LeetCode //C - 941. Valid Mountain Array
Summary: The problem checks if an array is a valid mountain array, which must: Have length ≥ 3 Strictly increase to a peak (not at the ends) Strictly decrease after the peak End at the last element Ap
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941. Valid Mountain Array
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 < = a r r . l e n g t h < = 1 0 4 1 <= arr.length <= 10^4 1<=arr.length<=104
- 0 < = a r r [ i ] < = 1 0 4 0 <= arr[i] <= 10^4 0<=arr[i]<=104
From: LeetCode
Link: 941. Valid Mountain Array
Solution:
Ideas:
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Walk up while strictly increasing, stop at the peak.
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Peak can’t be at ends.
-
Walk down while strictly decreasing.
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If you end exactly at the last index, it’s a valid mountain.
Code:
bool validMountainArray(int* arr, int arrSize){
if (arrSize < 3) return false;
int i = 0;
// climb up (strictly increasing)
while (i + 1 < arrSize && arr[i] < arr[i + 1]) i++;
// peak cannot be first or last
if (i == 0 || i == arrSize - 1) return false;
// go down (strictly decreasing)
while (i + 1 < arrSize && arr[i] > arr[i + 1]) i++;
// must finish exactly at the end
return i == arrSize - 1;
}
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