interactive-coding-challenges/linked-lists/kth-to-last-elem.ipynb

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"## Problem: Find the kth to last element of a linked list\n",
"\n",
"* [Clarifying Questions](#Clarifying-Questions)\n",
"* [Test Cases](#Test-Cases)\n",
"* [Algorithm](#Algorithm)\n",
"* [Code](#Code)"
]
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"## Clarifying Questions\n",
"\n",
"* Is k an integer?\n",
" * Yes\n",
"* If k = 0, does this return the last element?\n",
" * Yes\n",
"* What happens if k is greater than or equal to the length of the linked list?\n",
" * Return None\n",
"* Can you use additional data structures?\n",
" * No"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Test Cases\n",
"\n",
"* Empty list\n",
"* k is not an integer\n",
"* k is >= the length of the linked list\n",
"* One element, k = 0\n",
"* General case with many elements, k < length of linked list"
]
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"## Algorithm\n",
"\n",
"* Check for edge cases above, returning None for errors\n",
"* Setup two pointers, current and previous\n",
"* Give current a headstart, incrementing it once for k = 1, twice for k = 2, etc\n",
"* Increment both pointers until current reaches the end\n",
"* Return the value of previous\n",
"\n",
"Complexity:\n",
"* Time: O(n)\n",
"* Space: In-place"
]
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"cell_type": "markdown",
"metadata": {},
"source": [
"## Code"
]
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"%run linked_list.py"
]
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"source": [
"class MyLinkedList(LinkedList):\n",
" def kth_to_last_elem(self, k):\n",
" if self.head is None:\n",
" return\n",
" if type(k) != int:\n",
" raise ValueError('')\n",
" if k >= len(self):\n",
" return\n",
" curr = self.head\n",
" prev = self.head\n",
" counter = 0\n",
" while counter < k:\n",
" curr = curr.next\n",
" counter += 1\n",
" if curr is None:\n",
" return\n",
" while curr.next is not None:\n",
" curr = curr.next\n",
" prev = prev.next\n",
" return prev.data"
]
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"source": [
"# Empty list\n",
"linked_list = MyLinkedList(None)\n",
"print(linked_list.kth_to_last_elem(0))\n",
"# k is not an integer\n",
"print(linked_list.kth_to_last_elem('a'))\n",
"# k is >= the length of the linked list\n",
"print(linked_list.kth_to_last_elem(100))\n",
"# One element, k = 0\n",
"head = Node(2)\n",
"linked_list = MyLinkedList(head)\n",
"print(linked_list.kth_to_last_elem(0))\n",
"# General case with many elements, k < length of linked list\n",
"linked_list.insert_to_front(1)\n",
"linked_list.insert_to_front(3)\n",
"linked_list.insert_to_front(5)\n",
"linked_list.insert_to_front(7)\n",
"print(linked_list.kth_to_last_elem(2))"
]
}
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