Interview Classic 150 Questions P162 Find Peak Element

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Timeline

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2025-12-03

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divide and conquer

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Title:

P162 Find Peak Element

https://leetcode.cn/problems/find-peak-element/description/?envType=study-plan-v2&envId=top-interview-150

• If nums[mid] < nums[mid + 1] ⇒ peak must be in the right half
  Because from left to right it is increasing, and the rightmost is -∞, the right half must have a peak (even if monotonically increasing, the last one is a peak).
• If nums[mid] > nums[mid + 1] ⇒ peak must be in the left half
  Because from right to left it is increasing, and the leftmost is -∞, the left half must have a peak.

#include <vector>
using std::vector;

class Solution {
    private:
        bool isPeak(vector<int> &nums, int mid)
        {
                int n = nums.size();
                if (mid != 0 && nums[mid] <= nums[mid - 1]) {
                        return false;
                }
                if (mid != n - 1 && nums[mid] <= nums[mid + 1]) {
                        return false;
                }
                return true;
        }

    public:
        int findPeakElement(vector<int> &nums)
        {
                int n = nums.size();
                int left = 0, right = n - 1, mid;

                while (left <= right) {
                        mid = left + (right - left) / 2;
                        if (isPeak(nums, mid)) {
                                return mid;
                        }

                        if (nums[mid] < nums[mid + 1]) {
                                // If nums[mid] < nums[mid + 1] ⇒ peak must be in the right half
                                // Because from left to right it is increasing, and the rightmost is -∞, the right half must have a peak.

                                left = mid + 1;
                        } else {
                                // If nums[mid] > nums[mid + 1] ⇒ peak must be in the left half
                                // Since it is increasing from right to left, and the leftmost is -∞, the left half must have a peak.

                                right = mid - 1;
                        }
                }

                return mid;
        }
};