Formatted question description: https://leetcode.ca/all/845.html

# 845. Longest Mountain in Array (Medium)

Let's call any (contiguous) subarray B (of A) a mountain if the following properties hold:

• B.length >= 3
• There exists some 0 < i < B.length - 1 such that B < B < ... B[i-1] < B[i] > B[i+1] > ... > B[B.length - 1]

(Note that B could be any subarray of A, including the entire array A.)

Given an array A of integers, return the length of the longest mountain

Return 0 if there is no mountain.

Example 1:

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


Note:

1. 0 <= A.length <= 10000
2. 0 <= A[i] <= 10000

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

Related Topics:
Two Pointers

## Solution 1.

// OJ: https://leetcode.com/problems/longest-mountain-in-array/

// Time: O(N)
// Space: O(1)
class Solution {
public:
int longestMountain(vector<int>& A) {
int dir = 0, N = A.size(), start = -1, ans = 0;
for (int i = 1; i < N; ++i) {
int d = A[i] == A[i - 1] ? 0 : (A[i] > A[i - 1] ? 1 : -1); // the new direction
if (d == dir) continue; // if the direction is the same, skip
if (dir == -1 && start != -1) ans = max(ans, i - start); // if we have a valid starting point and we are going downwards, we can try to update the answer
dir = d;
if (dir == 1) start = i - 1; // if new direction is upwards, set the starting point
else if (dir == 0) start = -1; // going horizontal will invalidate the starting point.
}
if (dir == -1 && start != -1) ans = max(ans, N - start); // handle the case where the downward section reaches the end.
return ans;
}
};


Java

class Solution {
public int longestMountain(int[] A) {
int length = A.length;
int[] ascending = new int[length];
for (int i = 1; i < length; i++) {
if (A[i] > A[i - 1])
ascending[i] = ascending[i - 1] + 1;
}
int[] descending = new int[length];
for (int i = length - 2; i >= 0; i--) {
if (A[i] > A[i + 1])
descending[i] = descending[i + 1] + 1;
}
int longest = 0;
int start = 1, end = length - 1;
for (int i = start; i < end; i++) {
if (ascending[i] > 0 && descending[i] > 0) {
int mountainLength = ascending[i] + descending[i] + 1;
longest = Math.max(longest, mountainLength);
}
}
return longest;
}
}