add article

Author: neetcode-gh

articles/duplicate-integer.md

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+## 1. Brute Force
+
+```python
+class Solution:
+    def hasDuplicate(self, nums: List[int]) -> bool:
+        for i in range(len(nums)):
+            for j in range(i + 1, len(nums)):
+                if nums[i] == nums[j]:
+                    return True
+        return False
+```
+
+### Time & Space Complexity
+
+* Time complexity: $O(n^2)$
+* Space complexity: $O(1)$
+
+## 2. Sorting
+
+```python
+class Solution:
+    def hasDuplicate(self, nums: List[int]) -> bool:
+        nums.sort()
+        for i in range(1, len(nums)):
+            if nums[i] == nums[i - 1]:
+                return True
+        return False
+```
+
+### Time & Space Complexity
+
+* Time complexity: $O(n \log n)$
+* Space complexity: $O(1)$ or $O(n)$ depending on the sorting algorithm.
+
+## 3. Hash Set
+
+```python
+class Solution:
+    def hasDuplicate(self, nums: List[int]) -> bool:
+        seen = set()
+        for num in nums:
+            if num in seen:
+                return True
+            seen.add(num)
+        return False
+```
+
+### Time & Space Complexity
+
+* Time complexity: $O(n)$
+* Space complexity: $O(n)$
+
+## 4. Hash Set Length
+
+```python
+class Solution:
+    def hasDuplicate(self, nums: List[int]) -> bool:
+        return len(set(nums)) < len(nums)
+```
+
+### Time & Space Complexity
+
+* Time complexity: $O(n)$
+* Space complexity: $O(n)$
+