You can put all word arrays into one array and use a recursive function like this: function concat(array $array) { $current = array_shift($array); if(count($array) > 0) { $results = array(); $temp = concat($array); foreach($current as $word) { foreach($temp as $value) { $results[] = $word . ‘ ‘ . $value; } } return $results; } else … Read more
OK, here’s your code (and btw, thanks for posting such an interesting and challenging problem – at least for me… :-)) – using recursion for all possible permutations (by N) given an array of elements) Code : <?php function permutations($arr,$n) { $res = array(); foreach ($arr as $w) { if ($n==1) $res[] = $w; else … Read more
Here is a snippet that will do the work. #include <iostream> int main() { const int n = 10; const int k = 5; int combination[k] = {2, 5, 7, 8, 10}; int index = 0; int j = 0; for (int i = 0; i != k; ++i) { for (++j; j != combination[i]; … Read more
I have a marvelous followup to SilentGhost’s proposal – posting a separate answer since the margins of a comment would be too narrow to contain code 🙂 itertools.permutations is built in (since 2.6) and fast. We just need a filtering condition that for every (perm, perm[::-1]) would accept exactly one of them. Since the OP … Read more
I start with a simple but not quite linear-time algorithm. We choose some value between array1[0]+array2[0] and array1[N-1]+array2[N-1]. Then we determine how many pair sums are greater than this value and how many of them are less. This may be done by iterating the arrays with two pointers: pointer to the first array incremented when … Read more
google-sheets It is a hard task for native functions. Try a script and use it as a custom function: function getTournament(teams_from_range) { // teams_from_range — 2D Array var teams = []; // convert to list teams_from_range.forEach(function(row) { row.forEach(function(cell) { teams.push(cell); } ); } ); return getTournament_(teams); } function getTournament_(teams) { var start = 0; var … Read more
First convert the string to a set of unique characters and occurrence numbers e.g. BANANA -> (3, A),(1,B),(2,N). (This could be done by sorting the string and grouping letters). Then, for each letter in the set, prepend that letter to all permutations of the set with one less of that letter (note the recursion). Continuing … Read more
The limitations described below are because of lambda functions. The first solution can be successfully implemented without lambda: =ARRAYFORMULA(QUERY(BASE(SEQUENCE(PERMUTATIONA(7,7)),7,7),”where not Col1 matches ‘.*((“&JOIN(“)|(“,SEQUENCE(7,1,0)&”.*”&SEQUENCE(7,1,0))&”)).*'”,0)) The trick here is to use regex to find unique elements using query…match. The only problem with this is memory size needed will exceed 10 million for 8 items PERMUTATIONA(8,8). But that … Read more
If this isn’t simply a learning exercise, then it’s not necessary for you to roll your own algorithm to generate the partitions: Python’s standard library already has most of what you need, in the form of the itertools.combinations function. From Theorem 2 on the Wikipedia page you linked to, there are n+k-1 choose k-1 ways … Read more