Given a tree, and the weights of all the nodes, the task is to count the number of nodes whose weight is prime.
Only the weights of the nodes 1 and 3 are prime.
Approach: Perform dfs on the tree and for every node, check if it’s weight is prime or not.
Below is the implementation of above approach:
- Time Complexity: O(N*sqrt(V)), where V is the maximum weight of a node in the given tree.
In DFS, every node of the tree is processed once and hence the complexity due to the DFS is O(N) when there are N total nodes in the tree. Also, while processing every node, in order to check if the node value is prime or not, a loop up to sqrt(V) is being run, where V is the weight of the node. Hence for every node, there is an added complexity of O(sqrt(V)). Therefore, the time complexity is O(N*sqrt(V)).
- Auxiliary Space: O(1).
Any extra space is not required, so the space complexity is constant.
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- Print the nodes with a prime degree in given Prufer sequence of a Tree
- Count of Nodes whose both immediate children are its prime factors
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