Java Code Compile Issue

HackerRank says:

The first line contains two space-separated integers describing the respective values of n (the number of dirty plates) and k (the number of plates Harold has time to wash).

  • 1 <= n <= 20000, 1 <= k <= 20000

However, your code is doing something totally different:

String s =;
int l = s.length();
int n = s.charAt(0);
int k = s.charAt(l - 1);

Since next() only reads one token, s would be the String value of the HackerRank value n.

Let’s say the first line is:

7000 20000

Your code will read 7000 into s, then assign n = '7' and k = '0'.

The character '7' has the ASCII/Unicode numeric value 55, and the character '0' has value 48. So what you really got was n = 55 and k = 48.

What you should have done, is simply this:

int n = in.nextInt();
int k = in.nextInt();


Since the contest is over, you can read the editorial page to see a solution. Java doesn’t have a multiset (ordered list), but you could use a PriorityQueue.

Below is an alternate solution with a lower memory footprint for high values of k.

First, assume you can wash all the plates, so sum all the p values. If k >= n, you’re done. Now, for every plate you can’t wash, subtract p again and also subtract d, e.g. subtract p + d from the total.

The goal then is to wash the plates with the highest p + d value first, so we’ll be subtracting smaller values from the total. To do that, build an array of p + d values, sort it, and wash/remove/skip the k plates with the highest values.

Finally, remember to not return a negative value.

Here it is in compact form:

java.util.Scanner in = new java.util.Scanner(;
int n = in.nextInt(), k = in.nextInt(), pd[] = new int[n];
long total = 0;
for (int i = 0; i < n; i++) {
    int p = in.nextInt(), d = in.nextInt();
    total += p;
    pd[i] = p + d;
for (int i = n - k - 1; i >= 0; i--)
    total -= pd[i];
System.out.println(Math.max(0, total));

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