Peak Index in a Mountain Array

Problem

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.

Solution

/**
 * @param {number[]} arr
 * @return {number}
 */
var peakIndexInMountainArray = function(arr) {
    let lo = 0;
    let hi = arr.length - 1;

    while (lo <= hi) {
        const mid = Math.floor((lo + hi) / 2);
        if (arr[mid] < arr[mid + 1]) { // peak in range (mid, hi]
            lo = mid + 1;
        } else if (arr[mid - 1] > arr[mid]) { // peak in range [lo, mid)
            hi = mid - 1;
        } else { // peak in range [mid, mid]
            return mid;
        }
    }
};

We will implement a binary search solution. Since we are trying to find the peak of the mountain, we only need to search on the half that is greater than arr[mid] (ie. by checking arr[mid - 1] and arr[mid + 1]). If both are less, then arr[mid] must already be the peak.