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In this post, you will find the solution for the **Find Peak Element** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Find Peak Element– LeetCode Problem

Find Peak Element– LeetCode Problem

**Problem:**

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:2Explanation: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:5Explanation: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`

`-2`

^{31}<= nums[i] <= 2^{31}- 1`nums[i] != nums[i + 1]`

for all valid`i`

.

Find Peak Element– LeetCode Solutions

Find Peak Element Solution in C++:

class Solution { public: int findPeakElement(vector& 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; } };

Find Peak Element Solution in 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; } }

Find Peak Element Solution in 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