diff --git a/0015/3Sum.py b/0015/3Sum.py
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--- /dev/null
+++ b/0015/3Sum.py
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+class Solution:
+    def threeSum(self, nums: List[int]) -> List[List[int]]:
+        nums.sort()
+        ans = set()
+        for i in range(len(nums) - 2):
+            j = i + 1
+            k = len(nums) - 1
+            while k > j:
+                s = nums[i] + nums[j] + nums[k]
+                if s == 0:
+                    ans.add((nums[i], nums[j], nums[k]))
+                    j += 1
+                elif s < 0:
+                    j += 1
+                else:
+                    k -= 1
+        return list(ans)
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diff --git a/0015/README.md b/0015/README.md
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index 0000000..e3d690f
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+++ b/0015/README.md
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+# 3Sum
+Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.
+
+Notice that the solution set must not contain duplicate triplets.
+
+ 
+
+## Example 1:
+
+Input: nums = [-1,0,1,2,-1,-4]
+
+Output: [[-1,-1,2],[-1,0,1]]
+
+## Example 2:
+
+Input: nums = []
+
+Output: []
+
+## Example 3:
+
+Input: nums = [0]
+
+Output: []
+ 
+
+## Constraints:
+
+0 <= nums.length <= 3000
+-105 <= nums[i] <= 105
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