Find Peak Element LeetCode Solution

Problem – Find Peak Element

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.

Find Peak Element LeetCode Solution in Java

public int findPeakElement(int[] nums) {
    int N = nums.length;
    if (N == 1) {
        return 0;
    }
   
    int left = 0, right = N - 1;
    while (right - left > 1) {
        int mid = left + (right - left) / 2;
        if (nums[mid] < nums[mid + 1]) {
            left = mid + 1;
        } else {
            right = mid;
        }
    }
    
    return (left == N - 1 || nums[left] > nums[left + 1]) ? left : right;
}

Find Peak Element LeetCode Solution in Python

def findPeakElement(self, nums):
    left = 0
    right = len(nums)-1

    # handle condition 3
    while left < right-1:
        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
            
    #handle condition 1 and 2
    return left if nums[left] >= nums[right] else right

Find Peak Element LeetCode Solution in C++

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

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