Added Longest Common Subsequence Algorithm (#91)

s-sanyal · OmkarPathak · commit 0be25396340f · 2019-10-06T06:50:57.000+05:30
diff --git a/pygorithm/dynamic_programming/lcs.py b/pygorithm/dynamic_programming/lcs.py
@@ -0,0 +1,48 @@
+"""
+A subsequence is a sequence that can be derived from another
+sequence by deleting some or no elements without changing the
+order of the remaining elements.
+
+For example, 'abd' is a subsequence of 'abcd' whereas 'adc' is not
+
+Given 2 strings containing lowercase english alphabets, find the length
+of the Longest Common Subsequence (L.C.S.).
+
+Example:
+    Input:  'abcdgh'
+            'aedfhr'
+    Output: 3
+
+    Explanation: The longest subsequence common to both the string is "adh"
+
+Time Complexity : O(M*N)
+Space Complexity : O(M*N), where M and N are the lengths of the 1st and 2nd string
+respectively.
+
+"""
+
+
+def longest_common_subsequence(s1, s2):
+    """
+    :param s1: string
+    :param s2: string
+    :return: int
+    """
+    m = len(s1)
+    n = len(s2)
+
+    dp = [[0] * (n + 1) for i in range(m + 1)]
+    """
+    dp[i][j] : contains length of LCS of s1[0..i-1] and s2[0..j-1]
+    """
+
+    for i in range(m + 1):
+        for j in range(n + 1):
+            if i == 0 or j == 0:
+                dp[i][j] = 0
+            elif s1[i - 1] == s2[j - 1]:
+                dp[i][j] = dp[i - 1][j - 1] + 1
+            else:
+                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
+
+    return dp[m][n]
