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+"""
+162. Find Peak Element
+Difficulty: Medium
+https://leetcode.com/problems/find-peak-element/
+
+──────────────────────────────────────────────────
+
+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.
+"""
+
+
+class Solution:
+    def findPeakElement(self, nums: list[int]) -> int:
+        left, right = 0, len(nums) - 1
+        while left < right:
+            mid = (left + right) // 2
+            if nums[mid] < nums[mid + 1]:
+                left = mid + 1
+            else:
+                right = mid
+
+        return left