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_posts/2025-01-02-Leetcode-14.md

@@ -1,3 +1,4 @@
+---
 title: Leetcode 14. Longest Common Prefix
 description: Explanation for Leetcode 14 - Longest Common Prefix, and its solution in Python.
 date: 2025-01-02

_posts/2025-01-02-Leetcode-27.md

@@ -0,0 +1,74 @@
+---
+title: Leetcode 27. Remove Element
+description: Explanation for Leetcode 27 - Remove Element, and its solution in Python.
+date: 2025-01-02
+categories: [Leetcode, Arrays & Hashing, Easy]
+tags: [Leetcode, Python, Study, Arrays & Hashing, Easy]
+math: true
+---
+
+## Problem
+[Leetcode 27 - Remove Element](https://leetcode.com/problems/remove-element/description/)
+
+Example:
+
+```
+Input: nums = [3,2,2,3], val = 3
+Output: 2, nums = [2,2,_,_]
+Explanation: Your function should return k = 2, with the first two elements of nums being 2.
+It does not matter what you leave beyond the returned k (hence they are underscores).
+
+Input: nums = [0,1,2,2,3,0,4,2], val = 2
+Output: 5, nums = [0,1,4,0,3,_,_,_]
+Explanation: Your function should return k = 5, with the first five elements of nums containing 0, 0, 1, 3, and 4.
+Note that the five elements can be returned in any order.
+It does not matter what you leave beyond the returned k (hence they are underscores).
+```
+
+## Approach
+
+Since the question states that "It does not matter what you leave beyond the returned k" for the nums array, we can initialize the k to 0 then increment everytime we encounter an element that is not equal to the val. Meaning that once we encounter an element, we should put that element in nums[k], then increment the k value. Once we are done iterating through the array, we can simply return the k.
+
+Visualization of the Appraoch:
+```
+val = 3
+
+nums = [3, 2, 2, 3]
+k = 0
+
+Since nums[0] == 3, we can ignore
+
+Since nums[1] != 3, we put it in k then increment k by 1
+nums = [2, 2, 2, 3]
+k = 1
+
+Since nums[2] != 3, we put it in k then increment k by 1
+nums = [2, 2, 2, 3]
+k = 2
+
+Since nums[3] == 3, we can ignore
+
+Thus k = 2, we do not care about rest of the array as in the question it states that "It does not matter what you leave beyond the returned k"
+
+```
+
+Here is the Python code for the solution:
+```python
+class Solution:
+    def removeElemnt(self, nums: List[int], val: int) -> int:
+        k = 0
+
+        for i in range(len(nums)):
+            if nums[i] != val:
+                nums[k] = nums[i]
+                k += 1
+
+        return k
+```
+
+
+## Time Complexity and Space Complexity
+
+Time Complexity: $O(n)$ since we are iterating through the list.
+
+Space Complexity: $O(1)$ since we aren't using any extra space.