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162. Find Peak Element

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
  • -231 <= nums[i] <= 231 - 1
  • nums[i] != nums[i + 1] for all valid i.

Solutions

Solution 1: Binary Search

We define the left boundary of binary search as $left=0$ and the right boundary as $right=n-1$, where $n$ is the length of the array. In each step of binary search, we find the middle element $mid$ of the current interval, and compare the values of $mid$ and its right neighbor $mid+1$:

  • If the value of $mid$ is greater than the value of $mid+1$, there exists a peak element on the left side, and we update the right boundary $right$ to $mid$.
  • Otherwise, there exists a peak element on the right side, and we update the left boundary $left$ to $mid+1$.
  • Finally, when the left boundary $left$ is equal to the right boundary $right$, we have found the peak element of the array.

The time complexity is $O(\log n)$, where $n$ is the length of the array $nums$. Each step of binary search can reduce the search interval by half, so the time complexity is $O(\log n)$. The space complexity is $O(1)$.

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

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

  • class Solution:
    def findPeakElement(self, nums: List[int]) -> int:
        left, right = 0, len(nums) - 1
        while left < right:
            mid = (left + right) >> 1
            if nums[mid] > nums[mid + 1]:
                right = mid
            else:
                left = mid + 1
        return left


  • func findPeakElement(nums []int) int {
	left, right := 0, len(nums)-1
	for left < right {
		mid := (left + right) >> 1
		if nums[mid] > nums[mid+1] {
			right = mid
		} else {
			left = mid + 1
		}
	}
	return left
}

  • function findPeakElement(nums: number[]): number {
    let [left, right] = [0, nums.length - 1];
    while (left < right) {
        const mid = (left + right) >> 1;
        if (nums[mid] > nums[mid + 1]) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return left;
}

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