Problem 590. Greed is good - Simple partition P[n].

Created by Muthu Annamalai

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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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Last Solution submitted on Jul 27, 2024

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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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Problem 590. Greed is good - Simple partition P[n].

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Last 200 Solutions

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Problem 590. Greed is good - Simple partition P[n].

Solution Stats

Last 200 Solutions

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Solution Comments

Problem Recent Solvers27

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More from this Author10

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