Merge pull request #38 from miguelHx/miguel_2.2_return_kth_to_last

Miguel 2.2 - Kth To Last [Python]

Author: lhchavez

Python/chapter02/2.2 - Return Kth to Last/miguel_2.2_soln.py

@@ -0,0 +1,124 @@
+"""
+Python version 3.7.0
+2.2 - Return Kth to Last
+Implement an algorithm to find the kth
+to last element of a singly linked list
+Example:
+    input: ll = a -> b -> c -> d -> e -> f
+            k = 3
+    output: d because d is 3rd to last
+"""
+import unittest
+
+
+class Node:
+    def __init__(self, d: int):
+        self.data = d
+        self.next = None
+
+    def __repr__(self):
+        return self.__str__()
+
+    def __str__(self):
+        return '<Node Value: {}>'.format(self.data)
+
+    def __eq__(self, other: object):
+        if not isinstance(other, Node):
+            return NotImplemented
+        return self.data == other.data
+
+
+class LinkedList:
+    def __init__(self, *numbers: int):
+        self.head = None
+        self.tail = None
+        self.size = 0
+        for num in numbers:
+            self.append_to_tail(num)
+
+    def append_to_tail(self, d: int) -> None:
+        if self.head is None:
+            self.head = Node(d)
+            self.tail = self.head
+        else:
+            end = Node(d)
+            self.tail.next = end
+            self.tail = end
+        self.size += 1
+
+    def append_to_head(self, d: int) -> None:
+        new_head = Node(d)
+        new_head.next = self.head
+        self.head = new_head
+        self.size += 1
+
+    def __repr__(self):
+        return self.__str__()
+
+    def __str__(self):
+        if self.head is None:
+            return '<empty>'
+        ll = []
+        n = self.head
+        while n.next is not None:
+            ll.append('{} -> '.format(n.data))
+            n = n.next
+        ll.append(str(n.data))
+        return ''.join(ll)
+
+    def __eq__(self, other: object):
+        if not isinstance(other, LinkedList):
+            return NotImplemented
+        a = self.head
+        b = other.head
+        while a is not None and b is not None:
+            if a.data != b.data:
+                return False
+            # otherwise, advance both pointers
+            a = a.next
+            b = b.next
+        return a is None and b is None
+
+
+def kth_to_last(ll: LinkedList, k: int) -> Node:
+    """
+    kth_to_last will return the kth to last node
+    from the input linked list.
+    Going to reverse the linked list and then
+    count k steps.
+    Runtime: O(N)
+    Space Complexity: O(1)
+    :param ll: a linked list
+    :param k:
+        an integer where k > 0 and k < ll size
+    :return:
+        kth to last Node from the linked list
+        or None
+    """
+    if k <= 0 or k > ll.size:
+        raise IndexError('list index out of range')
+    # go size - k steps
+    n = ll.head
+    for i in range(1, ll.size - k + 1):
+        n = n.next
+    return n
+
+
+class TestKthToLast(unittest.TestCase):
+
+    def test_kth_to_last(self):
+        l = list(range(1, 6))
+        ll = LinkedList(*l)
+        for k in range(1, len(l)):
+            self.assertEqual(kth_to_last(ll, k), Node(l[-k]), msg=(ll, l[-k]))
+
+    def test_kth_to_last_index_error(self):
+        k_values = [-1, 7, 0]
+        ll = LinkedList(1, 2, 3)
+        for k in k_values:
+            with self.assertRaises(IndexError, msg=(ll, k)):
+                kth_to_last(ll, k)
+
+
+if __name__ == '__main__':
+    unittest.main()