g1901_2000.s1944_number_of_visible_people_in_a_queue

Class Solution

java.lang.Object

g1901_2000.s1944_number_of_visible_people_in_a_queue.Solution

public class Solution
extends Object

1944 - Number of Visible People in a Queue.

Hard

There are n people standing in a queue, and they numbered from 0 to n - 1 in left to right order. You are given an array heights of distinct integers where heights[i] represents the height of the ith person.

A person can see another person to their right in the queue if everybody in between is shorter than both of them. More formally, the ith person can see the jth person if i < j and min(heights[i], heights[j]) > max(heights[i+1], heights[i+2], ..., heights[j-1]).

Return an array answer of length n where answer[i] is the number of people the ith person can see to their right in the queue.

Example 1:

Input: heights = [10,6,8,5,11,9]

Output: [3,1,2,1,1,0]

Explanation:

Person 0 can see person 1, 2, and 4.

Person 1 can see person 2.

Person 2 can see person 3 and 4.

Person 3 can see person 4.

Person 4 can see person 5.

Person 5 can see no one since nobody is to the right of them.

Example 2:

Input: heights = [5,1,2,3,10]

Output: [4,1,1,1,0]

Constraints:

• n == heights.length
• 1 <= n <= 105
• 1 <= heights[i] <= 105
• All the values of heights are unique.

Constructor Summary

Constructors

Constructor and Description
Solution()

Method Summary

Modifier and Type	Method and Description
int[]	canSeePersonsCount(int[] heights)

Methods inherited from class java.lang.Object

clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait

Constructor Detail

Solution

public Solution()

Method Detail

canSeePersonsCount

public int[] canSeePersonsCount(int[] heights)