Return ALL intersections array javascript

Question

My current code:

    function intersect(first, second) {
    var temp = [];
    for(var i = 0; i < first.length; i++){
        for(var k = 0; k < second.length; k++){
            if(first[i] == second[k]){
                temp.push( first[i]);
                break;
            }
        }
    }

  return temp;
}

How can I change this so it returns ALL the intersections indices?

Rias · Accepted Answer · 2015-03-10 17:55:22Z

3

You need to add the indices to your result;

function intersect(first, second) {
  var temp = [];
  for(var i = 0; i < first.length; i++){
    for(var k = 0; k < second.length; k++){
      if(first[i] == second[k]){
        temp.push([i, k]); // push i and k as an array
      }
    }
  }

  return temp;
}

Also remove the break; if you want recurring exact intersections to be selected also.

Find a running example here: http://jsfiddle.net/0tL9sk5w/1

edited Mar 10, 2015 at 17:55

answered Mar 10, 2015 at 17:47

Rias

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3 Comments

Oceanity Over a year ago

Correct me if I'm wrong, but won't that break statement cause it to only return the first match in the second array?

Rias Over a year ago

Indeed, if in the second array a second exact same intersection occurs it is filtered by the break, so it has to be removed. I will edit it accordingly, thanks.

Y. Zhang Over a year ago

thank you... can't believe i didn't catch this part in all my attemtps temp.push([i, k])

Khalid · Accepted Answer · 2015-03-10 17:49:08Z

2

All you need to do is to push the two indexes as an array

instead of :

temp.push( first[i]);

you need to do this :

temp.push([i,k]);

answered Mar 10, 2015 at 17:49

Khalid

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