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Find missing element by comparing 2 arrays in Javascript

Asked 2012-03-16T11:58:49.907
15

For some reason I'm having some serious difficulty wrapping my mind around this problem. I need this JS function that accepts 2 arrays, compares the 2, and then returns a string of the missing element. E.g. Find the element that is missing in the currentArray that was there in the previous array.

function findDeselectedItem(CurrentArray, PreviousArray){

var CurrentArrSize = CurrentArray.length;
var PrevousArrSize = PreviousArray.length;

// Then my brain gives up on me...
// I assume you have to use for-loops, but how do you compare them??

return missingElement;

}

Thank in advance! I'm not asking for code, but even just a push in the right direction or a hint might help...

javascript arrays compare
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2 Answers

4

This should work. You should also consider the case where the elements of the arrays are actually arrays too. The indexOf might not work as expected then. 

function findDeselectedItem(CurrentArray, PreviousArray) {

   var CurrentArrSize = CurrentArray.length;
   var PreviousArrSize = PreviousArray.length;

   // loop through previous array
   for(var j = 0; j < PreviousArrSize; j++) {

      // look for same thing in new array
      if (CurrentArray.indexOf(PreviousArray[j]) == -1)
         return PreviousArray[j];

   }

   return null;

}
answered 2012-03-16T12:07:25.950
1

I know this is code but try to see the difference examples to understand the way:

var current = [1, 2, 3, 4],
    prev = [1, 2, 4],
    isMatch = false,
    missing = null;

var i = 0, y = 0,
    lenC = current.length,
    lenP = prev.length;

for ( ; i < lenC; i++ ) {
    isMatch = false;
    for ( y = 0; y < lenP; y++ ) {
        if (current[i] == prev[y]) isMatch = true;
    }
    if ( !isMatch ) missing = current[i]; // Current[i] isn't in prev
}

alert(missing);

Or using ECMAScript 5 indexOf:

var current = [1, 2, 3, 4],
    prev = [1, 2, 4],
    missing = null;

var i = 0,
    lenC = current.length;

for ( ; i < lenC; i++ ) {
    if ( prev.indexOf(current[i]) == -1 ) missing = current[i]; // Current[i] isn't in prev
}

alert(missing);

And with while

var current = [1, 2, 3, 4],
    prev = [1, 2, 4],
    missing = null,
    i = current.length;

while(i) {
    missing = ( ~prev.indexOf(current[--i]) ) ? missing : current[i];
}

alert(missing);
answered 2012-03-16T12:04:53.823

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