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Peak Index in a Mountain Array

Problem statement​

Let's call an array arr a mountain if the following properties hold:

  • arr.length >= 3
  • There exists some i with 0 < i < arr.length - 1 such that:
    • arr[0] < arr[1] < ... arr[i-1] < arr[i]
    • arr[i] > arr[i+1] > ... > arr[arr.length - 1]

Given an integer array arr that is guaranteed to be a mountain, return any i such that arr[0] < arr[1] < ... arr[i - 1] < arr[i] > arr[i + 1] > ... > arr[arr.length - 1].

Example 1:

Input: arr = [0,1,0]Output: 1

Example 2:

Input: arr = [0,2,1,0]Output: 1

Example 3:

Input: arr = [0,10,5,2]Output: 1

Constraints:

  • 3 <= arr.length <= 104
  • 0 <= arr[i] <= 106
  • arr is guaranteed to be a mountain array.
Follow up: Finding the O(n) is straightforward, could you find an O(log(n)) solution?

My solution​

/**
 * @param {number[]} arr
 * @return {number}
 */
var peakIndexInMountainArray = function(arr) {
    let peak = Number.MIN_SAFE_INTEGER;
    
    for (let i = 0; i < arr.length; i ++) {
        let curr = arr[i];
        let left = arr[i - 1] || Number.MIN_SAFE_INTEGER;
        let right = arr[i + i] || Number.MIN_SAFE_INTEGER;
        
        // console.log(curr, left, right)
        
        if (curr > left && curr > right) {
            peak = Math.max(peak, i)
        }
    }
    
    return peak
};