Welcome to Subscribe On Youtube

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
    }
    

All Problems

All Solutions