162.find-peak-element.py

#
# @lc app=leetcode id=162 lang=python3
#
# [162] Find Peak Element
#
# https://leetcode.com/problems/find-peak-element/description/
#
# algorithms
# Medium (46.04%)
# Likes:    8906
# Dislikes: 4252
# Total Accepted:    995.4K
# Total Submissions: 2.2M
# Testcase Example:  '[1,2,3,1]'
#
# A peak element is an element that is strictly greater than its neighbors.
# 
# Given a 0-indexed integer array nums, find a peak element, and return its
# index. If the array contains multiple peaks, return the index to any of the
# peaks.
# 
# You may imagine that nums[-1] = nums[n] = -∞. In other words, an element is
# always considered to be strictly greater than a neighbor that is outside the
# array.
# 
# You must write an algorithm that runs in O(log n) time.
# 
# 
# Example 1:
# 
# 
# Input: nums = [1,2,3,1]
# Output: 2
# Explanation: 3 is a peak element and your function should return the index
# number 2.
# 
# Example 2:
# 
# 
# Input: nums = [1,2,1,3,5,6,4]
# Output: 5
# Explanation: Your function can return either index number 1 where the peak
# element is 2, or index number 5 where the peak element is 6.
# 
# 
# Constraints:
# 
# 
# 1 <= nums.length <= 1000
# -2^31 <= nums[i] <= 2^31 - 1
# nums[i] != nums[i + 1] for all valid i.
# 
# 
#

# @lc code=start
class Solution:
    def findPeakElement(self, nums: List[int]) -> int:
        left, right = -1, len(nums)
        while left + 1 != right:
            mid = (left + right) // 2
            if mid == len(nums) - 1:
                return mid
            if nums[mid] < nums[mid + 1]:
                left = mid
            else:
                right = mid
        return right
# @lc code=end