852. Peak Index in a Mountain Array

Increasing                    decreasing 
1 2 3 4         5             4 3 2  1 

Peak is 5 

Increasing            decreasing 
123456     7         321

Mid  < mid + 1 
left = mid + 1 


Mid = 5 
Left = 6 
Mid = 3 
Mid > mid + 1  
Right = mid = 3 
Mid = 7 
7 < mid + 1 
Right = mid = 7 
Left = 6 
Mid = 6 
6 < mid + 1 
Left = mid + 1 = 7 
left  < right 

Return left 



// correct 
class Solution {
    public int peakIndexInMountainArray(int[] array) {
        int left = 0;
        int right = array.length - 1 ; 
        while(left < right){
            int mid = left + (right - left) / 2;
            if(array[mid] < array[mid + 1]){
                left = mid + 1;
            }else{
                right = mid;
            }
        }
        return left;
        
    }
}

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