2813. Max Elegance of K-Length Subseq: Swift code for elegantly selecting unique k-length subsequences with profit and categories. Solution uses sorting and iteration. Time: O(nlogn), Space: O(n).
The algorithm follows a breadth-first search (BFS) approach to determine the minimum number of reverse operations needed to bring the 1 to each position in the array.
The code uses dynamic programming to solve the problem. It maintains a dictionary dp, where the keys represent the possible height differences between the two billboards, and the values represent the maximum sum of heights achieved for each height difference.
Очередная задачка уровня Easy, но с довольно низким показателем Acceptance (44.4%). Что выражается в достаточно обширном наборе граничных случаев, некоторые из которых делают больно 🙂
Приятная задачка выпала в поиске – хоть и easy, а сделать интересно. Какая-то “практическая применимость” в ней видится.
The Swift solution uses dynamic programming to solve the problem. The rec function recursively calculates the count of stepping numbers based on certain conditions. The dp array is used to store previously computed results, which helps avoid redundant calculations and improves efficiency.
Overflow checks have been taken into consideration. The maximum time to move a box is at most 4 * 1000 (four steps to move the box, each taking 1000 time). With at most 1e4 boxes, the total time is at most 4e7, ensuring the solution is safe.
The time complexity of this solution is dominated by the sorting step, making it O(N logN), where N is the length of the input array usageLimits. The rest of the operations involve simple arithmetic and comparisons, which take linear time. Therefore, the overall time complexity of the function is O(N logN).