SDE - 1

There are ‘N’ people numbered from 0 to N - 1, standing in a queue. You are given two arrays ‘Height’ and ‘Infront‘ consisting of ‘N’ non-negative integers. ‘Height[i]’ gives the height of the ith person, and ‘Infront[i]’ gives the number of persons who are taller than the ith person and standing in front of the ith person.

Your task is to find out the actual order of people in a queue. You should print ‘N’ integers where the ‘ith’ integer is the height of the person who should be at the ith position from the start of the queue.

Note :

1. Consider that all elements in array ‘Height’ are unique.
2. It is guaranteed that a valid order always exists for the given array ‘Height’ and ‘Infront’. 

Example :

Let there are 6 people, their heights are given by array  ‘Height’ :  [5, 3, 2, 6, 1, 4],  and the number of people in front of them is given by array ‘Infront’: [0, 1, 2, 0, 3, 2]

Thus the actual order of people’s height in the queue will be [5, 3, 2, 1, 6, 4]

In this order, the first person in a queue i.e a person with a height of 5, has no person in front of them who is taller than him.
The second person in a queue i.e a person with a height of 3 has 1 person (person with height 5) in front of them who is taller than him.
The third person in a queue i.e a person with a height of 2 has 2 people (people with height 5 and 3) in front of them who are taller than him.
The fourth person in a queue i.e a person with a height of 1 has 3 people (people with height 5, 3, 2) in front of them who are taller than him.
The fifth person in a queue i.e a person with a height of 6 has no person in front of them who is taller than him.
The sixth person in a queue i.e a person with a height of 4 has 2 people (people with height 5, and 6) in front of them who are taller than him.

We can observe this is the only possible order that is possible according to the array ‘Infront’.

You have been given a Binary Search Tree and a key value ‘X’, find if a node with value ‘X’ is present in the BST or not.

Note:

You may assume that duplicates do not exist in the tree.

For example :

For the given tree shown below:

For the binary tree shown in the figure, if ‘X’ = 6, the output will be 1 as node value 6 is present in the BST.