], m_results[GetIndex(i, j + 1)]); } } } return GetText(first, second); }

The disadvantage of the bottom-up approach over memoizing is that this fills in the entire array even if the problem could be solved by computing a fraction of the array.

**Bottom Up Approach with Dynamic Programming(optimised)**

We can optimise the above solution since once we have computed the row ** i** of array

*, we no longer need the values of i*

**m_results***. However, we cannot recreate the subsequence using this approach and hence I won’t be showing it here.*

**+ 1****Conclusion**

If you would like to look at the code or run the benchmarks or tests yourself, you can find the code here

In the next article in the series, we will look at another problem that can be solved by Dynamic Programming.

Have you tried Dynamic Programming before? How was your experience? Let me know in the comments.

Interesting, but article isn’t done yet, is it? Two sections at the end seem to be missed.

Hi Aleksey,

Thanks for pointing that out. Somehow I published an earlier version. The article has been updated with the missing sections.

Ankur

Pingback: Dynamic Programming Series - Introduction