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+You are given the `head` of a linked list, and an integer `k`.
+
+Return _the head of the linked list after **swapping** the values of the_ `kth` _node from the beginning and the_ `kth` _node from the end (the list is **1-indexed**)._
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2020/09/21/linked1.jpg)
+
+    Input: head = [1,2,3,4,5], k = 2
+    Output: [1,4,3,2,5]
+    
+
+**Example 2:**
+
+    Input: head = [7,9,6,6,7,8,3,0,9,5], k = 5
+    Output: [7,9,6,6,8,7,3,0,9,5]
+    
+
+**Constraints:**
+
+*   The number of nodes in the list is `n`.
+*   `1 <= k <= n <= 105`
+*   `0 <= Node.val <= 100`
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diff --git a/0528_swapping-nodes-in-a-linked-list/python3/solution.py b/0528_swapping-nodes-in-a-linked-list/python3/solution.py
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diff --git a/0528_swapping-nodes-in-a-linked-list/python3/using_dict.py b/0528_swapping-nodes-in-a-linked-list/python3/using_dict.py
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+# NOTE: Not the best approach, since we use O(N) space
+# Time: O(N)
+class Solution:
+    def swapNodes(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
+        # Figure out length of LL and cache nodes by the index
+        length = 0
+        cache = {}
+        
+        ptr = head
+        while ptr != None:
+            cache[length] = ptr
+            length += 1
+            ptr = ptr.next
+        
+        cache[k - 1].val, cache[length - k].val = cache[length - k].val, cache[k - 1].val
+        
+        return head