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`<[T]>::sort_by_index` and `<[T]>::sort_by_key_and_index`

Rust Internals [Unofficial] April 16, 2026
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The O(N2) way is what get_disjoint_mut does, comparing every index against duplicates (and in range), which as you say is also a permutation if N == len(). But of course the assumption in that case is that N is usually small.

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