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Commit 2349dca

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Added Leetcode easy and medium problems
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"""
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# LOWEST COMMON ANCESTOR OF A BINARY SEARCH TREE
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Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.
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According to the definition of LCA on Wikipedia: "The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself)."
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Example 1:
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Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
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Output: 6
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Explanation: The LCA of nodes 2 and 8 is 6.
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Example 2:
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Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
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Output: 2
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Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.
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Example 3:
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Input: root = [2,1], p = 2, q = 1
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Output: 2
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Constraints:
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The number of nodes in the tree is in the range [2, 105].
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-109 <= Node.val <= 109
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All Node.val are unique.
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p != q
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p and q will exist in the BST.
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"""
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# Definition for a binary tree node.
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class TreeNode:
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def __init__(self, x):
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self.val = x
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self.left = None
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self.right = None
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class Solution:
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def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
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if root.val == p.val or root.val == q.val:
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return root
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if p.val < root.val < q.val or q.val < root.val < p.val:
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return root
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if p.val > root.val and q.val > root.val:
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return self.lowestCommonAncestor(root.right, p, q)
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else:
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return self.lowestCommonAncestor(root.left, p, q)
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"""
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# PALINDROME LINKED LIST
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Given a singly linked list, determine if it is a palindrome.
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Example 1:
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Input: 1->2
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Output: false
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Example 2:
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Input: 1->2->2->1
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Output: true
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Follow up:
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Could you do it in O(n) time and O(1) space?
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"""
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# Definition for singly-linked list.
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class ListNode:
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def __init__(self, val=0, next=None):
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self.val = val
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self.next = next
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class Solution:
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def isPalindrome(self, head: ListNode) -> bool:
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slow = head
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fast = head
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while fast and fast.next:
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slow = slow.next
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fast = fast.next.next
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parent = None
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cur = head
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while cur != slow:
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temp = cur.next
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cur.next = parent
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parent = cur
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cur = temp
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if fast and not fast.next:
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slow = slow.next
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while parent:
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if parent.val != slow.val:
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return False
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parent = parent.next
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slow = slow.next
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return True
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‎Leetcode/medium/elimination-game.py‎

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"""
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# ELIMINATION GAME
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There is a list of sorted integers from 1 to n. Starting from left to right, remove the first number and every other number afterward until you reach the end of the list.
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Repeat the previous step again, but this time from right to left, remove the right most number and every other number from the remaining numbers.
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We keep repeating the steps again, alternating left to right and right to left, until a single number remains.
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Find the last number that remains starting with a list of length n.
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Example:
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Input:
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n = 9,
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1 2 3 4 5 6 7 8 9
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2 4 6 8
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2 6
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6
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Output:
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6
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"""
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class Solution:
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def lastRemaining(self, n: int) -> int:
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k = 0
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start = 1
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left = True
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while n > 1:
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# print("k:",k,"start:",start,"n:",n)
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if left:
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start = start + pow(2, k)
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left = False
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else:
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if n % 2 != 0:
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start = start + pow(2, k)
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left = True
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k += 1
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n = n // 2
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return start

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