A peak element is an element that is greater than its neighbors.
Given an input array nums, where nums[i] ≠ nums[i+1], find a peak element and return its index.
The array may contain multiple peaks, in that case return the index to any one of the peaks is fine.
You may imagine that nums[-1] = nums[n] = -∞.
Define peak A[P] > A[P-1] && A[P] > A[P+1]
def findPeakElement(self, nums: List[int]) -> int:
l, r = 0, len(nums)-1
while l < r:
mid = (l+r)//2
if nums[mid] > nums[mid+1]:
r = mid
else:
l = mid+1
return l
222. Count Complete Tree Nodes
Given a complete binary tree, count the number of nodes.
Can you solve in O(d) time
"""
The node number in last level: 1 ~ 2^d-1
Use 2 binary search, first search if idx exist in last level, second in 'exist function'
Time O(2d)
1. find depth of root
2. return 1 if d == 0
3. Use 'exist' function to find if left idx exist
4. Implement exist(idx, root, d)
"""
class Solution:
def countNodes(self, root: TreeNode) -> int:
if not root: return 0
d = self.getDepth(root)
if d == 0: return 1
left, right = 1, 2**d
while left < right:
mid = (left + right)//2
if self.exist(mid, root, d):
left = mid + 1
else:
right = mid
return (2**d-1) + left
def getDepth(self, node):
d = 0
while node.left:
node = node.left
d += 1
return d
def exist(self, idx, node, d):
left, right = 0, 2**d-1
for _ in range(d):
mid = (left + right)//2
if idx <= mid:
node = node.left
right = mid
else:
node = node.right
left = mid
return node is not None