Table of contents
Given problem
Below is the description of our problem:
A peak element is an element that is greater than its neighbors.
Given an input array nums, where nums[i] ≠ nums[i+1], find a peak element and return its index.
The array may contain multiple peaks, in that case return the index to any one of the peaks is fine.
You may imagine that nums[1] = nums[n] = ∞.
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: 1 or 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.
Solution
In the above section, we have the definition of the peak element:
A peak element is an element that is greater than its neighbors.
So, it means that:
nums[i  1] < nums[i] > nums[i + 1] for 0 <= i <= N  1
nums[i  1] < nums[i] if i = N  1
nums[i] > nums[i + 1] if i = 0
In this our case, we can find that nums[i] and nums[i + 1] is not equal, and to get the peak element of this array, we only compare between nums[i] and nums[i + 1].
If we care an element nums[i  1], then we always check multiple conditions for that. And checking element nums[i  1] and nums[i] is redundancy. Because checking elements nums[i] and nums[i + 1] makes us know the relationship between nums[i] and nums[i  1] in the loop of an array.
Using bruteforce way
public static int findPeakElementBruteForce(int[] nums) {
List<Integer> peaks = new ArrayList<>();
peaks.add(nums[nums.length  1]);
for (int i = 0; i < nums.length  1; ++i) {
if (nums[i] > nums[i + 1]) {
peaks.add(nums[i]);
}
}
return peaks.get(peaks.size()  1);
}
The complexity of this solution:
 Time complexity: O(n).
 Space complexity: O(1).
Using binary search
Below is the source code about iterative version and recursive version.

The iterative version
public static int findPeakElement(int[] nums) { int left = 0, right = nums.length  1; while (left < right) { int mid = (left + right) / 2; if (nums[mid] > nums[mid + 1]) right = mid; else left = mid + 1; } return left; }
The complexity of this solution:
 Time complexity: O(log(n)).
 Space complexity: O(1).

The recursive version
public int findPeakElement(int[] nums) { return search(nums, 0, nums.length  1); } public int search(int[] nums, int l, int r) { if (l == r) return l; int mid = (l + r) / 2; if (nums[mid] > nums[mid + 1]) return search(nums, l, mid); return search(nums, mid + 1, r); }
The complexity of solution:
 Time complexity: O(log(n)).
 Space complexity: O(log(n)).
Wrapping up
 Understanding this variant of Binary Search when comparing two adjacency elements.
Refer: