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.

Test Cases

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:

Solution

class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        int l = 0, r = nums.size() - 1;
        while(l < r) {
            int mid = l + (r-l)/2;
            if (nums[mid] > nums[mid+1]) {
                r = mid;
            } else {
                l = mid+1;
            }
        }
        return l;
    }
};
public class Solution {
    public int findPeakElement(int[] nums) {
        int l = 0, r = nums.length - 1;
        while (l < r) {
            int mid = (l + r) / 2;
            if (nums[mid] > nums[mid + 1])
                r = mid;
            else
                l = mid + 1;
        }
        return l;
    }
}
class Solution:
    def findPeakElement(self, A: List[int]) -> int:
        nums = A
        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
Time Complexity: O(log n)
Space Complexity: O(1)