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]
![](https://spcdn.pages.dev/leetcode/problems/0941.Valid%20Mountain%20Array/images/hint_valid_mountain_array.png)
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 <= 104
0 <= arr[i] <= 104
Solutions#
Solution 1#
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| class Solution:
def validMountainArray(self, arr: List[int]) -> bool:
n = len(arr)
if n < 3:
return False
l, r = 0, n - 1
while l + 1 < n - 1 and arr[l] < arr[l + 1]:
l += 1
while r - 1 > 0 and arr[r] < arr[r - 1]:
r -= 1
return l == r
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| class Solution {
public boolean validMountainArray(int[] arr) {
int n = arr.length;
if (n < 3) {
return false;
}
int l = 0, r = n - 1;
while (l + 1 < n - 1 && arr[l] < arr[l + 1]) {
++l;
}
while (r - 1 > 0 && arr[r] < arr[r - 1]) {
--r;
}
return l == r;
}
}
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| class Solution {
public:
bool validMountainArray(vector<int>& arr) {
int n = arr.size();
if (n < 3) return 0;
int l = 0, r = n - 1;
while (l + 1 < n - 1 && arr[l] < arr[l + 1]) ++l;
while (r - 1 > 0 && arr[r] < arr[r - 1]) --r;
return l == r;
}
};
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| func validMountainArray(arr []int) bool {
n := len(arr)
if n < 3 {
return false
}
l, r := 0, n-1
for l+1 < n-1 && arr[l] < arr[l+1] {
l++
}
for r-1 > 0 && arr[r] < arr[r-1] {
r--
}
return l == r
}
|