Count of nodes in a Binary Tree whose child is its prime factors
Input: 1 / \ 15 20 / \ / \ 3 5 4 2 \ / 2 3 Output: 3 Explanation: Children of 15 (3, 5) are prime factors of 15 Child of 20 (2) is prime factors of 20 Child of 4 (2) is prime factors of 4 Input: 7 / \ 210 14 / \ \ 70 14 30 / \ / \ 2 5 3 5 / 23 Output: 2 Explanation: Children of 70 (2, 5) are prime factors of 70 Children of 30 (3, 5) are prime factors of 30
- Traverse the given Binary Tree and for each node:
- Check if children exist or not.
- If the children exist, check if each child is a prime factor of this node or not.
- Keep the count of such nodes and print it at the end.
- In order to check if a factor is prime, we will use Sieve of Eratosthenes to precompute the prime numbers to do the checking in O(1).
Below is the implementation of the above approach:
- Time Complexity : O(N).
In dfs, every node of the tree is processed once and hence the complexity due to the bfs is O(N) if there are total N nodes in the tree. Also, for processing each node the SieveOfEratosthenes() function is used which has a complexity of O(sqrt(N)) too but since this function is executed only once, it does not affect the overall time complexity. Therefore, the time complexity is O(N).
- Auxiliary Space : O(N).
Extra space is used for the prime array, so the space complexity is O(N).
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