Skip to content

845. Longest Mountain in Array

Difficulty Topics

Description

You may recall that an array arr is a mountain array if and only if:

  • arr.length >= 3
  • There exists some index i (0-indexed) 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]

Given an integer array arr, return the length of the longest subarray, which is a mountain. Return 0 if there is no mountain subarray.

 

Example 1:

Input: arr = [2,1,4,7,3,2,5]
Output: 5
Explanation: The largest mountain is [1,4,7,3,2] which has length 5.

Example 2:

Input: arr = [2,2,2]
Output: 0
Explanation: There is no mountain.

 

Constraints:

  • 1 <= arr.length <= 104
  • 0 <= arr[i] <= 104

 

Follow up:

  • Can you solve it using only one pass?
  • Can you solve it in O(1) space?

Solution

longest-mountain-in-array.py
class Solution:
    def longestMountain(self, A):
        up, down = [0] * len(A), [0] * len(A)
        for i in range(1, len(A)):
            if A[i] > A[i - 1]: up[i] = up[i - 1] + 1
        for i in range(len(A) - 1)[::-1]:
            if A[i] > A[i + 1]: down[i] = down[i + 1] + 1
        return max([u + d + 1 for u, d in zip(up, down) if u and d] or [0])