Finding unique pairs in lottery tickets

Finding unique pairs in lottery tickets

I was trying to solve one of the hackerrank practice questions. Could anyone please let me know how can I make this code much effective, in terms of time and space complexity The Super Bowl Lottery is about to commence, and there are several lottery tickets being sold, and each ticket is identified with a ticket ID. In one of the many winning scenarios in the Superbowl lottery, a winning pair of tickets is: Concatenation of the two ticket IDs in the pair, in any order, contains each digit from 0 to 9 at least once. For example, if there are 2 distinct tickets with ticket ID 12930455 and 56789, 129300455,56789 is a winning pair. NOTE: The ticket IDs can be concantenated in any order. Digits in the ticket ID can occur in any order. Your task is to find the number of winning pairs of distinct tickets, such that concatenation of their ticket IDs in any order makes for a winning scenario. Complete the function winningLotteryTicket which takes a string array of ticket IDs as input, and return the number of winning pairs. Input Format The first line contains n denoting the total number of lottery tickets in the Super Bowl. Each of the next n lines contains a string, where string on a ith line denotes the ticket id of the ith ticket. Constraints 1 pretty much everything input 10 Each ticket id consists of digits from 0, 9 Output Format Print the number of pairs in a new line. Sample Input 5 129300455 5559948277 012334556 56789 123456879 Sample Output 5 import java.io.; import java.util.; import java.text.; import java.math.; import java.util.regex.; public class Solution static Long winningLotteryTicketString tickets int count 0; forint i 0 ; i tickets.length-1 ; i forint j i1 ; j tickets.length ;j ifticketsi.equalsticketsj getStatusticketsi,ticketsj count; return Long.valueOfcount; public static boolean getStatusString a, String b String c ab; ifc.length 10 return false; SetCharacter charSet new HashSet; char arr c.toCharArray; forint i 0; i arr.length; i charSet.addarri; if charSet.size 10 return true; else return false; public static void mainString args Scanner in new ScannerSystem.in; int n in.nextInt; String tickets new Stringn; forint tickets_i 0; tickets_i n; tickets_i ticketstickets_i in.next; Long result winningLotteryTickettickets; System.out.printlnresult; in.close;

try-with-resources Since Java 7, you can use try-with-resources for safe and efficient handling of the underlying IO resource. return boolean if condition return true; else return false; This kind of code can be simplified as return condition. Method names Your naming can be better refined to reflect what they are doing. For example, getStatus can be renamed as hasUniqueNumerals, following the standard ishas prefix for methods returning a boolean. winningLotteryTicket can be renamed as countWinningPairs. for-each loop Your loop on c.toCharArray can also be written as: for char current : c.toCharArray charSet.addcurrent; Whats nice You checked if the concatenation of the two inputs will give you 10 or more digits, returning false first if not. You declared charSet as a Set rather than a HashSet and relied on generic type inference.

Listless additions to h.j.k.s answer: no doc comments the name getStatus is not hinting in a useful direction the number of winning pairs may easily exceed Integer.MAX_VALUE there may have been a reason for Long code alternative for getStatus without concatenation, with one more early out: Adds codecharcodes from codeacode to codecharSetcode return size of the resulting codeSetlt;Charactercode static int accumulateCharsSetCharacter charSet, String a for char c: a.toCharArray charSet.addc; return charSet.size; static boolean all10final String a, final String b if a.lengthb.length 10 return false; final SetCharacter charSet new HashSet; return accumulateCharscharSet, a 10 10 accumulateCharscharSet, b; In each iteration of winningLotteryTickets outer loop, ticketsi stays the same for all the iterations of the inner loop, as does its contribution to the set. If you didnt literally concatenate strings, it was apparent that the same sets were created for the 2nd tickets digits time and again - and the sets checked for completeness were the unions of the digitchar sets of tickets i and j: It looks advantageous to create each tickets set once and for all and think hard about what can be done to reduce the number of set unions to evaluate - if ticket i consists of 2 distinct digits, only, theres no need to pair with any ticket consisting of no more than 7 distinct digits. If one ticket contains every digit, it is a winner paired with every other ticket If and when coding that proves not fast enough, note that the digit sets are quite small and reconsider their representation.