Longest Mountain in Array

 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:

  1. 0 <= A.length <= 10000
  2. 0 <= A[i] <= 10000

Follow up:

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


 class Solution {
    public int longestMountain(int[] A) {
        int result = 0, up = 0, down = 0;
        for (int i = 1; i < A.length; ++i) {
            if (down > 0 && A[i - 1] < A[i] || A[i - 1] == A[i]) up = down = 0;
            if (A[i - 1] < A[i]) up++;
            if (A[i - 1] > A[i]) down++;
            if (up > 0 && down > 0 && up + down + 1 > result) result = up + down + 1;
        }
        return result;
    }
}

In this solution up defines count up mountain length and down defines count down mountain length. Both up and down length are clear to 0 when A[i - 1] == A[i] or down > 0 && A[i - 1] < A[i].

Click Follow button to receive latest updates from us instantly.


Please let us know about your views in comment section.

Like us? Help us to grow...


                                                  :-)  ..Happy Coding... :-)

The intention behind this blog is to help all my fellow programmers. Also to provide solutions for various coding platform problems with solution and explanation. I will add many tutorials in future. I am not a professional blogger but whenever time permits I will post. I hope you’ll enjoy the content and find it useful! Check out my Writer blog for a more valuable quotes. For any queries or suggestions, please feel free to reach out to me. Happy coding!

Comments

Post a Comment

Popular posts from this blog

Balanced Binary Tree

Given a binary tree, determine if it is height-balanced. For this problem, a height-balanced binary tree is defined as: a binary tree in which the left and right subtrees of  every  node differ in height by no more than 1. Example 1: Input: root = [3,9,20,null,null,15,7] Output: true Example 2: Input: root = [1,2,2,3,3,null,null,4,4] Output: false Example 3: Input: root = [] Output: true   Constraints: The number of nodes in the tree is in the range  [0, 5000] . -10 4  <= Node.val <= 10 4 Solution: /**  * Definition for a binary tree node.  * public class TreeNode {  *     int val;  *     TreeNode left;  *     TreeNode right;  *     TreeNode() {}  *     TreeNode(int val) { this.val = val; }  *     TreeNode(int ...
Read more

First Unique Character in a String

Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1. Examples: s = "leetcode" return 0. s = "loveleetcode", return 2. Note: You may assume the string contain only lowercase letters. class Solution {     public int firstUniqChar(String s) {         int length = s.length();         String str;         for(int i=0;i<length;i++){             if(i==length-1){                 str = s.substring(0, i);             }             else{                 str = s.substring(0,i)+s.sub...
Read more

Majority Element

Given an array of size n , find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times. You may assume that the array is non-empty and the majority element always exist in the array. Example 1: Input: [3,2,3] Output: 3 Example 2: Input: [2,2,1,1,1,2,2] Output: 2   class Solution { public int majorityElement(int[] nums) { Arrays.sort(nums); return nums[nums.length/2]; } } Here, we are sorting the entire array using sort() of Arrays class. Then, number at index position n/2 would be a majority element. There is no chance of wrong values in odd or even length.   Below is another method using HashMap. METHOD 2: class Solution { public int majorityElement(int[] nums) { HashMap<Integer, Integer>list = new HashMap<Integer, Integer>(); int ...
Read more

Smallest Range II

Given an array   A   of integers, for each integer   A[i]   we need to choose   either  x = -K  or  x = K , and add   x   to   A[i]  (only once) . After this process, we have some array  B . Return the smallest possible difference between the maximum value of  B  and the minimum value of  B . Example 1: Input: A = [1] , K = 0 Output: 0 Explanation : B = [1] Example 2: Input: A = [0,10] , K = 2 Output: 6 Explanation : B = [2,8] Example 3: Input: A = [1,3,6] , K = 3 Output: 3 Explanation : B = [4,6,3] Note: 1 <= A.length <= 10000 0 <= A[i] <= 10000 0 <= K <= 10000 Solution: class Solution {     public int smallestRangeII(int[] A, int K) { ...
Read more