845

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[0] < B[1] < ... 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

Follow up:

• 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/
// Author: github.com/lzl124631x
// 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;
    }
};