c – Sort an array which is partially sorted – Education Career Blog

I am trying to sort an array which has properties like

it increases upto some extent then it starts decreasing, then increases and then decreases and so on. Is there any algorithm which can sort this in less then nlog(n) complexity by making use of it being partially ordered?

array example = 14,19,34,56,36,22,20,7,45,56,50,32,31,45……… upto n

Thanks in advance

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Any sequence of numbers will go up and down and up and down again etc unless they are already fully sorted (May start with a down, of course). You could run through the sequence noting the points where it changes direction, then then merge-sort the sequences (reverse reading the backward sequences)

In general the complexity is N log N because we don’t know how sorted it is at this point. If it is moderately well sorted, i.e. there are fewer changes of direction, it will take fewer comparisons.

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You could find the change / partition points, and perform a merge sort between pairs of partitions. This would take advantage of the existing ordering, as normally the merge sort starts with pairs of elements.

Edit Just trying to figure out the complexity here. Merge sort is n log(n), where the log(n) relates to the number of times you have to re-partition. First every pair of elements, then every pair of pairs, etc… until you reach the size of the array. In this case you have n elements with p partitions, where p < n, so I’m guessing the complexity is p log(p), but am open to correction. e.g. merge each pair of paritions, and repeat based on half the number of partitions after the merge.

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See Topological sorting

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If you know for a fact that the data are “almost sorted” and the set size is reasonably small (say an array that can be indexed by a 16-bit integer), then Shell is probably your best bet. Yes, it has a basic time complexity of O(n^2) (which can be reduced by the sequence used for gap sizing to a current best-worst-case of O(n*log^2(n))), but the performance improves with the sortedness of the input set to a best-case of O(n) on an already-sorted set. Using Sedgewick’s sequence for gap size will give the best performance on those occasions when the input is not as sorted as you expected it to be.

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Strand Sort might be close to what you’re looking for. O(n sqrt(n)) in the average case, O(n) best case (list already sorted), O(n^2) worst case (list sorted in reverse order).

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