LeetCode 162: Find Peak Element Solution

Problem Explanation

Explanation:

To find a peak element in an array, we can use a binary search approach to achieve O(log n) time complexity. The key idea is to compare the middle element with its neighbors to determine if the peak lies to the left or right of it.

Initialize low to 0 and high to the length of the array - 1.
While low is less than high:
- Calculate mid as (low + high) / 2.
- Compare nums[mid] with its neighbors nums[mid-1] and nums[mid+1].
- If nums[mid] is greater than both neighbors, return mid as the peak.
- If nums[mid+1] is greater, update low = mid + 1 to search in the right half.
- If nums[mid-1] is greater, update high = mid - 1 to search in the left half.
At the end of the loop, low will be at the peak element index.

Solution Code

class Solution {
    public int findPeakElement(int[] nums) {
        int low = 0, high = nums.length - 1;
        while (low < high) {
            int mid = low + (high - low) / 2;
            if (nums[mid] > nums[mid + 1]) {
                high = mid;
            } else {
                low = mid + 1;
            }
        }
        return low;
    }
}

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