Native solution with Linear Scan
The naive solution is scan the array, check whether a element is peak.
Time complexity: $O(N)$.

Because we can return any peak, so there are 3 cases:
arr[0] is peak
arr[arr.size()1] is peak
peak is in the range of
1...arr.size()2
In this case, we just need to find the first index, which meets the condition of
arr[index] > arr[index + 1]
. So we can make the code more simpler:class Solution {
public:
int findPeakElement(vector<int>& nums) {
for(int i=0; i < nums.size()  1; i++) {
if(nums[i] > nums[i+1])
return i;
}
return nums.size()  1;
}
};
Approach with binary search
Using recursive binary search, the time complexity is $O(logN)$.

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