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] < A[i]
arr[i] > arr[i + 1] > ... > arr[arr.length - 1]
Example 1:
Input: arr = [2,1]
Output: false
From the Leetcode solution:
Intuition
If we walk along the mountain from left to right, we have to move strictly up, then strictly down.
Algorithm
Let's walk up from left to right until we can't: that has to be the peak. We should ensure the peak is not the first or last element. Then, we walk down. If we reach the end, the array is valid, otherwise its not.
classSolution(object):defvalidMountainArray(self,A): N =len(A) i =0# walk upwhile i+1< N and A[i]< A[i+1]: i +=1# peak can't be first or lastif i ==0or i == N-1:returnFalse# walk downwhile i+1< N and A[i]> A[i+1]: i +=1return i == N-1
Two people climb from left and from right separately. If they are climbing the same mountain, they will meet at the same point.
classSolution:defvalidMountainArray(self,arr: List[int]) ->bool: A = arr i, j, n =0,len(A)-1,len(A)while i +1< n and A[i]< A[i +1]: i +=1while j >0and A[j -1]> A[j]: j -=1return0< i == j < n -1