941. Valid Mountain Array
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 <= 10^40 <= 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.