Update longest common subsequence.py

dynamic_programming/longest common subsequence.py

@@ -3,16 +3,46 @@ LCS Problem Statement: Given two sequences, find the length of longest subsequen
 A subsequence is a sequence that appears in the same relative order, but not necessarily continious.
 Example:"abc", "abg" are subsequences of "abcdefgh". 
 """
-def LCS(s1, s2):
-    m = len(s1)
-    n = len(s2)
+def LCS(x,y):
+    b=[[] for j in range(len(x)+1)]
+    c=[[] for i in range(len(x))]
+    for i in range(len(x)+1):
+        b[i].append(0)
+    for i in range(1,len(y)+1):
+        b[0].append(0)
+    for i in range(len(x)):
+        for j in range(len(y)):
+            if x[i]==y[j]:
+                b[i+1].append(b[i][j]+1)
+                c[i].append('/')
+            elif b[i][j+1]>=b[i+1][j]:
+                b[i+1].append(b[i][j+1])
+                c[i].append('|')
+            else :
+                b[i+1].append(b[i+1][j])
+                c[i].append('-')            
+    return b,c
 
-    arr = [[0 for i in range(n+1)]for j in range(m+1)]
 
-    for i in range(1,m+1):
-        for j in range(1,n+1):
-            if s1[i-1] == s2[j-1]:
-                arr[i][j] = arr[i-1][j-1]+1
-            else:
-                arr[i][j] = max(arr[i-1][j], arr[i][j-1])
-    return arr[m][n]
+def print_lcs(x,c,n,m):
+    n,m=n-1,m-1
+    ans=[]
+    while n>=0 and m>=0:
+        if c[n][m]=='/':
+            ans.append(x[n])
+            n,m=n-1,m-1
+        elif c[n][m]=='|':
+            n=n-1
+        else:
+            m=m-1 
+    ans=ans[::-1]
+    return ans                   
+                       
+
+if __name__=='__main__':
+    x=['a','b','c','b','d','a','b']
+    y=['b','d','c','a','b','a']
+    b,c=LCS(x,y)
+    print('Given \nX : ',x)
+    print('Y : ',y)
+    print('LCS : ',print_lcs(x,c,len(x),len(y)))