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845. Longest Mountain in Array

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?

Solutions

  • class Solution {
    public int longestMountain(int[] arr) {
        int n = arr.length;
        int ans = 0;
        for (int l = 0, r = 0; l + 2 < n; l = r) {
            r = l + 1;
            if (arr[l] < arr[r]) {
                while (r + 1 < n && arr[r] < arr[r + 1]) {
                    ++r;
                }
                if (r + 1 < n && arr[r] > arr[r + 1]) {
                    while (r + 1 < n && arr[r] > arr[r + 1]) {
                        ++r;
                    }
                    ans = Math.max(ans, r - l + 1);
                } else {
                    ++r;
                }
            }
        }
        return ans;
    }
}

  • class Solution {
public:
    int longestMountain(vector<int>& arr) {
        int n = arr.size();
        int ans = 0;
        for (int l = 0, r = 0; l + 2 < n; l = r) {
            r = l + 1;
            if (arr[l] < arr[r]) {
                while (r + 1 < n && arr[r] < arr[r + 1]) {
                    ++r;
                }
                if (r + 1 < n && arr[r] > arr[r + 1]) {
                    while (r + 1 < n && arr[r] > arr[r + 1]) {
                        ++r;
                    }
                    ans = max(ans, r - l + 1);
                } else {
                    ++r;
                }
            }
        }
        return ans;
    }
};

  • class Solution:
    def longestMountain(self, arr: List[int]) -> int:
        n = len(arr)
        ans = l = 0
        while l + 2 < n:
            r = l + 1
            if arr[l] < arr[r]:
                while r + 1 < n and arr[r] < arr[r + 1]:
                    r += 1
                if r < n - 1 and arr[r] > arr[r + 1]:
                    while r < n - 1 and arr[r] > arr[r + 1]:
                        r += 1
                    ans = max(ans, r - l + 1)
                else:
                    r += 1
            l = r
        return ans


  • func longestMountain(arr []int) (ans int) {
	n := len(arr)
	for l, r := 0, 0; l+2 < n; l = r {
		r = l + 1
		if arr[l] < arr[r] {
			for r+1 < n && arr[r] < arr[r+1] {
				r++
			}
			if r+1 < n && arr[r] > arr[r+1] {
				for r+1 < n && arr[r] > arr[r+1] {
					r++
				}
				ans = max(ans, r-l+1)
			} else {
				r++
			}
		}
	}
	return
}

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