Last Updated: 23 Mar, 2021
BALLOONS GAME
Easy
Asked in company
Ninja went to an amusement park to enjoy his weekends. There is an offer running in the amusement park for free passes for the park. For that, a visitor has to win a game of balloons. In this game, the visitor will be provided with a gun and he has to hit a balloon.
He has to select a balloon by taking into consideration some conditions. Each balloon is marked with some number and balloons were arranged in the form of a binary search tree so the visitor has to hit the balloon which has the largest repeating frequency.
Ninja is good with hitting the target with the gun but not in algorithms so help our Ninja in choosing the balloon with repeated frequency.
Your task is to write a code that can find the element with a maximum frequency in the given binary search tree you will be provided with the root of the tree.
Example:
Consider the following binary search tree so the ‘7’ is the element with the highest repeated frequency i.e ‘3’ so ‘7’ is our answer.
Note :
If there is more than one element with the same maximum frequency you have to return a minimum of them.
Input Format:
The first line contains an integer ‘T' which denotes the number of test cases or queries to be run. Then the test cases follow.
The first line of each test case contains the elements of the tree in the level order form separated by a single space.
If any node does not have a left or right child, take -1 in its place. Refer to the example below.
Example:
Elements are in the level order form. The input consists of values of nodes separated by a single space in a single line. In case a node is null, we take -1 in its place.
For example, the input for the tree depicted in the below image would be :
1
2 3
4 -1 5 6
-1 7 -1 -1 -1 -1
-1 -1
Explanation :
Level 1 :
The root node of the tree is 1
Level 2 :
Left child of 1 = 2
Right child of 1 = 3
Level 3 :
Left child of 2 = 4
Right child of 2 = null (-1)
Left child of 3 = 5
Right child of 3 = 6
Level 4 :
Left child of 4 = null (-1)
Right child of 4 = 7
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 6 = null (-1)
Right child of 6 = null (-1)
Level 5 :
Left child of 7 = null (-1)
Right child of 7 = null (-1)
The first not-null node (of the previous level) is treated as the parent of the first two nodes of the current level. The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on.
The input ends when all nodes at the last level are null (-1).
Note :
The above format was just to provide clarity on how the input is formed for a given tree.
The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as:
1 2 3 -1 5 6 7 -1 -1 -1 -1
Output Format :
For each test case, print the node with maximum repeated frequency.
Print output of each test case in a separate line.
Note:
You are not required to print anything explicitly. It has already been taken care of. Just implement the function.
Constraints:
1 <= T <= 100
1 <= N <= 3000
0 <= DATA <= 10^5
Time Limit: 1 sec
Approaches
We have to find the maximum frequency node’s values, suppose we have an array instead of a tree then we can easily store the count of each element.
Suppose ‘ARRAY' = { 3, 1, 3, 6, 6, 6} , sort the ‘ARRAY’ ={ 1, 3, 3, 6, 6, 6 }.
So maximum frequency is by updating our max variable if the maximum value comes.
We want to sort the values of nodes in the tree. For this we need a vector for storing all the values then simply we can sort them and find the minimum absolute value.
- So we declare a vector of ‘INT’ type named as ‘ARR’.
- Then with the help of the queue, we do the level order traversal of our tree and fill the values of the nodes in our vector ‘ARR’.
- Now we use a queue ‘Q’ for the level order traversal of our tree and fill our vector ‘ARR’ accordingly.
- Now we sort our vector ‘ARR’ and run a loop to iterate over the vector ‘arr’ and check
- if ‘ARR[i]' = 'ARRR[i+1]’
- COUNT++
- Else
- MAX(COUNT, ANS)
- if ‘ARR[i]' = 'ARRR[i+1]’
- Now we simply return ‘ANS’.
To find the node, we need to find the Inorder Traversal of the BST because its Inorder Traversal will be in sorted order.
Suppose level order of tree is = { 3, 1, 3, 6, 6, 6}’, but if we do inorder traversal, the tree will look like in sorted order ={ 1, 3, 3, 6, 6, 6 }’
so maximum frequency is by updating our max variable if the maximum value comes.
- So, in this approach, we do a recursive Inorder traversal and we keep the track of the previous node.
- If the current node value is equal to the previous value we increase the current count
- Else we again initialized the count as 1
- Now we update our maximum after each in order recursion.
- Hence at last we return our answer.