# Strand Sort

Strand sort is a recursive sorting algorithm that sorts items of a list into increasing order. It has O(n²) worst time complexity which occurs when the input list is reverse sorted. It has a best case time complexity of O(n) which occurs when the input is a list that is already sorted.

Given a list of items, sort them in increasing order.

Examples:

Input: ip[] = {10, 5, 30, 40, 2, 4, 9}
Output: op[] = {2, 4, 5, 9, 10, 30, 40}

Input: ip[] = {1, 10, 7}
Output: op[] = {1, 7, 10}

Illustrations:

Let, input[] = {10, 5, 30, 40, 2, 4, 9}

Initialize: output[] = {}, sublist[] = {}

Move first item of input to sublist.
sublist[] = {10}

Traverse remaining items of input and if current element is greater than last item of sublist, move this item from input to sublist.
Now, sublist[] = {10, 30, 40}, input[] = {5, 2, 4, 9}

Merge sublist into output.
op = {10, 30, 40}

Next recursive call: Move first item of input to sublist. sublist[] = {5}

Traverse remaining items of input and move elements greater than last inserted.
input[] = {2, 4}
sublist[] = {5, 9}

Merge sublist into op.
output = {5, 9, 10, 30, 40}

Last Recursive Call:

{2, 4} are first moved to sublist and then merged into output.
output = {2, 4, 5, 9, 10, 30, 40}

Below are simple steps used in the algorithm:

• Let ip[] be input list and op[] be output list.
• Create an empty sublist and move first item of ip[] to it.
• Traverse remaining items of ip. For every item x, check if x is greater than last inserted item to sublist. If yes, remove x from ip and add at the end of sublist. If no, ignore x (Keep it it in ip)
• Merge sublist into op (output list)
• Recur for remaining items in ip and current items in op.

Below is the implementation of above algorithm. The C++ implementation uses list in C++ STL

## CPP

 `// CPP program to implement Strand Sort``#include ``using` `namespace` `std;`` ` `// A recursive function to implement Strand``// sort.``// ip is input list of items (unsorted).``// op is output list of items (sorted)``void` `strandSort(list<``int``> &ip, list<``int``> &op)``{``    ``// Base case : input is empty``    ``if` `(ip.empty())``        ``return``;`` ` `    ``// Create a sorted sublist with``    ``// first item of input list as``    ``// first item of the sublist``    ``list<``int``> sublist;``    ``sublist.push_back(ip.front());``    ``ip.pop_front();``      ` `    ``// Traverse remaining items of ip list``    ``for` `(``auto` `it = ip.begin(); it != ip.end(); ) {`` ` `        ``// If current item of input list``        ``// is greater than last added item``        ``// to sublist, move current item``        ``// to sublist as sorted order is``        ``// maintained.``        ``if` `(*it > sublist.back()) {``            ``sublist.push_back(*it);`` ` `            ``// erase() on list removes an``            ``// item and returns iterator to``            ``// next of removed item.``            ``it = ip.erase(it);``        ``}`` ` `        ``// Otherwise ignore current element``        ``else``            ``it++;``    ``}`` ` `    ``// Merge current sublist into output``    ``op.merge(sublist);`` ` `    ``// Recur for remaining items in``    ``// input and current items in op.``    ``strandSort(ip, op);``}`` ` `// Driver code``int` `main(``void``)``{``    ``list<``int``> ip{10, 5, 30, 40, 2, 4, 9};`` ` `    ``// To store sorted output list``    ``list<``int``> op;`` ` `    ``// Sorting the list``    ``strandSort(ip, op);`` ` `    ``// Printing the sorted list``    ``for` `(``auto` `x : op)``        ``cout << x << ``" "``;``    ``return` `0;``}`

Output:

`2 4 5 9 10 30 40`

Time complexity: O(N2)
Auxiliary Space: O(N)

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