162. Find Peak Element
Medium LeetCodeA 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.
How to solve the problem
Binary Search with O(log n) time complexity. If nums[mid] > nums[mid + 1], peak is equal to mid or on the left, so move
right = mid
. If nums[mid] < nums[mid + 1], nums[mid] can never the peak, so we moveleft = mid + 1
instead of left = mid. CodeBinary Search
class Solution:
def findPeakElement(self, nums: List[int]) -> int:
left = 0
right = len(nums) - 1
while left < right:
mid = (left + right) // 2
if nums[mid] > nums[mid + 1]: # if nums[mid] > nums[mid + 1], peak is equal to mid or on the left, so move `right = mid`
right = mid
else: # if nums[mid] < nums[mid + 1], nums[mid] can never the peak, so we move `left = mid + 1` instead of left = mid
left = mid + 1
return left
Complexity
Time complexity: O(log n)
Space complexity: O(1)
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