Source: https://leetcode.com/problems/peak-index-in-a-mountain-array/

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].

/**
* @param {number[]} arr
* @return {number}
*/
// Binary Search Approach
// left = 0;
// right = arr.length - 1
// while left <= right
// mid = left + Math.floor(right - left / 2)
// if arr[mid] > arr[mid+1], it's decreasing so we check if the peak can start more in the left half aka right = mid - 1, peakIndex = mid
// else it's still increasing so we check if the peak is in the right half aka left = mid + 1
// O(logN) time, O(1) space
var peakIndexInMountainArray = function(arr) {
 let left = 0;
 let right = arr.length - 1;
 let peakIndex = -1;


 // Binary search to find the peak index based on the arr[mid] > arr[mid+1]
 while (left <= right) {
   const mid = left + Math.floor((right - left) / 2);
   // It's decreasing though the peak can start more on the left half so we check there and set the peakIndex to mid
   if (arr[mid] > arr[mid + 1]) {
     right = mid - 1;
     peakIndex = mid;
   // It's increasing, so we check if the peak is in the right half
   } else {
     left = mid + 1;
   }
 }
 
 return peakIndex;
};