Practice/Meta/Leetcode 162. Find Peak Element

Leetcode 162. Find Peak Element

CodingMust

Problem

You are given an array of integers where no two adjacent elements are equal. A peak element is defined as an element that is strictly greater than its neighbors. For elements at the boundaries of the array, consider the out-of-bounds neighbor as negative infinity.

Your task is to find and return the index of any peak element in the array. If multiple peaks exist, you may return the index of any one of them.

The algorithm must run in O(log n) time complexity.

Requirements

1. Return the index (not the value) of a peak element
2. A peak at index i satisfies: nums[i] > nums[i-1] and nums[i] > nums[i+1]
3. For boundary elements, treat out-of-bounds positions as negative infinity
4. If multiple peaks exist, any valid peak index is acceptable
5. The solution must achieve O(log n) time complexity

Constraints

• 1 ≤ nums.length ≤ 1000
• -2³¹ ≤ nums[i] ≤ 2³¹ - 1
• nums[i] ≠ nums[i + 1] for all valid i
• Must achieve O(log n) time complexity
• O(1) space complexity preferred

Examples

Example 1:

Input: nums = [1, 3, 2] Output: 1 Explanation: The element at index 1 has value 3, which is greater than both neighbors (1 and 2).

Example 2:

Input: nums = [5, 3, 1] Output: 0 Explanation: The element at index 0 has value 5, greater than its right neighbor 3. The left side is treated as negative infinity.

Example 3:

Input: nums = [1, 5, 3, 6, 4] Output: 1 or 3 Explanation: Both index 1 (value 5) and index 3 (value 6) are peaks. Either answer is correct.