> For the complete documentation index, see [llms.txt](https://maksimdan.gitbook.io/interview-practice-problems/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://maksimdan.gitbook.io/interview-practice-problems/leetcode_sessions/162-find-peak-element.md).

# 162 Find Peak Element

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

Given an input array where`num[i] ≠ num[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`num[-1] = num[n] = -∞`.

For example, in array`[1, 2, 3, 1]`, 3 is a peak element and your function should return the index number 2.

**The Idea:** Binary search until we converge onto a peak. We can detect whether we are on either the left or right side of the peak. On the left side of a peak, the element to the right is greater. On the left, the opposite is true. If we are on the left side, we want to move right, and on the right left we want to conversely move left. Eventually we will converge to the top of *some* local peak.

**Complexity:** O(logn) time and space

```python
class Solution:
    def findPeakElement(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        return self.b_search(0, len(nums) - 1, nums)

    def b_search(self, left, right, nums):
        if left == right:
            return right # or left
        else:
            mid1 = int(left + (right - left) / 2)
            mid2 = mid1 + 1

        # left end of the peak -> move right
        if (nums[mid1] < nums[mid2]):
            return self.b_search(mid1 + 1, right, nums)

        # otherwise on the right end of the peak
        else:
            return self.b_search(left, mid1, nums)
```


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