Interview Classic 150 Problem P162: Find Peak Element

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Timeline

Timeline

2025-12-03

init

Divide and Conquer

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Problem:

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

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

                                right = mid - 1;
                        }
                }

                return mid;
        }
};