Find Peak Element

Problem

A peak is an index i with nums[i] > nums[i−1] and nums[i] > nums[i+1] (treating out-of-bounds as −∞). Return the index of any peak. Aim for O(log n).

Constraints

• 1 ≤ nums.length ≤ 1000
• nums[i] ≠ nums[i+1]
• −2³¹ ≤ nums[i] ≤ 2³¹−1

Approach — Binary Search

This is a Binary Search problem. The idea: repeatedly halve the search space, discarding the half that can't contain the answer. Work through the reference code below line by line, then re-derive it yourself in the editor — that's how the pattern sticks.

Complexity: O(log n) time.

Solution code

Python

class Solution:
    def findPeakElement(self, nums):
        lo, hi = 0, len(nums) - 1
        while lo < hi:
            mid = (lo + hi) // 2
            if nums[mid] < nums[mid + 1]:
                lo = mid + 1
            else:
                hi = mid
        return lo

Java

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

Practice it

Reading a solution isn't the same as being able to write it under pressure. Open this problem in the in-browser editor, hide the solution, and solve it from scratch — your code runs against real test cases instantly.