845. Longest Mountain in Array #

Problem #

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:

• 0 <= A.length <= 10000
• 0 <= A[i] <= 10000

Follow up:

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

Main Idea #

This problem tests the sliding window technique.

Given an array, the task is to find the length of the longest “mountain” in the array. A “mountain” means starting from a number, gradually increasing, and after reaching the peak, gradually decreasing.

Solution Approach #

The solution approach for this problem is also sliding window, except that during the sliding process, you only need to additionally determine the increasing and decreasing states.

Code #

package leetcode

func longestMountain(A []int) int {
	left, right, res, isAscending := 0, 0, 0, true
	for left < len(A) {
		if right+1 < len(A) && ((isAscending == true && A[right+1] > A[left] && A[right+1] > A[right]) || (right != left && A[right+1] < A[right])) {
			if A[right+1] < A[right] {
				isAscending = false
			}
			right++
		} else {
			if right != left && isAscending == false {
				res = max(res, right-left+1)
			}
			left++
			if right < left {
				right = left
			}
			if right == left {
				isAscending = true
			}
		}
	}
	return res
}