This can be done in O(1) space and O(N2) time. First lets solve a simpler problem: Given two arrays A and B pick one element from each so that their sum is equal to given number K. Sort both the arrays which takes O(NlogN). Take pointers i and j so that i points to the … Read more
in your for loop use a Redim on the array like here: For i = 0 to 3 ReDim Preserve test(i) test(i) = 3 + i Next i
I can, of course, only speculate. However when I run your test with the JIT compiler enabled vs disabled, I get the following results: % with JIT no JIT 0.1677 0.0011 %# init 0.0974 0.0936 %# #1 I added an assigment before this line to avoid issues with deferring 0.4005 0.4028 %# #2 0.4047 0.4005 … Read more
I believe you want imgs.0. eg, given your example document, you want to say: db.products.find({“imgs.0”: “http://foo.jpg”}) Be aware that referencing array indexes only works for the first-level array. Mongo doesn’t support searching array indexes any deeper.
You can use linear indexing to access each element. for idx = 1:numel(array) element = array(idx) …. end This is useful if you don’t need to know what i,j,k, you are at. However, if you don’t need to know what index you are at, you are probably better off using arrayfun()
You can use zip(), which creates a sequence of pairs from the two given sequences: let strArr1 = [“Some1”, “Some2”] let strArr2 = [“Somethingelse1”, “Somethingelse2”] for (e1, e2) in zip(strArr1, strArr2) { print(“\(e1) – \(e2)”) } The sequence enumerates only the “common elements” of the given sequences/arrays. If they have different length then the additional … Read more
You could do: if [[ ” ${arr[*]} ” == *” d “* ]]; then echo “arr contains d” fi This will give false positives for example if you look for “a b” — that substring is in the joined string but not as an array element. This dilemma will occur for whatever delimiter you choose. … Read more
$0 is the first parameter passed into the closure. $1 is the second parameter, etc. That closure you showed is shorthand for: let sortedNumbers = numbers.sort { (firstObject, secondObject) in return firstObject > secondObject }
This works for me: var test = [1, 2, 3] var n = 2 var test2 = test[0..<n] Your issue could be with how you’re declaring your array to begin with. EDIT: To fix your function, you have to cast your Slice to an array: func aFunction(numbers: Array<Int>, position: Int) -> Array<Int> { var newNumbers … Read more
List(1,2,3) :+ 4 Results in List[Int] = List(1, 2, 3, 4) Note that this operation has a complexity of O(n). If you need this operation frequently, or for long lists, consider using another data type (e.g. a ListBuffer).