# 852. Peak Index in a Mountain Array

## Description

An array arr is 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 a mountain array arr, return the index i such that arr[0] < arr[1] < ... < arr[i - 1] < arr[i] > arr[i + 1] > ... > arr[arr.length - 1].

You must solve it in O(log(arr.length)) time complexity.

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 <= 105
• 0 <= arr[i] <= 106
• arr is guaranteed to be a mountain array.

## Solutions

Binary search.

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

• class Solution {
public:
int peakIndexInMountainArray(vector<int>& arr) {
int left = 1, right = arr.size() - 2;
while (left < right) {
int mid = (left + right) >> 1;
if (arr[mid] > arr[mid + 1])
right = mid;
else
left = mid + 1;
}
return left;
}
};

• class Solution:
def peakIndexInMountainArray(self, arr: List[int]) -> int:
left, right = 1, len(arr) - 2
while left < right:
mid = (left + right) >> 1
if arr[mid] > arr[mid + 1]:
right = mid
else:
left = mid + 1
return left


• func peakIndexInMountainArray(arr []int) int {
left, right := 1, len(arr)-2
for left < right {
mid := (left + right) >> 1
if arr[mid] > arr[mid+1] {
right = mid
} else {
left = mid + 1
}
}
return left
}

• function peakIndexInMountainArray(arr: number[]): number {
let left = 1,
right = arr.length - 2;
while (left < right) {
const mid = (left + right) >> 1;
if (arr[mid] > arr[mid + 1]) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
}


• /**
* @param {number[]} arr
* @return {number}
*/
var peakIndexInMountainArray = function (arr) {
let left = 1;
let right = arr.length - 2;
while (left < right) {
const mid = (left + right) >> 1;
if (arr[mid] > arr[mid + 1]) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
};


• impl Solution {
pub fn peak_index_in_mountain_array(arr: Vec<i32>) -> i32 {
let mut left = 1;
let mut right = arr.len() - 2;
while left < right {
let mid = left + (right - left) / 2;
if arr[mid] > arr[mid + 1] {
right = mid;
} else {
left = left + 1;
}
}
left as i32
}
}