NINJA: The Exam Instructor
Last Updated: 5 Mar, 2021
Difficulty: Moderate
Ninja went to a school as an invigilator for an exam the seats in the room were mentioned with the number and they form a tree-like structure so he started thinking of a solution so that he allocated seats to students in such a way that no students would sit adjacent to each other.
So help our ninja write a code for the problem so he is able to count the maximum number of seats available for the students that no student can sit adjacent to each other.
So your task is to write a code that can find the maximum subset of a node in a given binary tree where there is no edge between any two nodes of the subset.
For example, consider the following binary tree. The largest subset set is {1, 5, 6, 7} and the size of the subset is 4.
Input Format:
The first line contains an integer 'T' which denotes the number of test cases or queries to be run. Then the 'T' test cases are as 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 maximum number of elements in a subset satisfying the given condition. 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
-10^9 <= DATA <= 10^9
Where ‘T’ is the total number of test cases, ‘N’ is the number of benches(nodes of the tree) in the given class, and 'DATA' is the value mentioned on the bench(value of nodes).
Time Limit: 1sec
Approach 1
Approach: We use recursion by using simple two possibilities for every node x, either X is a member of the set or not a member:
- If X is a member, then the value of the maximum subset(X) is 1 plus the maximum subset of all grandchildren.
- If X is not a member, then the value is the sum of the maximum subset of all children.
Here is the algorithm:
- The logic behind this approach is whether we can choose the child of a node or the grandchild of a node.
- So for this, we take two variables ‘SIZEEXCL’ and ‘SIZEINCL’, ‘SIZEEXCL’ store the count of a child of the node, and ‘SIZEINCL’ stores the count of the grandchild of the node and recursively we check all the possibility of that node.
- After checking all the possibilities we return maximum from ‘SIZEINCL’ and ‘SIZEEXCL’.
Approach 2
Approach: We use recursion by using simple two possibilities for every node X, either X is a member of the set or not a member and we now store the solution or we can say we take an extra integer for every node or we can say bench and for every node, we store the value of maximum subset value for its child node so when we go to that node we simply return the value that we store so in this manner we don’t need to call the node once again and we call our nodes in the fashion:
- If X is a member, then the value of the maximum subset(X) is 1 plus the maximum subset of all grandchildren.
- If X is not a member, then the value is the sum of the maximum subset of all children.
Here is the algorithm:
- As if we use simple recursion we have to call the same node again and again so to avoid that situation now we make use of a map so to store that value.
- This map is capable of storing a root address and for that root its maximum subset size. Now we declare two variables ‘SIZEINCL’ and ‘SIZEEXCL’ ‘SIZEINCL’ is for storing grandchild count and ‘SIZEEXCL’ is for storing child count.
- So when a recursive call is called first we check if that matches in our map we simply return the value else a recursive call is to make.
- So this is an optimized method rather than simple recursion.
- After checking all the possibilities we return the maximum from ‘SIZEINCL’ or ‘SIZEEXCL’.
