From 32aa7ff0819dba3d2166b8a66317999a7790e510 Mon Sep 17 00:00:00 2001
From: Maxwell Aladago <maxwell.aladago@gmail.com>
Date: Mon, 19 Aug 2019 03:40:36 -0400
Subject: [PATCH] ENH: refactored longest common subsequence, also fixed a bug
 with the sequence returned (#1142)

* function for the knapsack problem which returns one of the optimal subsets

* function for the knapsack problem which returns one of the optimal subsets

* function for the knapsack problem which returns one of the optimal subsets

* function for the knapsack problem which returns one of the optimal subsets

* function for the knapsack problem which returns one of the optimal subsets

* some pep8 cleanup too

* ENH: refactored longest common subsequence, also fixed a bug with the sequence returned

* renamed function
---
 .../longest_common_subsequence.py             | 88 ++++++++++++++-----
 1 file changed, 67 insertions(+), 21 deletions(-)

diff --git a/dynamic_programming/longest_common_subsequence.py b/dynamic_programming/longest_common_subsequence.py
index 0a4771cb2..7836fe303 100644
--- a/dynamic_programming/longest_common_subsequence.py
+++ b/dynamic_programming/longest_common_subsequence.py
@@ -1,37 +1,83 @@
 """
 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 continious.
+A subsequence is a sequence that appears in the same relative order, but not necessarily continuous.
 Example:"abc", "abg" are subsequences of "abcdefgh".
 """
 from __future__ import print_function
 
-try:
-    xrange          # Python 2
-except NameError:
-    xrange = range  # Python 3
 
-def lcs_dp(x, y):
+def longest_common_subsequence(x: str, y: str):
+    """
+    Finds the longest common subsequence between two strings. Also returns the
+    The subsequence found
+
+    Parameters
+    ----------
+
+    x: str, one of the strings
+    y: str, the other string
+
+    Returns
+    -------
+    L[m][n]: int, the length of the longest subsequence. Also equal to len(seq)
+    Seq: str, the subsequence found
+
+    >>> longest_common_subsequence("programming", "gaming")
+    (6, 'gaming')
+    >>> longest_common_subsequence("physics", "smartphone")
+    (2, 'ph')
+    >>> longest_common_subsequence("computer", "food")
+    (1, 'o')
+    """
     # find the length of strings
+
+    assert x is not None
+    assert y is not None
+
     m = len(x)
     n = len(y)
 
     # declaring the array for storing the dp values
-    L = [[None] * (n + 1) for i in xrange(m + 1)]
-    seq = []
+    L = [[0] * (n + 1) for _ in range(m + 1)]
 
-    for i in range(m + 1):
-        for j in range(n + 1):
-            if i == 0 or j == 0:
-                L[i][j] = 0
-            elif x[i - 1] == y[ j - 1]:
-                L[i][j] = L[i - 1][j - 1] + 1
-                seq.append(x[i -1])
+    for i in range(1, m + 1):
+        for j in range(1, n + 1):
+            if x[i-1] == y[j-1]:
+                match = 1
             else:
-                L[i][j] = max(L[i - 1][j], L[i][j - 1])
-    # L[m][n] contains the length of LCS of X[0..n-1] & Y[0..m-1]
+                match = 0
+
+            L[i][j] = max(L[i-1][j], L[i][j-1], L[i-1][j-1] + match)
+
+    seq = ""
+    i, j = m, n
+    while i > 0 and i > 0:
+        if x[i - 1] == y[j - 1]:
+            match = 1
+        else:
+            match = 0
+
+        if L[i][j] == L[i - 1][j - 1] + match:
+            if match == 1:
+                seq = x[i - 1] + seq
+            i -= 1
+            j -= 1
+        elif L[i][j] == L[i - 1][j]:
+            i -= 1
+        else:
+            j -= 1
+
     return L[m][n], seq
 
-if __name__=='__main__':
-    x = 'AGGTAB'
-    y = 'GXTXAYB'
-    print(lcs_dp(x, y))
+
+if __name__ == '__main__':
+    a = 'AGGTAB'
+    b = 'GXTXAYB'
+    expected_ln = 4
+    expected_subseq = "GTAB"
+
+    ln, subseq = longest_common_subsequence(a, b)
+    assert expected_ln == ln
+    assert expected_subseq == subseq
+    print("len =", ln, ", sub-sequence =", subseq)
+