diff --git a/Python/10_Regular_Expression_Matching.py b/Python/10_Regular_Expression_Matching.py
new file mode 100644
index 0000000..3200adb
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+++ b/Python/10_Regular_Expression_Matching.py
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+'''
+Link: https://leetcode.com/problems/regular-expression-matching/
+
+10. Regular Expression Matching (Hard)
+
+Given an input string s and a pattern p, implement regular expression matching with support for '.' and '*' where:
+
+'.' Matches any single character.​​​​
+'*' Matches zero or more of the preceding element.
+The matching should cover the entire input string (not partial).
+'''
+
+
+class Solution:
+    def isMatch(self, s: str, p: str) -> bool:
+        if len(re.findall(p,s))==0:
+            return False
+        if re.findall(p,s)[0]==s:
+            return True
+        else:
+            return False
diff --git a/Python/315_Count_of_Smaller_Numbers_After_Self.py b/Python/315_Count_of_Smaller_Numbers_After_Self.py
new file mode 100644
index 0000000..fcff936
--- /dev/null
+++ b/Python/315_Count_of_Smaller_Numbers_After_Self.py
@@ -0,0 +1,19 @@
+'''
+Link: https://leetcode.com/problems/count-of-smaller-numbers-after-self/
+
+315. Count of Smaller Numbers After Self (Hard)
+
+You are given an integer array nums and you have to return a new counts array. The counts array has the property where counts[i] is the number of smaller elements to the right of nums[i].
+'''
+
+
+class Solution:
+    def countSmaller(self, nums: List[int]) -> List[int]:
+        nums = nums[::-1]
+        out = []
+        sorted = []
+        for num in nums:
+            c = bisect.bisect_left(sorted, num)
+            out.append(c)
+            bisect.insort(sorted, num)
+        return out[::-1]