658

_posts/2025-01-10-Leetcode-209.md

@@ -2,8 +2,8 @@
 title: Leetcode 209. Minimum Size Subarray Sum
 description: Explanation for Leetcode 209 - Minimum Size Subarray Sum, and its solution in Python.
 date: 2025-01-10
-categories: [Leetcode, Sliding Windows, Medium]
-tags: [Leetcode, Python, Study, Sliding Windows, Medium]
+categories: [Leetcode, Sliding Window, Medium]
+tags: [Leetcode, Python, Study, Sliding Window, Medium]
 math: true
 ---
 

_posts/2025-01-10-Leetcode-219.md

@@ -2,13 +2,13 @@
 title: Leetcode 219. Contains Duplicate II
 description: Explanation for Leetcode 219 - Contains Duplicate II, and its solution in Python.
 date: 2025-01-10
-categories: [Leetcode, Sliding Windows, Easy]
-tags: [Leetcode, Python, Study, Sliding Windows, Easy]
+categories: [Leetcode, Sliding Window, Easy]
+tags: [Leetcode, Python, Study, Sliding Window, Easy]
 math: true
 ---
 
 ## Problem
-[Leetcode 219 - Contains Duplicate II](https://leetcode.com/problems/contains-duplicate-ii/submissions/1504764844/)
+[Leetcode 219 - Contains Duplicate II](https://leetcode.com/problems/contains-duplicate-ii/description/)
 
 Example:
 ```

_posts/2025-01-11-Leetcode-658.md

@@ -0,0 +1,70 @@
+---
+title: Leetcode 658. Find K Closest Elements
+description: Explanation for Leetcode 658 - Find K Closest Elements, and its solution in Python.
+date: 2025-01-11
+categories: [Leetcode, Sliding Window, Medium]
+tags: [Leetcode, Python, Study, Sliding Window, Medium]
+math: true
+---
+
+## Problem
+[Leetcode 658 - Find K Closest Elements](https://leetcode.com/problems/find-k-closest-elements/description/)
+
+
+Example:
+```
+Input: arr = [1,2,3,4,5], k = 4, x = 3
+Output: [1,2,3,4]
+
+Input: arr = [1,1,2,3,4,5], k = 4, x = -1
+Output: [1,1,2,3]
+```
+
+## Approach
+
+Since we need to return k elements taht are closest to x, and the input array is sorted. We can use binary search to find the window that is closest to x. 
+
+Our initial left = 0, while right = len(arr) - k because we need extra k space for the return array
+
+If arr[mid] is farther from target than arr[mid+k] which is k places ahead of mid then we need to pull left to mid with 1 offset; otherwise we can pull right at mid
+
+Visualization of the Approach:
+```
+arr = [1,2,3,4,5], k = 2, x = 5
+       l     r
+mid = 1, nums[mid] = 2
+x - arr[mid] > arr[mid+k] - x -> 5 - 1 > 4 - 5 -> 4 > -1. Thus left = mid+1
+
+arr = [1,2,3,4,5]
+           l r
+mid = 2, nums[mid] = 3
+x - arr[mid] > arr[mid+k] - x -> 5 - 3 > 5 - 5 -> 2 > 0. Thus left = mid+1
+
+arr = [1,2,3,4,5]
+             lr
+left == right stop while loop
+
+arr[left: left+k] = [4,5]
+```
+
+Here is the Python code for the solution:
+```python
+class Solution:
+    def findClosestElements(self, arr: List[int], k: int, x: int) -> List[int]:
+        left, right = 0, len(arr)-k
+
+        while left < right:
+            mid = (left + right) // 2
+
+            if x - arr[mid] > arr[mid+k] - x:
+                left = mid+1
+            else:
+                right = mid
+
+        return arr[left: left+k]
+```
+## Time Complexity and Space Complexity
+
+Time Complexity: $O(log(n-k)+k)$
+
+Space Complexity: $O(k)$