# LC 941. Valid Mountain Array

### [Problem link](https://leetcode.com/problems/valid-mountain-array/)

###### tags: `leedcode` `python` `c++` `easy` `Two Pointer`

Given an array of integers <code>arr</code>, return <code>true</code> if and only if it is a valid mountain array.

Recall that arr is a mountain array if and only if:

- <code>arr.length >= 3</code>
- There exists some <code>i</code> with <code>0 < i < arr.length - 1</code> such that:

- <code>arr[0] < arr[1] < ... < arr[i - 1] < arr[i] </code>
- <code>arr[i] > arr[i + 1] > ... > arr[arr.length - 1]</code>

<img src="https://assets.leetcode.com/uploads/2019/10/20/hint_valid_mountain_array.png" width="500" />

**Example 1:**

Output: false

Output: false

Output: true

**Constraints:**

- <code>1 <= arr.length <= 10⁴</code>
- <code>0 <= arr[i] <= 10⁴</code>

## Solution 1 - Two Pointer
#### Python

class Solution:
    def validMountainArray(self, arr: List[int]) -> bool:
        n = len(arr)
        left, right = 0, n - 1

        while left <= n - 2 and arr[left] < arr[left + 1]:
            left += 1

        while right >= 1 and arr[right - 1] > arr[right]:
            right -= 1

        if left == n - 1 or right == 0:
            return False
        return left == right

#### C++

class Solution {
public:
    bool validMountainArray(vector<int>& arr) {
        if (arr.size() < 3) {
            return false;
        }
        int left = 0;
        int right = arr.size() - 1;
        while (left < arr.size() - 1 && arr[left] < arr[left + 1]) {
            left++;
        }
        while (right > 0 && arr[right - 1] > arr[right]) {
            right--;
        }
        if (left == 0 || right == arr.size() - 1 || left != right) {
            return false;
        }
        return true;
    }
};

>### Complexity
>n = arr.length
>| | Time Complexity | Space Complexity |
>| ----------- | --------------- | ---------------- |
>| Solution 1 | O(n) | O(1) |

## Note
x