The bit-twiddling hacks page has a number of options. Of course, you could argue that iterating over all 32 possible bits is O(N) in that it’s the same cost every time 🙂 For simplicity, I’d consider the lookup-table-per-byte approach, or Brian Kernighan’s neat idea which iterates as many times as there are bits set, which … Read more
No, it is not random. It is “arbitrarily ordered”, which means that you cannot depend on it being either ordered or random.
HashSet consumes about 5.5 times more memory than ArrayList for the same number of elements (although they’re both still linear), and has significantly slower iteration (albeit with the same asymptotics); a quick Google search suggests a 2-3x slowdown for HashSet iteration versus ArrayList. If you don’t care about uniqueness or the performance of contains, then … Read more
Use the symmetric difference operator for sets (aka the XOR operator): >>> set([1,2,3]) ^ set([3,4,5]) set([1, 2, 4, 5])
I used a function like void setHighlighted(boolean highlight) { button.setBackgroundResource( highlight ? R.drawable.bbg_pressed : R.drawable.button_background); } where button_background is a selector defined in button_backgroung.xml: <?xml version=”1.0″ encoding=”utf-8″?> <selector xmlns:android=”http://schemas.android.com/apk/res/android” > <item android:state_pressed=”true” android:drawable=”@drawable/bbg_pressed”></item> <item android:state_focused=”true” android:drawable=”@drawable/bbg_selected”></item> <item android:drawable=”@drawable/bbg_normal”></item> </selector> That is, this code does not interfere with the pressed state used by the Android framework; … Read more
When a new object is being added to a python set, the hash code of the object is first computed and then, if one or more objects with the same hash code is/are already in the set, these objects are tested for equality with the new object. The upshot of this is that you need … Read more
They are not randomly ordered, they are arbitrarily ordered. It means you should not count on the order of insertions being maintained as the actual internal implementation details determine the order instead. The order depends on the insertion and deletion history of the set. In CPython, sets use a hash table, where inserted values are … Read more
Usage: var results = new[] { 1, 2, 3, 4, 5, 6, 7, 8, 9 }.DifferentCombinations(3); Code: public static class Ex { public static IEnumerable<IEnumerable<T>> DifferentCombinations<T>(this IEnumerable<T> elements, int k) { return k == 0 ? new[] { new T[0] } : elements.SelectMany((e, i) => elements.Skip(i + 1).DifferentCombinations(k – 1).Select(c => (new[] {e}).Concat(c))); } }
Seems like you want itertools.combinations: >>> list(itertools.combinations((1, 2, 3), 2)) [(1, 2), (1, 3), (2, 3)] If you want sets you’ll have to convert them explicitly. If you don’t mind an iterable instead of a list, and you’re using Python 3, you can use map: >>> s = set((1, 2, 3)) >>> map(set, itertools.combinations(s, 2)) … Read more
I suspect the restrictions should be taken as referring to the relation’s behavior on the values actually used as keys, not necessarily on all values of the type. Don’t have time at the moment to go through the standard looking for “smoking gun” language that refers to actual container elements rather than all values of … Read more