Problem statement

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

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

• 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]

Example 1

Input: arr = [2,1]
Output: false

Example 2

Input: arr = [3,5,5]
Output: false

Example 3

Input: arr = [0,3,2,1]
Output: true

Constraints

• 1 <= arr.length <= 10^4.

• 0 <= arr[i] <= 10^4.

Solution

Following the conditions, we have the following implementation.

Code

#include <vector>
#include <iostream>
using namespace std;
bool validMountainArray(vector<int>& arr) {
    if (arr.size() < 3) {
        return false;
    }
    const int N = arr.size() - 1;
    int i = 0;
    while (i < N && arr[i] < arr[i + 1]) {
        i++;
    }
    if (i == 0 || i == N) {
        return false;
    }
    while (i < N && arr[i] > arr[i + 1]) {
        i++;
    }
    return i == N;
}
int main() {
    vector<int> arr{2,1};
    cout << validMountainArray(arr) << endl;
    arr = {3,5,5};   
    cout << validMountainArray(arr) << endl;
    arr = {0,3,2,1};   
    cout << validMountainArray(arr) << endl;
    arr = {9,8,7,6,5,4,3,2,1,0};
    cout << validMountainArray(arr) << endl;
}

Output:
0
0
1
0

Complexity

• Runtime: O(N), where N = arr.length.

• Extra space: O(1).

References

• https://leetcode.com/problems/valid-mountain-array/

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