875. Longest Mountain In Array¶

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875. Longest Mountain in Array

Medium

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?

Solution¶

class Solution {
    public int longestMountain(int[] arr) {
        int n = arr.length;
        if (n < 3)
            return 0;
        int count[] = new int[n];
        int cnt = 1;
        for (int i = 1; i < n; i++) {
           if (arr[i] > arr[i - 1]) cnt++;
           else cnt = 1;
           count[i] += cnt; 
        }  
        cnt = 1;
        for (int i = n - 2; i >= 0; i--) {
            if (arr[i] > arr[i + 1]) cnt++;
            else cnt = 1;
            count[i] += cnt;
        }
        int maxi = 0;
        for (int i = 1; i < n - 1; i++) {
            if (count[i] > 2 && arr[i - 1] < arr[i] && arr[i] > arr[i + 1]) {
                maxi = Math.max(maxi, count[i] - 1);
            }
        } 
        return maxi;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here