In this problem, we have a list of entries where each cow produced some amount of milk, and we need to use the entries to determine which cow produced the second-smallest amount of milk.

There are a couple parts to the problem. The first part is to read in the list of entries and determine the total amount of milk each cow produced. There are a few ways to maintain this information. In the provided model solution, we keep an array of the names of all the cows and maintain a parallel array with the amount of milk that the cow has produced.

After we have computed how much milk each cow has produced, we need to compute the second smallest amount of milk that any cow has produced. We can start by computing the minimum amount of milk that any cow has produced. The second smallest amount of milk that any cow has produced is therefore the next smallest amount of milk present.

After we have computed that quantity, we need to determine which cows have produced that amount of milk. We can loop over all the cows - if a cow produces that amount, then if we haven't found any other cows, then that cow is considered to be the only cow that has produced that amount of milk. Otherwise, there is a tie.

import java.io.*; import java.util.*; public class notlast { static String[] cows; public static void main(String[] args) throws IOException { // initialize file I/O BufferedReader br = new BufferedReader(new FileReader("notlast.in")); PrintWriter pw = new PrintWriter(new BufferedWriter(new FileWriter("notlast.out"))); // initialize an array with all the cow names cows = new String[]{ "Bessie", "Elsie", "Daisy", "Gertie", "Annabelle", "Maggie", "Henrietta"}; // allocate an array to count how much milk each cow produces int[] amount = new int[cows.length]; // read in the number of log entries int n = Integer.parseInt(br.readLine()); // loop over the entries for(int i = 0; i < n; i++) { StringTokenizer st = new StringTokenizer(br.readLine()); // read the name of the cow and the amount of milk String name = st.nextToken(); int milk = Integer.parseInt(st.nextToken()); // update the amount of milk for the given cow amount[findCowIndex(name)] += milk; } // compute the minimum amount of milk produced int minimumAmount = 1000000; for(int i = 0; i < amount.length; i++) { if(amount[i] < minimumAmount) { minimumAmount = amount[i]; } } // compute the second smallest amount of milk produced int secondSmallestAmount = 1000000; for(int i = 0; i < amount.length; i++) { // the second smallest amount of milk must be larger than the smallest amount of milk if(amount[i] > minimumAmount && amount[i] < secondSmallestAmount) { secondSmallestAmount = amount[i]; } } // determine the cow that milked the second-smallest amount of milk final int NOT_FOUND = -1; final int MORE_THAN_ONE = -2; int indexOfSecondSmallest = NOT_FOUND; for(int i = 0; i < amount.length; i++) { if(amount[i] == secondSmallestAmount) { if(indexOfSecondSmallest == NOT_FOUND) { // we haven't found any cow yet, so that cow has milked the correct amount indexOfSecondSmallest = i; } else { // we have found at least two cows, so there is a tie indexOfSecondSmallest = MORE_THAN_ONE; } } } // print the answer if(indexOfSecondSmallest >= 0) { pw.println(cows[indexOfSecondSmallest]); } else { pw.println("Tie"); } // close the file pw.close(); } public static int findCowIndex(String s) { // This function takes in the name of a cow and returns the index where // that name can be found in the array of names. for(int i = 0; i < cows.length; i++) { if(cows[i].equals(s)) { return i; } } return -1; } }