# Peak Index in a Mountain Array

Easy

Let's call an array`A`amountain 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:

1. `3 <= A.length <= 10000`
2. `0 <= A[i] <= 10^6`
3. A is a mountain, as defined above.

## Solution

Linear Scan

``````class Solution {
public int peakIndexInMountainArray(int[] A) {
for (int i = 0; i < A.length; i++) {
if (i > 0 && i < A.length - 1 && A[i] > A[i - 1] && A[i] > A[i + 1]) {
return i;
}
}
return -1;
}
}
``````

Linear Scan II

``````class Solution {
public int peakIndexInMountainArray(int[] A) {
int i = 0;
while (A[i] < A[i+1]) i++;
return i;
}
}
``````

Binary Search

BS Template #2

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

BS Template #1

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