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One of my sons asked an interesting question: what number raised to itself equals to 25? In other words, he wanted to find X such that X^X = 25.
I decided to write a simple program (no GAI for a change, just vintage coding) to do that. The key observations here are:
1/ X^X is continuous in the real domain
2/ It grows monotonically
3/ It grows very fas...
Solution here is a simple sliding window approach, the caveat is to cache the mapping number -> list of unique factors which you can use across test cases. Reasonably fast prime factorization is important too, mine isn't the most optimal but fast enough to pass all test cases. Code is down below, cheers, ACC.
Another problem whose a solution can be accomplished via Depth-First Search (DFS). Map the tree to a hash table. Calculate the height of the tree separately (also a DFS). Then perform a DFS to calculate the weighted sum. Code is down below, cheers, ACC.