Merge pull request #203 from manthan0227/main

Added Longest Common Subsequence

Author: prathimacode-hub

Dynamic Programming/Longest Common Subsequence/Images/TEST_1.png

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+
+# Longest Common Subsequence 
+### Aim 
+- The aim of the script is to find out the maximum length of the common sequence between two strings.
+
+### Purpose
+- The main purpose of this script is to show the implementaion of recursion to solve the Longest Common Subsequence.
+
+### Description
+Given two string and determine the length of the string which is common subsequence.
+
+### Workflow
+`Input:`
+s1 = "ABCDEFG", 
+s2 = "DEFGHI"
+
+`output:`
+Length of the LCS: 4
+
+### Explanation
+- Find the number of subsequences with lengths ranging from 1,2,..n-1. In Recursion we first passed the two string and length of the two string as parameters.
+- We first find the size of the both string if they are both empty string then we return 0.
+- And if we matching the charcter of two string and those two charcters are matched then we add one.
+- Otherwise we check, if next charcter is match to the this charcter than we go to that side and decrease the size of the string and add one to our answer.
+- At the last we get the length of the longest common subsequence.
+
+### Time and Space Complexity
+- Time Complexity: `O(len_of_string1*len_of_string2)`
+- Space Complexity:`O(len_of_the_lcs)` 
+
+## 💻 Input and Output 
+- **Test Case 1 :**
+
+![](https://github.com/manthan0227/PyAlgo-Tree/blob/main/Dynamic%20Programming/Longest%20Common%20Subsequence/Images/TEST_1.png)
+
+- **Test Case 2 :**
+
+![](https://github.com/manthan0227/PyAlgo-Tree/blob/main/Dynamic%20Programming/Longest%20Common%20Subsequence/Images/TEST_2.png)

Dynamic Programming/Longest Common Subsequence/Images/TEST_2.png

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+# LCS: There are two strings in which we have to find the length of the longest common subsequence
+# Subarray: Continuous sequence in an array
+# Subsequence: Need not to be contiguous, but maintains order
+# Subset: Same as subsequence except it has empty set
+
+
+def lcs(n, m, s1, s2):
+    if n == 0 or m == 0: # if the one of the string size 0 then we returned 0
+        return 0
+    if s1[n - 1] == s2[m - 1]: # if string have the same letter then we add one and go to the next letter
+        return 1 + lcs(n - 1, m - 1, s1, s2);
+    else:                       # otherwise we do maximum of both strings 
+        return max(lcs(n, m - 1, s1, s2), lcs(n - 1, m, s1, s2))
+
+if __name__ == "__main__":
+    print("Longest Common Subsequence Problem")
+    s1 = input("First String: ") # input the first string
+    s2 = input("Second String: ") # input the second string
+    print("Length of the Longest Common Subsequence: ", lcs(len(s1), len(s2), s1, s2)) # len(s) is length of the string s