Find Peak Element LeetCode Solution | Easy Approach

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

Given an 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] = -∞.

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.

Find Peak Element Solutions

✅Time: O( log n)
✅Space: O(1)

C++

class Solution {
 public:
  int findPeakElement(vector<int>& nums) {
    int l = 0;
    int r = nums.size() - 1;

    while (l < r) {
      const int m = l + (r - l) / 2;
      if (nums[m] >= nums[m + 1])
        r = m;
      else
        l = m + 1;
    }

    return l;
  }
};

Java

 
class Solution {
  public int findPeakElement(int[] nums) {
    int l = 0;
    int r = nums.length - 1;

    while (l < r) {
      final int m = l + (r - l) / 2;
      if (nums[m] >= nums[m + 1])
        r = m;
      else
        l = m + 1;
    }

    return l;
  }
}

Python

 class Solution:
  def findPeakElement(self, nums: List[int]) -> int:
    l = 0
    r = len(nums) - 1

    while l < r:
      m = (l + r) // 2
      if nums[m] >= nums[m + 1]:
        r = m
      else:
        l = m + 1

    return l

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