Problem

Let’s call an array A a mountain if the following properties hold:

A.length >= 3 There exists some 0 < i < A.length - 1 such that A[0] < A[1] < … A[i-1] < A[i] > A[i+1] > … > A[A.length - 1] Given an array that is definitely a mountain, return any i such that A[0] < A[1] < … A[i-1] < A[i] > A[i+1] > … > A[A.length - 1].

Example 1:

Input: [0,1,0]

Output: 1

Example 2:

Input: [0,2,1,0]

Output: 1

Note:

3 <= A.length <= 10000
0 <= A[i] <= 10^6
A is a mountain, as defined above.

Pre analysis

I always thought of problem like this when seeing stock graphs involving global maxima-minima and local maxima-minima. Since there is a definite mountain, whenever there is a number which is changing curve would be the result.

Post analysis

A better approach was trying to find mid index satisfying the condition rather than looping and findind the first index satisfying the condition. On large dataset, former approach is better I think.

Another solution

Another JS specific better solution

var peakIndexInMountainArray = function(nums) {
    let low = 0;
    let high = nums.length - 1;
    let midPoint = Math.floor(nums.length / 2);

    while(low <= high){
        if(nums[midPoint] < nums[midPoint - 1]){
            high = midPoint -1;
            midPoint = Math.floor((high+low)/2);
        }
        else if(nums[midPoint] < nums[midPoint + 1]){
            low = midPoint + 1;
            midPoint = Math.floor((high+low)/2);
        }
        else
            return midPoint;
    }

};