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Find Peak Element⚓︎

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Description⚓︎

A peak element is an element that is strictly greater than its neighbors.

Given a 0-indexed integer array nums, find a peak element, and return its index. If the array contains multiple peaks, return the index to any of the peaks.

You may imagine that nums[-1] = nums[n] = -∞. In other words, an element is always considered to be strictly greater than a neighbor that is outside the array.

You must write an algorithm that runs in \(O(\log n)\) time.

Example 1:

  • Input: nums = [1,2,3,1]
  • Output: 2
  • Explanation: 3 is a peak element and your function should return the index number 2.

Example 2:

  • Input: nums = [1,2,1,3,5,6,4]
  • Output: 5
  • Explanation: Your function can return either index number 1 where the peak element is 2, or index number 5 where the peak element is 6.

Constraints:

  • 1 <= nums.length <= 1000
  • -2^31 <= nums[i] <= 2^31 - 1
  • nums[i] != nums[i + 1] for all valid i.

Solution⚓︎

Way 1⚓︎

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class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        long l = -1, r = nums.size();
        while (l + 1 < r) {
            long mid = l + (r - l) / 2;
            long midVal = (mid == 0) ? LONG_MIN : nums[mid - 1];
            if (nums[mid] > midVal) l = mid;
            else r = mid;
        }
        return l;
    }
};

Way 2⚓︎

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class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        int l = -1, r = nums.size();
        while (l + 1 < r) {
            int mid = l + (r - l) / 2;
            bool isGreaterLeft = (mid == 0) || (nums[mid] > nums[mid - 1]);
            bool isGreaterRight = (mid == nums.size() - 1) || (nums[mid] > nums[mid + 1]);
            if (isGreaterLeft && isGreaterRight) return mid;
            else if (isGreaterLeft) l = mid;
            else r = mid;
        }
        return l;
    }
};

Better way of writing:

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class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        int left = -1, right = nums.size() - 1;  // interval: (-1, n - 1)
        while (left + 1 < right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] > nums[mid + 1]) right = mid;
            else left = mid;
        }
        return right;
    }
};

Way 3⚓︎

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class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        int l = 0, r = nums.size() - 1;
        while (l < r) {
            int mid = l + r >> 1;
            if (nums[mid] > nums[mid + 1]) r = mid;
            else l = mid + 1;
        }
        return r;
    }
};