# 162. Find Peak Element

## Intitution

A *peak* is an element that is greater than its immediate neighbors. Because we can assume `nums[-1]` and `nums[n]` are `-∞`, the array is guaranteed to have at least one peak.

To find a peak in **O(log n)** time, we can use **binary search**:

* If `nums[mid]` is greater than both neighbors → it's a peak.
* If the right neighbor is greater → the peak lies to the right.
* Else → the peak lies to the left.

This works because in a unimodal-like structure, there's always a path toward a peak.

## Complexity

| Space Complexity | Time Complexity       |
| ---------------- | --------------------- |
| $$\text{O}(1)$$  | $$\text{O}(\log{N})$$ |

## Code

```java
public int findPeakElement(int[] nums) {
    int left = 0, right = nums.length - 1;

    while (left <= right) {
        int mid = (left + right) >> 1;

        // Check if nums[mid] is greater than its left and right neighbors
        boolean isGreaterThanLeft = (mid - 1 < 0 || nums[mid] > nums[mid - 1]);
        boolean isGreaterThanRight = (mid + 1 >= nums.length || nums[mid] > nums[mid + 1]);

        // If both conditions hold, we found a peak
        if (isGreaterThanLeft && isGreaterThanRight) {
            return mid;
        }

        // If increasing toward the right, move right
        if (!isGreaterThanRight) {
            left = mid + 1;
        } else { // Else move left
            right = mid - 1;
        }
    }

    // Should never reach here if input is valid
    return -1;
}

```