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?

Answer 1

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

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

Answer 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]);