Problem description:

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

Given an input array nums, where nums[i] ≠ nums[i+1], find a peak element and return its index.

The array may contain multiple peaks, in that case return the index to any one of the peaks is fine.

You may imagine that nums[-1] = nums[n] = -∞.

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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.

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Example 2:

Input: nums = [1,2,1,3,5,6,4]
Output: 1 or 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.

Note:
Your solution should be in logarithmic complexity.

Solution:

  1. The question is asking for peaking element, if the sequence have multiple peak, anyone of them is fine.
  2. We can use sequential find or binary search

We can consider the following case:

  • Sequence is decreasing
    If we use a mid pointer to find, then the peak must be on the left side.

  • Sequence is increasing
    The peak is on the right part.

  • Peak is in the middle

  • sequential find
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    class Solution {
    public:
        int findPeakElement(const vector<int> &num) {
            for(int i = 1; i < num.size(); i ++)
            {
                if(num[i] < num[i-1])
                {// <
                    return i-1;
                }
            }
            return num.size()-1;
        }
    };

time complexity: O(n)
space complexity: O(1)

  • binary search, recursive
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    class Solution {
    public:
        int findPeakElement(vector<int>& nums) {
            return helper(nums, 0, nums.size()-1);
        }
        
        int helper(vector<int>& nums, int left, int right){
            if(left == right) return left;
            else{
                int mid = (left+right)/2;
                if(nums[mid] > nums[mid+1]) 
                    return helper(nums, left, mid);
                else 
                    return helper(nums, mid+1, right);
            }
            
        }
    };

time complexity: O(logn)
space complexity: O(logn). In recursion loop, the search space is reduced in half, so the depth tree is log_2(n)

  • binary search, iterative
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class Solution:
    def findPeakElement(self, nums: List[int]) -> int:
        left = 0
        right = len(nums)-1

        while left < right:
            mid = (left+right)//2
            if nums[mid] > nums[mid+1] and nums[mid] > nums[mid-1]:
                return mid

            if nums[mid] < nums[mid+1]:
                left = mid+1
            else:
                right = mid-1

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

time complexity: O(logn)
space complexity: O(1)