diff --git "a/leetcode3/\353\202\250\355\232\250\354\240\225/315. Count of Smaller Numbers After Self.py" "b/leetcode3/\353\202\250\355\232\250\354\240\225/315. Count of Smaller Numbers After Self.py"
new file mode 100644
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+++ "b/leetcode3/\353\202\250\355\232\250\354\240\225/315. Count of Smaller Numbers After Self.py"	
@@ -0,0 +1,44 @@
+# 풀이 실패
+# 병합 정렬할 때 오른쪽에서 넘어간 개수 카운트하면 구할 수 있음
+class Solution:
+    def countSmaller(self, nums: List[int]) -> List[int]:
+        n = len(nums)
+        ans = [0] * n
+        arr = [(nums[i], i) for i in range(n)]
+
+        def merge_sort(enum):
+            half = len(enum) // 2
+            if half:
+                left, right = merge_sort(enum[:half]), merge_sort(enum[half:])
+                m, n = len(left), len(right)
+                i = j = 0
+
+                # 투 포인터 사용해서 병합 진행한다
+                while i < m or j < n:
+                    # 오른쪽 다 썼거나 왼쪽 값이 더 작거나 같은 경우
+                    if j == n or (i < m and left[i][0] <= right[j][0]):
+                        ans[left[i][1]] += j
+                        enum[i + j] = left[i]
+                        i += 1
+                    else:
+                        enum[i + j] = right[j]
+                        j += 1
+
+            return enum
+            
+        merge_sort(arr)
+        return ans
+
+        # 시간 복잡도 초과
+        # ans = []
+        # for i in range(len(nums)):
+        #     if i == len(nums) - 1:
+        #         ans.append(0)
+        #         break
+        #     check = 0
+        #     for j in range(i + 1, len(nums)):
+        #         if nums[i] > nums[j]:
+        #             check += 1
+        #     ans.append(check)
+        
+        # return ans
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