Simplify CombinationsWithReplacement #487
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It is given by indices.len. (Similar to what we do in Combinations.)
Can be inferred from `pool`: * For `self.indices.len()==0`, the first iteration yields `Some(self.current())`, and afterwards we just go through, unnecessarily call `self.pool.get_next()`, `increment` stays `None` as the loop is not iterated, and we return `None`. * For `self.indices.len()>0`: * If `pool` is empty from the start, `pool.get_next` is always `false` and we will always return `None` (without even setting `first=false`). * Otherwise (i.e. if `pool` is non-empty), we return `Some(self.current())` on first it eration, and afterwards always call `pool.get_next`, possibly updating `self.max_index`. In the first iteration `max_index` is 0, thus equals `pool.len()-1` when we return `Some(self.current())`. Thus, we know that `self.max_index` is equal to `self.pool.len()-1` when we come to the line `let mut increment...`.
phimuemue
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k: It can be obtained viaindices.len.max_index: It can be obtained viapool.len()-1.