LCS py implementation

lcs.py

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+# Dynamic programming implementation of LCS problem 
+
+# Returns length of LCS for X[0..m-1], Y[0..n-1]  
+def lcs(X, Y, m, n): 
+    L = [[0 for x in range(n+1)] for x in range(m+1)] 
+
+    # Following steps build L[m+1][n+1] in bottom up fashion. Note 
+    # that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-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
+            else: 
+                L[i][j] = max(L[i-1][j], L[i][j-1])
+
+    # Following code is used to print LCS 
+    index = L[m][n] 
+
+    # Create a character array to store the lcs string 
+    lcs = [""] * (index+1) 
+    lcs[index] = "" 
+
+    # Start from the right-most-bottom-most corner and 
+    # one by one store characters in lcs[] 
+    i = m 
+    j = n 
+    while i > 0 and j > 0: 
+
+        # If current character in X[] and Y are same, then 
+        # current character is part of LCS 
+        if X[i-1] == Y[j-1]: 
+            lcs[index-1] = X[i-1] 
+            i-=1
+            j-=1
+            index-=1
+
+        # If not same, then find the larger of two and 
+        # go in the direction of larger value 
+        elif L[i-1][j] > L[i][j-1]: 
+            i-=1
+        else: 
+            j-=1
+
+    print("LCS of " + X + " and " + Y + " is " + "".join(lcs) )
+
+# Driver program 
+X = "acf"
+Y = "abcf"
+m = len(X) 
+n = len(Y) 
+lcs(X, Y, m, n)