Which is the most efficient way of taking input in Java?

Question

I am solving this question.

This is my code:

import java.io.IOException;
import java.util.Scanner;


public class Main {
    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        int[] t = new int[n];
        int count = 0;
        for (int i = 0; i < n; i++) {
            t[i] = sc.nextInt();
            if (t[i] % k == 0) {
                count++;
            }
        }
        System.out.println(count);

    }
}

But when I submit it, it get's timed out. Please help me optimize this to as much as is possible.

Example

Input:

Output:

They say :

You are expected to be able to process at least 2.5MB of input data per second at runtime.

Modified CODE

Thank you all...I modified my code and it worked...here it is....

 public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String[] input = br.readLine().split(" ");
        int n = Integer.parseInt(input[0]);
        int k = Integer.parseInt(input[1]);
        int count = 0;
        for (int i = 0; i < n; i++) {
            if (Integer.parseInt(br.readLine()) % k == 0) {
                count++;
            }
        }
        System.out.println(count);
    }

regards

shahensha

Why use `int[] t` if you are not using this array for any useful thing here? You could have use `int t` instead. — limc, Feb 10 '11 at 17:51
This looks like input code for an ACM-ICPC problem, is this the entire "program"? From where is it timing out? an online judge? Can you give us some sample input? — Argote, Feb 10 '11 at 17:53
yes that's the entire program...and they run the code on their servers....they have given us 8 seconds for this particular program....they say " You are expected to be able to process at least 2.5MB of input data per second at runtime." — shahensha, Feb 10 '11 at 17:58
is the input formatted like that? or do you have an unknown amount of whitespace between each int? or is each number o a different line? — Argote, Feb 10 '11 at 18:04
There is a space between n and k in the first line...and thereafter, each number is on a new line.... — shahensha, Feb 10 '11 at 18:08
@Argote - I've formatted the Input and Output sections to make them clearer. — Andrzej Doyle, Feb 10 '11 at 18:17
In this case you can use `BufferedReader` as I mentioned in the answer, get the first line, use `String.split` on the String and parse each element, then read each line and put it into an `Integer.parseInt` and use that in the modulo operation. — Argote, Feb 10 '11 at 18:17

score 1 · Answer 1 · answered Feb 10 '11 at 17:53

1

How about this?

Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int k = sc.nextInt();
int count = 0;
for (int i = 0; i < n; i++) {
    if (sc.nextInt() % k == 0) {
        count++;
    }
}
System.out.println(count);

answered Feb 10 '11 at 17:53

limc

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does this really change input speed? – Piotr Findeisen Feb 10 '11 at 17:56
1

The advantage to this (speed-wise) is that it doesn't instantiate the array (plus it uses less memory), the input speed should be the same. – Argote Feb 10 '11 at 17:59
1

It depends on his `n`. If the value for `n` is huge, then you are creating a huge `int` array in the memory for nothing. – limc Feb 10 '11 at 17:59

score 1 · Answer 2 · answered Feb 10 '11 at 18:14

You may consider reading big chunks of input and then get the numbers from there.

Other change is, you may use Integer.parseInt() instead of Scanner.nextInt() although I don't know the details of each one, somethings tells me Scanner version performs a bit more computation to know if the input is correct. Another alternative is to convert the number yourself ( although Integer.parseInt should be fast enough )

Create a sample input, and measure your code, change a bit here and there and see what the difference is.

Measure, measure!

score 1 · Accepted Answer · answered Feb 10 '11 at 18:14

This could be slightly faster, based on limc's solution, BufferedReader should be faster still though.

import java.io.IOException;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        int count = 0;
        while (true) {
            try {
                if (sc.nextInt() % k == 0) {
                    count++;
                }
            } catch (NoSuchElementException e) {
                break;
            }
        }
        System.out.println(count);

    }
}

score 0 · Answer 4 · answered Feb 10 '11 at 17:52

0

BufferedReader is supposed to be faster than Scanner. You will need to parse everything yourself though and depending on your implementation it could be worse.

answered Feb 10 '11 at 17:52

Argote

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Thanks a lot Argote! I implemented it with BufferedReader and without the array and it worked....Should I post the corrected code for others....??? – shahensha Feb 10 '11 at 18:12

Which is the most efficient way of taking input in Java?

4 Answers4

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