162. Find Peak Element

162. Find Peak Element

Tag: Binary Search

Difficulty: Medium

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.

---

Greedy

class Solution:
    def findPeakElement(self, nums: List[int]) -> int:
        peak = ans = -math.inf
        for idx, num in enumerate(nums):
            if peak < num:
                peak = num
                ans = idx
        return ans

Index()

class Solution:
    def findPeakElement(self, nums: List[int]) -> int:
        return nums.index(max(nums))

Binary Search (Template II)

Iterative Binary Search

• Time complexity: O(log(n))
• Space complexity: O(1)

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

        while lo < hi:
            mi = lo + (hi - lo) // 2

            if nums[mi] > nums[mi + 1]:
                hi = mi
            else:
                lo = mi + 1

        return lo 

Recursive Binary Search

• Time complexity: O(log(n))
• Space complexity: O(log(n))

class Solution:
    def findPeakElement(self, nums: List[int]) -> int:
        def binary_search(lo, hi):
            # Base case
            if lo == hi:
                return lo

            mi = lo + (hi - lo) // 2

            if nums[mi] < nums[mi + 1]:
                return binary_search(mi + 1, hi)
            else:
                return binary_search(lo, mi)

        return binary_search(0, len(nums) - 1)

Binary Search (Template III)

class Solution:
    def findPeakElement(self, nums: List[int]) -> int:
        n = len(nums)

        lo, hi = 0, n - 1
        while lo + 1 < hi:
            mi = lo + (hi - lo) // 2

            # Found a peak
            if nums[mi] > nums[mi + 1] and nums[mi] > nums[mi - 1]:
                return mi
            # Otherwise, if not a peak, climb up to the higher side
            elif nums[mi] < nums[mi + 1]:
                lo = mi
            else:
                hi = mi
        
        return hi if nums[hi] > nums[lo] else lo