day6:LCS in python

Author: mandliya

README.md

@@ -7,10 +7,10 @@
 | Current Status|     Stats     |
 | :------------: | :----------: |
 | Total C++ Problems | 188 |
-| Total Python Problems | 10 |
-| Current Daily Streak| 5 |
+| Total Python Problems | 11 |
+| Current Daily Streak| 6 |
 | Last Streak | 06/20/2019 - 06/21/2019|
-| Current Streak | 06/23/2019 - 06/27/2019|
+| Current Streak | 06/23/2019 - 06/28/2019|
 
 </center>
 
@@ -106,7 +106,7 @@ Include contains single header implementation of data structures and some algori
 | Problem | Solution |
 | :------------ | :----------: |
 | N<sup>th</sup> Fibonacci term using different memoization techniques | [fibonacci.cpp](dynamic_programming_problems/fibonacci.cpp)|
-| Longest Common Subsequence Problem | [lcs.cpp](dynamic_programming_problems/lcs.cpp) |
+| Longest Common Subsequence Problem | [lcs.cpp](dynamic_programming_problems/lcs.cpp), [longest_common_subsequence.py](dynamic_programming_problems/longest_common_subsequence.py) |
 | Maximum Value Contigous Subsequence Problem [wiki](https://en.wikipedia.org/wiki/Maximum_subarray_problem)| [max_subsequence.cpp](dynamic_programming_problems/max_subsequence.cpp)|
 | Catalan number - Count the number of possible Binary Search Trees with n keys | [catalan_number.cpp](dynamic_programming_problems/catalan_number.cpp)|
 | Calculate the number of unique paths from source origin (0, 0) to  destination (m-1, n-1) in a m x n grid. You can only move either in down or right direction.|[unique_paths.cpp](dynamic_programming_problems/unique_paths.cpp)|

dynamic_programming_problems/longest_common_subsequence.py

@@ -0,0 +1,33 @@
+"""
+Longest common Subsequence problem
+If we are given two sequences, find the length of longest common subsequences present in both of them.
+For example:
+String1: helloworld
+String2: ellorald
+Longest common substring should be "ellorld" of size 7
+"""
+
+def lcs(str1, str2):
+    """Function to calculate longest common substring between str1 and str2"""
+    m = len(str1)
+    n = len(str2)
+
+    dp = [[None] * (n+1) for i in range(m+1)]
+
+    for i in range(m+1):
+        for j in range(n+1):
+            if i == 0 or j == 0:
+                dp[i][j] = 0
+            elif str1[i-1] == str2[j-1]:
+                dp[i][j] = 1 + dp[i-1][j-1]
+            else:
+                dp[i][j] = max(dp[i-1][j], dp[i][j-1])
+
+    return dp[m][n]
+
+
+if __name__ == "__main__":
+    str1 = "helloworld"
+    str2 = "ellorald"
+
+    print(F"The length of longest common subsequence between '{str1}' and '{str2}' is:", lcs(str1, str2))