DP # LCS

Author: harrypotter0

CP/January Challenge 2018/Untitled Document

@@ -0,0 +1,19 @@
+def longest_common_subsequence(sequence1, sequence2):
+    cols = len(sequence1) + 1   # Add 1 to represent 0 valued column for DP
+    rows = len(sequence2) + 1   # Add 1 to represent 0 valued row for DP
+
+    T = [[0 for _ in range(cols)] for _ in range(rows)]
+
+    max_length = 0
+
+    for i in range(1, rows):
+        for j in range(1, cols):
+            if sequence2[i - 1] == sequence1[j - 1]:
+                T[i][j] = 1 + T[i - 1][j - 1]
+            else:
+                T[i][j] = max(T[i - 1][j], T[i][j - 1])
+
+            max_length = max(max_length, T[i][j])
+
+    return max_length
+

CP/January Challenge 2018/string.py

@@ -0,0 +1,19 @@
+def longest_common_subsequence(sequence1, sequence2):
+    cols = len(sequence1) + 1   # Add 1 to represent 0 valued column for DP
+    rows = len(sequence2) + 1   # Add 1 to represent 0 valued row for DP
+
+    T = [[0 for _ in range(cols)] for _ in range(rows)]
+
+    max_length = 0
+
+    for i in range(1, rows):
+        for j in range(1, cols):
+            if sequence2[i - 1] == sequence1[j - 1]:
+                T[i][j] = 1 + T[i - 1][j - 1]
+            else:
+                T[i][j] = max(T[i - 1][j], T[i][j - 1])
+
+            max_length = max(max_length, T[i][j])
+
+    return max_length
+

dynamic/longest_common_subsequence.py

@@ -5,11 +5,6 @@
 Given two sequences A = [A1, A2, A3,..., An] and B = [B1, B2, B3,..., Bm], find the length of the longest common
 subsequence.
 
-Video
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-
-* https://youtu.be/NnD96abizww
-
 Complexity
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