Now my algorithm is O(n) time and O(Dn) space where Dn is the total imblance in the list. This solution doesn’t modify the list. let D be the difference of 1s and 0s found in the list. First, let’s step linearily through the list and calculate D, just to see how it works: I’m gonna … Read more
If the structure stays the same, you can use this Decodable approach. First create a decodable Model like this: struct MyModel: Decodable { let firstString: String let stringArray: [String] init(from decoder: Decoder) throws { var container = try decoder.unkeyedContainer() firstString = try container.decode(String.self) stringArray = try container.decode([String].self) } } Or if you really want to … Read more
A simple stack based solution. Iterate over the array left to right, with the stack holding elements (technically, indexes, but use values for comparisons) that are either: The largest from the left (i.e. no larger or equal element between the start of the array and the element) The largest since the previous element on the … Read more
The reason we don’t use (12) for a one-element tuple (like [12] for one-element list) is that round parentheses also appear in formulas. E.g., in x = 2*(5+7) the part (5+7) is just a number, not a tuple. But what if we actually meant it to be a one-element tuple? The trailing comma is a … Read more
You can’t typically use OR or AND in “array formulas” because, as you have discovered here, they return only a single result (AND is TRUE only if all conditions are TRUE, OR is TRUE is any are TRUE, as you would expect, so in your example you’ll get TRUE as long as at least one … Read more
you can simply using Array#joinToString: val result: String = chars.joinToString(“”); OR convert chars to CharArray: val result: String = String(chars.toCharArray()); OR declaring a primitive CharArray by using charArrayOf: val chars = charArrayOf(‘A’, ‘B’, ‘C’); val result: String = String(chars);
There is a beautiful structure called an extendible array that has worst-case O(1) insertion and O(n) memory overhead (that is, it’s asymptotically comparable to a dynamic array, but has O(1) worst-case insertion). The trick is to take the approach that the vector uses – doubling and copying – but to make the copying lazy. For … Read more
If you want key and value, and based on How do i convert a json object to key=value format in JQ, you can do: $ jq -r “to_entries|map(\”\(.key)=\(.value|tostring)\”)|.[]” file SALUTATION=Hello world SOMETHING=bla bla bla Mr. Freeman In a more general way, you can store the values into an array myarray[key] = value like this, just … Read more
There’s a beautiful algorithm for solving this problem in time O(n + Sort), where Sort is the amount of time required to sort the input array. The idea behind the algorithm is to sort the array and then ask the following question: what is the smallest positive integer you cannot make using the first k … Read more
Adding values to an array is analogous to incrementing an integer or concatenating a string and is signified the same way, in your case (let c be your [c:contains {id:[12345]}]) c.id = c.id + 1111 // [12345,1111] c.id = c.id + 14567 // [12345,1111,14567] or c.id = c.id + [1111,14567] // [12345,1111,14567]