# Find Peak Element LeetCode Solution | Easy Approach Share:

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`.

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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