Checking if an element sastifying a condition in a list comes before another of a different condition
Currently, you are in fact looping through the entire list three times.


Collections2.filter(list, condition1).size()
Collections2.filter(list, condition2).size()
for (T t: list)


This is a bit inefficient, as ultimately, once should be enough.

To accomplish that, we need to check each item exactly once (or technically, at most once, as it is possible to return false early).

Also, to make the method slightly more useful, you can use Predicate&lt;? super T&gt;.

Which is easiest, checking if the list does adhere to your requirements or checking if it does not? In my opinion, it is easier to check if one of your requirements fail, which means that we can have the method return true; as the last statement and perform some early returns for false if something doesn't match your requirements.

What I would end up with is this, which is pretty much straight-forward:

private static &lt;T&gt; boolean isItemWithConditionBeforeAnother(Predicate&lt;T&gt; condition1, Predicate&lt;T&gt; condition2, List&lt;T&gt; list) {
    int found1 = 0;
    int found2 = 0;
    for (T t: list) {
        if (condition1.apply(t)) {
            found1++;
        }
        if (condition2.apply(t)) {
            found2++;
        }
        if (found1 &gt; 1 || found2 &gt; 1) {
            return false;
        }
        if (found2 &gt; found1) {
            return false;
        }
    }
    return found1 == 1 &amp;&amp; found2 == 1;
}


Using the found variables as int instead of boolean helps greatly here. It lets us check if any condition has been matched twice, and lets us use the trick found2 &gt; found1 to check if condition2 was matched before condition1.



Alternative approach using booleans, inspired by @janos.

boolean found1 = false;
boolean found2 = false;
for (T value : list) {
    if (found1 &amp;&amp; condition1.apply(value)) {
        return false;
    }
    if (found2 &amp;&amp; condition2.apply(value)) {
        return false;
    }
    found1 = found1 || condition1.apply(value);
    found2 = found2 || condition2.apply(value);
    if (found2 &amp;&amp; !found1) {
        return false;
    }
}
return found1 &amp;&amp; found2;


An important thing here is that once found1 is true, it will never be set to false again thanks to the short-circuiting || operator.