diff --git a/docs/solutions/(array)/1-two-sum.mdx b/docs/solutions/(array)/1-two-sum.mdx
new file mode 100644
index 0000000..6c7753f
--- /dev/null
+++ b/docs/solutions/(array)/1-two-sum.mdx
@@ -0,0 +1,47 @@
+---
+title: '1. Two Sum'
+description: You are given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target
+sidebar:
+  label: 'Two Sum'
+  badge: 'Easy'
+---
+
+<Badge variant="accent">Hash Table</Badge>
+
+::::warning
+Can you come up with an algorithm that is less than O(n^2) time complexity?
+::::
+
+### Example 1:
+- Input: `nums = [2,7,11,15], target = 9`
+- Output: `[0,1]`
+- Explanation: Because `nums[0] + nums[1]` == `9`, we return [0, 1].
+
+### Example 2:
+- Input: `nums = [3,2,4], target = 6`
+- Output: `[1,2]`
+
+### Example 3:
+- Input: `nums = [3,3], target = 6`
+- Output: `[0,1]`
+
+### Constraints:
+
+- `2 <= nums.length <= 10^4`
+- `-10^9 <= nums[i] <= 10^9`
+- `-10^9 <= target <= 10^9`
+- Only one valid answer exists.
+
+## Solution
+
+```py
+class Solution:
+    def twoSum(self, nums: List[int], target: int) -> List[int]:
+        seen = {}
+        for i, num in enumerate(nums):
+            complement = target - num
+            if complement in seen:
+                return [seen[complement], i]
+            seen[num] = i
+        return []
+```
diff --git a/docs/solutions/(array)/15-3sum.mdx b/docs/solutions/(array)/15-3sum.mdx
index f2baa05..9ca8ec4 100644
--- a/docs/solutions/(array)/15-3sum.mdx
+++ b/docs/solutions/(array)/15-3sum.mdx
@@ -34,12 +34,45 @@ Notice that the solution set must not contain duplicate triplets.
 
 ## Solution
 
-```js
-/**
- * @param {number[]} nums
- * @return {number[][]}
- */
-var threeSum = function(nums) {
-    
-};
+```py
+class Solution:
+    def threeSum(self, nums: list[int]) -> list[list[int]]:
+        nums.sort()
+        result = []
+        n = len(nums)
+
+        for i in range(n):
+            # skip all zero
+            if i > 0 and nums[i] == nums[i - 1]:
+                continue
+
+            # two pointers
+            left = i + 1
+            right = n - 1
+            target = -nums[i]
+
+            while left < right:
+                current = nums[left] + nums[right]
+
+                if current == target:
+                    result.append([nums[i], nums[left], nums[right]])
+
+                    # skip duplicates
+                    while left < right and nums[left] == nums[left + 1]:
+                        left += 1
+                    while left < right and nums[right] == nums[right - 1]:
+                        right -= 1
+
+                    # shift pointers
+                    left += 1
+                    right -= 1
+
+                # since sorted, if current < target, then move left
+                elif current < target:
+                    left += 1
+                # since sorted, if current > target, then move right
+                else:
+                    right -= 1
+
+        return result
 ```
diff --git a/docs/solutions/(array)/167-two-sum-ii-input-array-is-sorted.mdx b/docs/solutions/(array)/167-two-sum-ii-input-array-is-sorted.mdx
index 9731302..0853ffd 100644
--- a/docs/solutions/(array)/167-two-sum-ii-input-array-is-sorted.mdx
+++ b/docs/solutions/(array)/167-two-sum-ii-input-array-is-sorted.mdx
@@ -37,27 +37,19 @@ Your solution must use only constant extra space.
 
 ## Solution
 
-```js
-/**
- * @param {number[]} numbers
- * @param {number} target
- * @return {number[]}
- */
-var twoSum = function(numbers, target) {
-    let i = 0, j = numbers.length - 1;
+```py
+class Solution:
+    def twoSum(self, numbers: List[int], target: int) -> List[int]:
+        i = 0
+        j = len(numbers) - 1
 
-    while (i < j) {
-        const curr = numbers[i] + numbers[j];
-        if (curr === target){
-            return [i + 1, j + 1];
-        }else{
-            if (curr < target){
-                i++;
-            }else{
-                j--;
-            }
-        }
-    }
-    return [];
-};
+        while i < j:
+            c = numbers[i] + numbers[j]
+            if c == target:
+                return [i + 1, j + 1]
+            elif c < target:
+                i+=1
+            else:
+                j-=1
+        return []
 ```