Problem 590. Greed is good - Simple partition P[n].
Find a simple partition P[n]. E.g. P[10] = 4 + 3 + 2 + 1.
- There are many solutions, compute just one set.
- Don't repeat numbers.
- Be Greedy ;-)
- To check against trivial solutions, E.g. [x-k, k] etc; but I'll provide you with one to start.
- Show me how you write the whole solution.
Bonus points if you solve the general problem of producing all unique partitions of [n].
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1 Comment
Rafael S.T. Vieira
on 3 Oct 2020
Maybe you should say that the partition must have size greater than ceil(log2(x)/2)), since you test this at the test suite. The total number of partitions that a number has may be huge, even if we consider only unique sets (ignoring the order): a sum of binomial coefficients from 1 to the number-1 (where there is only one possible set of ones).
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