LeetCodee

941. Valid Mountain Array

Jump to Solution: Python Java C++ JavaScript C#

Problem Description

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

Python Solution

class Solution:
    def validMountainArray(self, arr: List[int]) -> bool:
        if len(arr) < 3:
            return False
            
        i = 0
        # Walk up
        while i + 1 < len(arr) and arr[i] < arr[i + 1]:
            i += 1
            
        # Peak can't be first or last
        if i == 0 or i == len(arr) - 1:
            return False
            
        # Walk down
        while i + 1 < len(arr) and arr[i] > arr[i + 1]:
            i += 1
            
        return i == len(arr) - 1

Time Complexity: O(n)

Where n is the length of the array.

Space Complexity: O(1)

We only use a single pointer variable.

Java Solution

class Solution {
    public boolean validMountainArray(int[] arr) {
        if (arr.length < 3) {
            return false;
        }
        
        int i = 0;
        // Walk up
        while (i + 1 < arr.length && arr[i] < arr[i + 1]) {
            i++;
        }
        
        // Peak can't be first or last
        if (i == 0 || i == arr.length - 1) {
            return false;
        }
        
        // Walk down
        while (i + 1 < arr.length && arr[i] > arr[i + 1]) {
            i++;
        }
        
        return i == arr.length - 1;
    }
}

Time Complexity: O(n)

Where n is the length of the array.

Space Complexity: O(1)

We only use a single pointer variable.

C++ Solution

class Solution {
public:
    bool validMountainArray(vector& arr) {
        if (arr.size() < 3) {
            return false;
        }
        
        int i = 0;
        // Walk up
        while (i + 1 < arr.size() && arr[i] < arr[i + 1]) {
            i++;
        }
        
        // Peak can't be first or last
        if (i == 0 || i == arr.size() - 1) {
            return false;
        }
        
        // Walk down
        while (i + 1 < arr.size() && arr[i] > arr[i + 1]) {
            i++;
        }
        
        return i == arr.size() - 1;
    }
};

Time Complexity: O(n)

Where n is the length of the array.

Space Complexity: O(1)

We only use a single pointer variable.

JavaScript Solution

/**
 * @param {number[]} arr
 * @return {boolean}
 */
var validMountainArray = function(arr) {
    if (arr.length < 3) {
        return false;
    }
    
    let i = 0;
    // Walk up
    while (i + 1 < arr.length && arr[i] < arr[i + 1]) {
        i++;
    }
    
    // Peak can't be first or last
    if (i === 0 || i === arr.length - 1) {
        return false;
    }
    
    // Walk down
    while (i + 1 < arr.length && arr[i] > arr[i + 1]) {
        i++;
    }
    
    return i === arr.length - 1;
};

Time Complexity: O(n)

Where n is the length of the array.

Space Complexity: O(1)

We only use a single pointer variable.

C# Solution

public class Solution {
    public bool ValidMountainArray(int[] arr) {
        if (arr.Length < 3) {
            return false;
        }
        
        int i = 0;
        // Walk up
        while (i + 1 < arr.Length && arr[i] < arr[i + 1]) {
            i++;
        }
        
        // Peak can't be first or last
        if (i == 0 || i == arr.Length - 1) {
            return false;
        }
        
        // Walk down
        while (i + 1 < arr.Length && arr[i] > arr[i + 1]) {
            i++;
        }
        
        return i == arr.Length - 1;
    }
}

Time Complexity: O(n)

Where n is the length of the array.

Space Complexity: O(1)

We only use a single pointer variable.