Given a weighted tree, the task is to count the number of nodes whose sum of digits of weights is a Prime Number.
Node 1: digitSum(144) = 1 + 4 + 4 = 9
Node 2: digitSum(1234) = 1 + 2 + 3 + 4 = 10
Node 3: digitSum(21) = 2 + 1 = 3
Node 4: digitSum(5) = 5
Node 5: digitSum(77) = 7 + 7 = 14
Only the sum of digits of the weights of nodes 3 and 4 are prime.
Approach: To solve the problem mentioned above we have to perform DFS on the tree and for every node and check if the sum of the digits of its weight is prime or not. If yes then increment the count. To check if the digit sum is prime or not, we will use Sieve Of Eratosthenes. Create a sieve which will help us to identify if the degree is prime or not in O(1) time.
Below is the implementation of the above approach:
- Count the nodes in the given tree whose weight is prime
- Count of all prime weight nodes between given nodes in the given Tree
- Count the nodes in the given tree whose weight is even
- Count the nodes in the given tree whose sum of digits of weight is odd
- Count the nodes in the given tree whose weight is a power of two
- Count the nodes in the given tree whose weight is even parity
- Count the nodes of the given tree whose weight has X as a factor
- Count the nodes in the given Tree whose weight is a Perfect Number
- Count nodes in the given tree whose weight is a fibonacci number
- Count the nodes in the given tree whose weight is a powerful number
- Count of nodes in a Binary Tree whose immediate children are co-prime
- Count of nodes in a Binary Tree whose child is its prime factors
- Count the nodes whose weight is a perfect square
- Query to find the maximum and minimum weight between two nodes in the given tree using LCA.
- Count the nodes of the tree which make a pangram when concatenated with the sub-tree nodes
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to email@example.com. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.