diff --git a/Data Structures/Dynamic Programming/longestpalindromicsubstring.py b/Data Structures/Dynamic Programming/longestpalindromicsubstring.py
new file mode 100644
index 00000000..e48b1c82
--- /dev/null
+++ b/Data Structures/Dynamic Programming/longestpalindromicsubstring.py	
@@ -0,0 +1,41 @@
+// Problem Statement link:- https://leetcode.com/problems/longest-palindromic-substring/	
+
+  def longestPalindrome(self, s: str) -> str:
+
+	  n=len(s)
+	  dp=[]
+	  for _ in range(n):
+		dp.append([None]*n)
+
+	  maxLen=1
+	  start,end=0,0 #start,end
+
+	  for i in range(n):
+		dp[i][i]=1 #main diag
+
+		if i+1<=n-1:#2nd diag
+		  if s[i]==s[i+1]: 
+			dp[i][i+1]=2 
+			maxLen=2
+			start,end=i,i+1 #start,end
+		  else:
+			dp[i][i+1]=0
+
+
+
+	  for d in range(2,n): #2,3,4,..,n
+		for rStart in range(n):
+		  cEnd=d+rStart
+		  if cEnd>n-1: break
+
+		  if s[rStart]==s[cEnd] and dp[rStart+1][cEnd-1]>0:
+			dp[rStart][cEnd]=dp[rStart+1][cEnd-1]+2
+
+			#update the longest palindromic substring
+			if (cEnd-rStart+1)>maxLen:
+			  maxLen=(cEnd-rStart+1)
+			  start,end=rStart,cEnd
+
+		  else:dp[rStart][cEnd]=0
+
+	  return s[start:end+1]
diff --git a/Leetcode/Python/Longest_Common_Subsequence.py b/Leetcode/Python/Longest_Common_Subsequence.py
new file mode 100644
index 00000000..fd1e2241
--- /dev/null
+++ b/Leetcode/Python/Longest_Common_Subsequence.py
@@ -0,0 +1,22 @@
+# Problem Description:- Longest_Common_Subsequence
+#Link:- https://leetcode.com/problems/longest-common-subsequence/
+
+
+class Solution:
+    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
+        
+        m = len(text2)
+        n = len(text1)
+        
+        memo = [[0 for j in range(m+1)] for i in range(n+1)]
+        
+        for i in range(n-1,-1,-1):
+            for j in range(m-1,-1,-1):
+                
+                if text1[i] == text2[j]:
+                    memo[i][j] = 1 + memo[i+1][j+1]
+                    
+                else:
+                    memo[i][j] = max(memo[i+1][j],memo[i][j+1])
+           
+        return memo[0][0]