Both Comparable and Comparator are interfaces. All they do is define a way of asking whether one object is greater than, equal to, or less than another. The interfaces don’t enforce what that means — that’s up to the class that implements the interface. So one Comparator<Employee> might say that Adam is “greater than” Bill … Read more
var result = string.trim().split(“, “).map(el=>el.split(“|”).map(n=>+n)) this splits the string into an array of these groups ( yet as string ) “1|2, 3|4” => [“1|2″,”3|4”] then maps this array to a new array containing the strings splitted and converted to numbers: => [[1,2],[3,4]]
The algorithm to sort this problem is simple, just like you said in your question description, sort characters in each word first, then sort these sorted-word again. Like this: $ echo heya64 this is21 a good89 day91 | perl -anE ‘say(join ” “, sort(map { join “”, sort split // } @F))’ 12is 19ady 46aehy … Read more
Pseudo code, find last O(log N) of each name O(M). auto it = vec.begin(); while (it != vec.end()) { // O(M) auto last = find_last_with_same_name(it, it->name); sum += last.value; it++; } Use exponential_search for O(log N) for finding the last and therefore largest value. For a total of O(M log N). If M, the number … Read more
The LINQ way is to just do this (assuming you want to end up with a sorted array, the ToArray bit isn’t needed): var sorted = values.OrderBy( value => value < 0 ).ThenBy( Math.Abs ).ToArray(); The only reason I’m showing that is so you have can compare it with the non-LINQ approach using a comparer: … Read more
The solution can be Linq query: int[] first = new int[] { 3, 5, 6, 9, 12, 14, 18, 20, 25, 28 }; int[] second = new int[] { 30, 32, 34, 36, 38, 40, 42, 44, 46, 48 }; int[] result = first .Concat(second) .OrderBy(x => x) .ToArray(); To test // 3, 5, 6, … Read more
You could take an offset for Under and + quantifier and take the value, depending on the sorting position. Then return the delta of values or the delta of the offset. Later, you could build a new object with sorted keys. var object = { “60+”: 0.1413972485314015, “18-29”: 0.0832178903621611, “40-49”: 0.1033361204013377, “30-39”: 0.0835906328864075, “Under 18”: … Read more
You just: std::sort(v.begin(), v.end()); std::pair is lexicographically compared. On the other hand if you want to sort them with respect the second element of the std::pair then you would have to defind a custom comparator in the following manner: std::sort(v.begin(), v.end(), [](std::pair<int, std::string> const &p1, std::pair<int, std::string> const &p2) { return (p1.second == p2.second)? p1.first … Read more
Lets start here: } catch (ParseException e) { return 0; } ParseException means parsing went wrong. You are suppressing that and turn it into: the two strings equal the same time. Simply don’t do that. Then I would also suggest to not store strings, but Date objects in that list. You have to understand that … Read more
Your code would do the following: arr[] = 1, 5, 2, 4, 3 // find the max from arr[0..4] // substitute it with the index 0, if it is not of that index already arr[] = 5, 1, 2, 4, 3 // find the max from arr[1..4] // substitute it with the index 1, if … Read more