README.md

845. Longest Mountain in Array

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?

Solutions (Rust)

1. Solution

impl Solution {
    pub fn longest_mountain(arr: Vec<i32>) -> i32 {
        let mut upcount = 0;
        let mut downcount = 0;
        let mut ret = 0;

        for i in 1..arr.len() {
            if arr[i] == arr[i - 1] {
                upcount = 0;
                downcount = 0;
            } else if arr[i] > arr[i - 1] && i > 1 && arr[i - 1] < arr[i - 2] {
                upcount = 1;
                downcount = 0;
            } else if arr[i] > arr[i - 1] {
                upcount += 1;
            } else {
                downcount += 1;
            }

            if upcount > 0 && downcount > 0 {
                ret = ret.max(upcount + downcount + 1);
            }
        }

        ret
    }
}