implement longest common subsequence

pyds_a/algorithms/dynamic_programming/longest_common_subsequence.py

@@ -0,0 +1,64 @@
+"""
+Author: Avijit-Dey
+
+longest_common_subsequence.py - A Python implementation of Longest Common Subsequence,
+a famous problem in Dynamic Programming.
+
+A longest common subsequence (LCS) is the longest subsequence common to all sequences
+in a set of sequences (often just two sequences). It differs from the longest common
+substring: unlike substrings, subsequences are not required to occupy consecutive
+positions within the original sequences. The problem of computing longest common 
+subsequences is a classic computer science problem, the basis of data comparison
+programs such as the diff utility, and has applications in computational linguistics
+and bioinformatics. It is also widely used by revision control systems such as Git for
+reconciling multiple changes made to a revision-controlled collection of files. 
+
+Resources used:
+--https://en.wikipedia.org/wiki/Longest_common_subsequence
+
+For doctests run the following command:
+python3 -m doctest -v longest_common_subsequence.py (linux/unix)
+python -m doctest -v longest_common_subsequence.py (windows)
+
+For manual testing run:
+python3 longest_common_subsequence.py (linux/unix)
+python longest_common_subsequence.py (windows)
+"""
+
+
+def longest_common_subsequence(first_string, second_string):
+    """
+
+    Examples:
+    >>> longest_common_subsequence("AGGTAB", "GXTXAYB")
+    4
+    >>> longest_common_subsequence("AAAAAA", "GGGGGG")
+    0
+    >>> longest_common_subsequence("ROFLABU", "DAFLADBU")
+    5
+    """
+    # length of the strings
+    len_first_str = len(first_string)
+    len_secong_str = len(second_string)
+
+    # declaring the lookup table list for storing the values
+    lookup_table = [[0] * (len_secong_str + 1) for i in range(len_first_str + 1)]
+
+    # Following steps build in bottom up fashion:
+    for i in range(len_first_str + 1):
+        for j in range(len_secong_str + 1):
+            if i == 0 or j == 0:
+                lookup_table[i][j] = 0
+            elif first_string[i - 1] == second_string[j - 1]:
+                lookup_table[i][j] = lookup_table[i - 1][j - 1] + 1
+            else:
+                lookup_table[i][j] = max(lookup_table[i - 1][j], lookup_table[i][j - 1])
+
+    # return the length of the longest common subsequence
+    return lookup_table[len_first_str][len_secong_str]
+
+
+if __name__ == "__main__":
+    import doctest
+
+    doctest.testmod()