Search for room selection options

Question

I have the following sample data in javascript:

var variations = [
  {group: 1, id: 1},
  {group: 1, id: 2},
  {group: 1, id: 3},
  {group: 1, id: 4},
  {group: 2, id: 5},
  {group: 2, id: 6},
  {group: 2, id: 7},
  {group: 3, id: 8},
  {group: 3, id: 9}
];

Say I already defined a choice using the following variables:

var selected_variation_groups = [1,2,3];
var selected_variation_group_ids = [1,2,3,4,5,6,7,8,9];

When I try to find the number selection options from the above data, then I will have 24 possibilities:

=> 1, 5, 8
=> 1, 5, 9
=> 1, 6, 8
=> 1, 6, 9
=> 1, 7, 8
=> 1, 7, 9

=> 2, 5, 8
=> 2, 5, 9
=> 2, 6, 8
=> 2, 6, 9
=> 2, 7, 8
=> 2, 7, 9

=> 3, 5, 8
=> 3, 5, 9
=> 3, 6, 8
=> 3, 6, 9
=> 3, 7, 8
=> 3, 7, 9

=> 4, 5, 8
=> 4, 5, 9
=> 4, 6, 8
=> 4, 6, 9
=> 4, 7, 8
=> 4, 7, 9

Is there anyone who can help me give an algorithm for this, or is there anyone who can help me provide javascript code to create these features?

groupand idmay be unlimited.

Answer 1

You compute all permutations to select one element from each group.

, , , 1 , 2 ..

, .

var groups = [
      [1,2,3,4], [5,6,7], [8,9]
];

var result = new Array();
appendPermutation(groups, 0, groups.length, "", result);

alert(result.length);
alert(result);


function appendPermutation(groups, start, end, currentResult, result)
{
   if (start==end)
   {

       result.push(currentResult);
        return;
   }

   var group = groups[start];
   for (var i=0; i<group.length; i++) {
      var value = group[i].toString();
      var nextResult;
      if (currentResult.length==0)
          nextResult = currentResult + value;
      else
          nextResult = currentResult + "," + value;      
      appendPermutation(groups, start+1, end, nextResult, result);
   }
}

Answer 2

mdma , . -, :

var groups = {};
for(var i = 0; i < variations.length; i++){
    var key = variations[i].group;
    var value = variations[i].id;

    if(!groups[key])
        groups[key] = [];
     groups[key].push(value);
}

groups = {}; groups = [];, , , ( ) ( ).

Answer 3

@mdma :

function permutation(options) {
    if(options.length == 1) {
        var permutations = [];
        for(var i = 0; i < options[0].length; i++) {
            permutations.push([options[0][i]]);
        }
        return permutations;
    }
    return addOptionWithPermutation(options[0], permutation(options.slice(1)));
}

function addOptionWithPermutation(option, permutations) {
    var newPermutations = [];
    for(var i = 0; i < option.length; i++) {
        for(var j = 0; j < permutations.length; j++) {
            var newPerm = permutations[j].slice(0); //just to take copy
            newPerm.splice(0, 0, option[i]); //insert in the beginning
            newPermutations.push(newPerm);
        }
    }
    return newPermutations;
}

var permutations = permutation([[1,2,3,4], [5,6,7], [8,9]]);

for(var i = 0; i < permutations.length; i++) {
  alert(permutations[i]);
}