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Java: Most Efficient Way to Get Input Integer Array

I'm working on a problem that requires me to store a very large amount of integers into an integer array. The input is formatted so that one line displays the amount of integers and the next displays all of the values meant to be stored. Ex:
3
12 45 67
In the problem there is closer to 100,000 integers to be stored. Currently I am using this method of storing the integers:
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int[] iVau = new int[n];
String[] temp = scanner.nextLine().split(" ");
for(int i = 0; i < n; i++) {
iVau[i] = Integer.parseInt(temp[i]);
}
This works fine, however the problem I am solving has a strict time limit and my current solution is exceeding it. I know that there is a more efficient way to store this input using buffered readers and input streams, but I don't know how to do it, can someone please show me.

The way you are using Scanner makes your program save a String containing the whole numbers at once, in memory. With 100000 numbers in the 2nd line of your input, it is not so efficient, you could read numbers one after the other without keeping the previous one in memory. So, this way, avoiding using Scanner.readLine() should make your program run faster. You will not have to read the whole line one time, and read a 2nd time this String to parse the integers from it: you will do both of these operations only once.
Here is an example. The method testing() does not use any Scanner. The method testing2() is the one you provided. The file tst.txt contains 100000 numbers. The output from this program, on my Mac Mini (Intel Core i5#2.6GHz) is:
duration without reading one line at a time, without using a Scanner instance: 140 ms
duration when reading one line at a time with a Scanner instance: 198 ms
As you can see, not using Scanner makes your program 41% faster (integer part of (198-140)/140*100 equals 41).
package test1;
import java.io.*;
import java.util.*;
public class Test {
// Read and parse an Int from the stream: 2 operations at once
private static int readInt(InputStreamReader ir) throws IOException {
StringBuffer str = new StringBuffer();
int c;
do { c = ir.read(); } while (c < '0' || c > '9');
do {
str.append(Character.toString((char) c));
c = ir.read();
} while (!(c < '0' || c > '9'));
return Integer.parseInt(str.toString());
}
// Parsing the input step by step
private static void testing(File f) throws IOException {
InputStreamReader ir = new InputStreamReader(new BufferedInputStream(new FileInputStream(f)));
int n = readInt(ir);
int [] iVau = new int[n];
for (int i = 0; i < n; i++) iVau[i] = readInt(ir);
ir.close();
}
// Your code
private static void testing2(File f) throws IOException {
Scanner scanner = new Scanner(f);
int n = scanner.nextInt();
int[] iVau = new int[n];
scanner.nextLine();
String[] temp = scanner.nextLine().split(" ");
for(int i = 0; i < n; i++)
iVau[i] = Integer.parseInt(temp[i]);
scanner.close();
}
// Compare durations
public static void main(String[] args) throws IOException {
File f = new File("/tmp/tst.txt");
// My proposal
long t = System.currentTimeMillis();
testing(f);
System.out.println("duration without reading one line at a time, without using a Scanner instance: " + (System.currentTimeMillis() - t) + " ms");
// Your code
t = System.currentTimeMillis();
testing2(f);
System.out.println("duration when reading one line at a time with a Scanner instance: " + (System.currentTimeMillis() - t) + " ms");
}
}
NOTE: creating the input file is done this way, with bash or zsh:
echo 100000 > /tmp/tst.txt
for i in {1..100000}
do
echo -n $i" " >> /tmp/tst.txt
done

I believe this is what you're looking for. A BufferedReader can only read a line at a time, so it is necessary to split the line and cast Strings to ints.
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
try {
int n = Integer.parseInt(br.readLine());
int[] arr = new int[n];
String[] line = br.readLine().split(" ");
for (int i = 0; i < n; i++) {
arr[i] = Integer.parseInt(line[i]);
}
} catch (IOException e) {
e.getStackTrace();
}

Just a thought, String.split returns an array of Strings. You say the input can be around 100,000 values. So in order to split the array in this way, String.split must be iterating through each element. Now in parsing the new array of strings to Integers you have iterated through the collection twice. You could do this in one iteration with a few small tweaks.
Scanner scanner = new Scanner(System.in);
String tmp = scanner.nextLine();
scanner = new Scanner(tmp);
for(int i = 0; scanner.hasNextInt(); i++) {
arr[i] = scanner.nextInt();
}
The reason for linking the scanner to a String instead of leaving it on System.in is so that it ends properly. It doesn't open System.in for user input on the last token. I believe in big O notation this is the difference between O(n) and O(2n) where the original snippet is O(2n)

I am not quite sure why OP has to use Integer.parseInt(s) here since Scanner can just do the parsing directly by new Scanner(File source).
Here is a demo/test for this idea:
public class NextInt {
public static void main(String... args) {
prepareInputFile(1000, 500); // create 1_000 arrays which each contains 500 numbers;
Timer.timer(() -> readFromFile(), 20, "NextInt"); // read from the file 20 times using Scanner.nextInt();
Timer.timer(() -> readTest(), 20, "Split"); // read from the file 20 times using split() and Integer.parseInt();
}
private static void readTest() {
Path inputPath = Paths.get(Paths.get("").toAbsolutePath().toString().concat("/src/main/java/io/input.txt"));
try (Scanner scanner = new Scanner(new File(inputPath.toString()))) {
int n = Integer.valueOf(scanner.nextLine());
int[] iVau = new int[n];
String[] temp = scanner.nextLine().split(" ");
for (int i = 0; i < n; i++) {
iVau[i] = Integer.parseInt(temp[i]);
}
} catch (IOException ignored) {
ignored.printStackTrace();
}
}
private static void readFromFile() {
Path inputPath = Paths.get(Paths.get("").toAbsolutePath().toString().concat("/src/main/java/io/input.txt"));
try (Scanner scanner = new Scanner(new File(inputPath.toString()))) {
while (scanner.hasNextInt()) {
int arrSize = scanner.nextInt();
int[] arr = new int[arrSize];
for (int i = 0; i < arrSize; ++i) {
arr[i] = scanner.nextInt();
}
// System.out.println(Arrays.toString(arr));
}
} catch (IOException ignored) {
ignored.printStackTrace();
}
}
private static void prepareInputFile(int arrCount, int arrSize) {
Path outputPath = Paths.get(Paths.get("").toAbsolutePath().toString().concat("/src/main/java/io/input.txt"));
List<String> lines = new ArrayList<>();
for (int i = 0; i < arrCount; ++i) {
int[] arr = new int[arrSize];
for (int j = 0; j < arrSize; ++j) {
arr[j] = new Random().nextInt();
}
lines.add(String.valueOf(arrSize));
lines.add(Arrays.stream(arr).mapToObj(String::valueOf).collect(Collectors.joining(" ")));
}
try {
Files.write(outputPath, lines);
} catch (IOException ignored) {
ignored.printStackTrace();
}
}
}
Locally tested it with 1_000 arrays while each array has 500 numbers, reading all the elements cost about: 340ms using Scanner.nextInt() while OP's method about 1.5ms.
NextInt: LongSummaryStatistics{count=20, sum=6793762162, min=315793916, average=339688108.100000, max=618922475}
Split: LongSummaryStatistics{count=20, sum=26073528, min=740860, average=1303676.400000, max=5724370}
So I really have doubt the issue lies in the input reading.

Since in your case you are aware of the total count of elements all that you have to do is to read X integers from the second line. Here is an example:
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int count = in.nextInt();
int array[] = new int[count];
for (int i = 0; i < count; i++) {
array[i] = in.nextInt();
}
}
If this is not fast enough, which I doubt, then you could switch to the use of a BufferedReader as follows:
public static void main(String[] args) throws IOException {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
int count = Integer.parseInt(in.readLine());
int array[] = new int[count];
for (int i = 0; i < count; i++) {
int nextInteger = 0;
int nextChar = in.read();
do {
nextInteger = nextInteger * 10 + (nextChar - '0');
nextChar = in.read();
} while (nextChar != -1 && nextChar != (int)' ');
array[i] = nextInteger;
}
}
In your case the input will be aways valid so this means that each of the integers will be separated by a single whitespace and the input will end up with EoF character.
If both are still slow enough for you then you could keep looking for more articles about Reading Integers in Java, Competative programming like this one: https://www.geeksforgeeks.org/fast-io-in-java-in-competitive-programming/
Still my favorite language when it comes to competitions will always be C :) Good luck and enjoy!

Java - NumberFormatException when using .parseInt(String)

I am trying to run a loop to see if an int is sorted. however the int has to be converted from a string. here is my code.
public static void main(String[] args) {
// TODO code application logic here
Scanner maxVal = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
System.out.println("enter the max value of ordered squares:");
int max = maxVal.nextInt();
for(int i = 0; i*i <= max; i++){
int L = String.valueOf(i*i).length();
String sq = String.valueOf(i*i);
String [] digits = new String[L];
for(int a = 0; a < L; a++){
digits [a] = Character.toString(sq.charAt(a));
if(L == 1){
System.out.print(sq + "");
}else if(Integer.parseInt(digits [a]) < Integer.parseInt(digits[a+1])){
System.out.print(sq);
}else{
}
}
}
}
when I run it, I get an error :
Exception in thread "main" java.lang.NumberFormatException: null
0149 at java.lang.Integer.parseInt(Integer.java:542)
at java.lang.Integer.parseInt(Integer.java:615)
why does Integer.parseInt() not work

Your problem is that digits[a+1] hasn't been defined yet. I see that on line 2 you have
digits[a] = Character.toString(sq.charAt(a));
and you're iterating over a in a for loop, so I daresay that digits[a+1] hasn't been assigned yet.
UPDATE 1
Check out this solution, it shows how to properly catch that exception and how to avoid it:
Java: Good way to encapsulate Integer.parseInt()
UPDATE 2
I decided to add a fixed version of your code:
public static void main(String[] args) {
// TODO code application logic here
Scanner maxVal = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
System.out.println("enter the max value of ordered squares:");
int max = maxVal.nextInt();
for(int i = 0; i*i <= max; i++){
int L = String.valueOf(i*i).length();
String sq = String.valueOf(i*i);
String [] digits = new String[L];
for(int a = 0; a < L; a++){
digits [a] = Character.toString(sq.charAt(a));
if(L == 1 || a == 0){
System.out.print(sq + "");
}else if(Integer.parseInt(digits [a]) < Integer.parseInt(digits[a+1])){
System.out.print(sq);
}else{
}
}
}
}
While I don't know the utility of your code, but this implementation might be simpler:
public static void main(String[] args) {
// TODO code application logic here
Scanner maxVal = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
System.out.println("enter the max value of ordered squares:");
int max = maxVal.nextInt();
for(int i = 0; i*i <= max; i++){
long sq = i*i;
if(sq > 9){
String[] digits = sq.toString().split("");
//Notice that I start at index 1, so I can do [a-1] safely
for(int a = 1; a < digits.length; a++){
if(Integer.parseInt(digits [a-1]) < Integer.parseInt(digits[a])){
System.out.print(sq);
//I guess we don't want a number like 169 (13*13) to be displayed twice, so:
break;
}
}
} else {
System.out.print(sq);
}
}
}

Java words reverse

I am new to Java and I found a interesting problem which I wanted to solve. I am trying to code a program that reverses the position of each word of a string. For example, the input string = "HERE AM I", the output string will be "I AM HERE". I have got into it, but it's not working out for me. Could anyone kindly point out the error, and how to fix it, because I am really curious to know what's going wrong. Thanks!
import java.util.Scanner;
public class Count{
static Scanner sc = new Scanner(System.in);
static String in = ""; static String ar[];
void accept(){
System.out.println("Enter the string: ");
in = sc.nextLine();
}
void intArray(int words){
ar = new String[words];
}
static int Words(String in){
in = in.trim(); //Rm space
int wc = 1;
char c;
for (int i = 0; i<in.length()-1;i++){
if (in.charAt(i)==' '&&in.charAt(i+1)!=' ') wc++;
}
return wc;
}
void generate(){
char c; String w = ""; int n = 0;
for (int i = 0; i<in.length(); i++){
c = in.charAt(i);
if (c!=' '){
w += c;
}
else {
ar[n] = w; n++;
}
}
}
void printOut(){
String finale = "";
for (int i = ar.length-1; i>=0;i--){
finale = finale + (ar[i]);
}
System.out.println("Reversed words: " + finale);
}
public static void main(String[] args){
Count a = new Count();
a.accept();
int words = Words(in);
a.intArray(words);
a.generate();
a.printOut();
}
}

Got it. Here is my code that implements split and reverse from scratch.
The split function is implemented through iterating through the string, and keeping track of start and end indexes. Once one of the indexes in the string is equivalent to a " ", the program sets the end index to the element behind the space, and adds the previous substring to an ArrayList, then creating a new start index to begin with.
Reverse is very straightforward - you simply iterate from the end of the string to the first element of the string.
Example:
Input: df gf sd
Output: sd gf df
import java.util.Scanner;
import java.util.ArrayList;
import java.util.Collections;
public class Count{
public static void main(String[] args)
{
Scanner scan = new Scanner(System.in);
System.out.println("Enter string to reverse: ");
String unreversed = scan.nextLine();
System.out.println("Reversed String: " + reverse(unreversed));
}
public static String reverse(String unreversed)
{
ArrayList<String> parts = new ArrayList<String>();
String reversed = "";
int start = 0;
int end = 0;
for (int i = 0; i < unreversed.length(); i++)
{
if (unreversed.charAt(i) == ' ')
{
end = i;
parts.add(unreversed.substring(start, end));
start = i + 1;
}
}
parts.add(unreversed.substring(start, unreversed.length()));
for (int i = parts.size()-1; i >= 0; i--)
{
reversed += parts.get(i);
reversed += " ";
}
return reversed;
}
}

There is my suggestion :
String s = " HERE AM I ";
s = s.trim();
int j = s.length() - 1;
int index = 0;
StringBuilder builder = new StringBuilder();
for (int i = j; i >= 0; i--) {
Character c = s.charAt(i);
if (c.isWhitespace(c)) {
index = i;
String r = s.substring(index+1, j+1);
j = index - 1;
builder.append(r);
builder.append(" ");
}
}
String r=s.substring(0, index);
builder.append(r);
System.out.println(builder.toString());

From adding debug output between each method call it's easy to determine that you're successfully reading the input, counting the words, and initializing the array. That means that the problem is in generate().
Problem 1 in generate() (why "HERE" is duplicated in the output): after you add w to your array (when the word is complete) you don't reset w to "", meaning every word has the previous word(s) prepended to it. This is easily seen by adding debug output (or using a debugger) to print the state of ar and w each iteration of the loop.
Problem 2 in generate() (why "I" isn't in the output): there isn't a trailing space in the string, so the condition that adds a word to the array is never met for the last word before the loop terminates at the end of the string. The easy fix is to just add ar[n] = w; after the end of the loop to cover the last word.

I would use the split function and then print from the end of the list to the front.
String[] splitString = str.split(" ");
for(int i = splitString.length() - 1; i >= 0; i--){
System.out.print(splitString[i]);
if(i != 0) System.out.print(' ');
}
Oops read your comment. Disregard this if it is not what you want.

This has a function that does the same as split, but not the predefined split function
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
System.out.println("Enter the string : ");
String input = sc.nextLine();
// This splits the string into array of words separated with " "
String arr[] = myOwnSplit(input.trim(), ' '); // ["I", "AM", "HERE"]
// This ll contain the reverse string
String rev = "";
// Reading the array from the back
for(int i = (arr.length - 1) ; i >= 0 ; i --) {
// putting the words into the reverse string with a space to it's end
rev += (arr[i] + " ");
}
// Getting rid of the last extra space
rev.trim();
System.out.println("The reverse of the given string is : " + rev);
}
// The is my own version of the split function
public static String[] myOwnSplit(String str, char regex) {
char[] arr = str.toCharArray();
ArrayList<String> spltedArrayList = new ArrayList<String>();
String word = "";
// splitting the string based on the regex and bulding an arraylist
for(int i = 0 ; i < arr.length ; i ++) {
char c = arr[i];
if(c == regex) {
spltedArrayList.add(word);
word = "";
} else {
word += c;
}
if(i == (arr.length - 1)) {
spltedArrayList.add(word);
}
}
String[] splitedArray = new String[spltedArrayList.size()];
// Converting the arraylist to string array
for(int i = 0 ; i < spltedArrayList.size() ; i++) {
splitedArray[i] = spltedArrayList.get(i);
}
return splitedArray;
}

Compression algorithm in java

My goal is to write a program that compresses a string, for example:
input: hellooopppppp!
output:he2l3o6p!
Here is the code I have so far, but there are errors.
When I have the input: hellooo
my code outputs: hel2l3o
instead of: he213o
the 2 is being printed in the wrong spot, but I cannot figure out how to fix this.
Also, with an input of: hello
my code outputs: hel2l
instead of: he2lo
It skips the last letter in this case all together, and the 2 is also in the wrong place, an error from my first example.
Any help is much appreciated. Thanks so much!
public class compressionTime
{
public static void main(String [] args)
{
System.out.println ("Enter a string");
//read in user input
String userString = IO.readString();
//store length of string
int length = userString.length();
System.out.println(length);
int count;
String result = "";
for (int i=1; i<=length; i++)
{
char a = userString.charAt(i-1);
count = 1;
if (i-2 >= 0)
{
while (i<=length && userString.charAt(i-1) == userString.charAt(i-2))
{
count++;
i++;
}
System.out.print(count);
}
if (count==1)
result = result.concat(Character.toString(a));
else
result = result.concat(Integer.toString(count).concat(Character.toString(a)));
}
IO.outputStringAnswer(result);
}
}

I would
count from 0 as that is how indexes work in Java. Your code will be simpler.
would compare the current char to the next one. This will avoid printing the first character.
wouldn't compress ll as 2l as it is no smaller. Only sequences of at least 3 will help.
try to detect if a number 3 to 9 has been used and at least print an error.
use the debugger to step through the code to understand what it is doing and why it doesn't do what you think it should.

I am doing it this way. Very simple:
public static void compressString (String string) {
StringBuffer stringBuffer = new StringBuffer();
for (int i = 0; i < string.length(); i++) {
int count = 1;
while (i + 1 < string.length()
&& string.charAt(i) == string.charAt(i + 1)) {
count++;
i++;
}
if (count > 1) {
stringBuffer.append(count);
}
stringBuffer.append(string.charAt(i));
}
System.out.println("Compressed string: " + stringBuffer);
}

You can accomplish this using a nested for loops and do something simial to:
count = 0;
String results = "";
for(int i=0;i<userString.length();){
char begin = userString.charAt(i);
//System.out.println("begin is: "+begin);
for(int j=i+1; j<userString.length();j++){
char next = userString.charAt(j);
//System.out.println("next is: "+next);
if(begin == next){
count++;
}
else{
System.out.println("Breaking");
break;
}
}
i+= count+1;
if(count>0){
String add = begin + "";
int tempcount = count +1;
results+= tempcount + add;
}
else{
results+= begin;
}
count=0;
}
System.out.println(results);
I tested this output with Hello and the result was He2lo
also tested with hellooopppppp result he2l3o6p

If you don't understand how this works, you should learn regular expressions.
public String rleEncodeString(String in) {
StringBuilder out = new StringBuilder();
Pattern p = Pattern.compile("((\\w)\\2*)");
Matcher m = p.matcher(in);
while(m.find()) {
if(m.group(1).length() > 1) {
out.append(m.group(1).length());
}
out.append(m.group(2));
}
return out.toString();
}

Try something like this:
public static void main(String[] args) {
System.out.println("Enter a string:");
Scanner IO = new Scanner(System.in);
// read in user input
String userString = IO.nextLine() + "-";
int length = userString.length();
int count = 0;
String result = "";
char new_char;
for (int i = 0; i < length; i++) {
new_char = userString.charAt(i);
count++;
if (new_char != userString.charAt(i + 1)) {
if (count != 1) {
result = result.concat(Integer.toString(count + 1));
}
result = result.concat(Character.toString(new_char));
count = 0;
}
if (userString.charAt(i + 1) == '-')
break;
}
System.out.println(result);
}

The problem is that your code checks if the previous letter, not the next, is the same as the current.
Your for loops basically goes through each letter in the string, and if it is the same as the previous letter, it figures out how many of that letter there is and puts that number into the result string. However, for a word like "hello", it will check 'e' and 'l' (and notice that they are preceded by 'h' and 'e', receptively) and think that there is no repeat. It will then get to the next 'l', and then see that it is the same as the previous letter. It will put '2' in the result, but too late, resulting in "hel2l" instead of "he2lo".
To clean up and fix your code, I recommend the following to replace your for loop:
int count = 1;
String result = "";
for(int i=0;i<length;i++) {
if(i < userString.length()-1 && userString.charAt(i) == userString.charAt(i+1))
count++;
else {
if(count == 1)
result += userString.charAt(i);
else {
result = result + count + userString.charAt(i);
count = 1;
}
}
}
Comment if you need me to explain some of the changes. Some are necessary, others optional.

Here is the solution for the problem with better time complexity:
public static void compressString (String string) {
LinkedHashSet<String> charMap = new LinkedHashSet<String>();
HashMap<String, Integer> countMap = new HashMap<String, Integer>();
int count;
String key;
for (int i = 0; i < string.length(); i++) {
key = new String(string.charAt(i) + "");
charMap.add(key);
if(countMap.containsKey(key)) {
count = countMap.get(key);
countMap.put(key, count + 1);
}
else {
countMap.put(key, 1);
}
}
Iterator<String> iterator = charMap.iterator();
String resultStr = "";
while (iterator.hasNext()) {
key = iterator.next();
count = countMap.get(key);
if(count > 1) {
resultStr = resultStr + count + key;
}
else{
resultStr = resultStr + key;
}
}
System.out.println(resultStr);
}

Java String Bubble Sorting

I need help sorting this array in alphabetical order using the bubble sort algorithm.
My code is:
public class Strings
{
public static void main(String[] args)
{
Scanner reader = new Scanner(System.in);
String tempStr;
System.out.print("Enter the strings > ");
String s1 = new String(reader.nextLine());
String[] t1 = s1.split(", ");
for (int t=0; t<t1.length-1; t++)
{
for (int i = 0; i<t1.length -1; i++)
{
if(t1[i+1].compareTo(t1[1+1])>0)
{
tempStr = t1[i];
t1[i] = t1[i+1];
t1[i+1] = tempStr;
}
}
}
for(int i=0;i<t1.length;i++)
{
System.out.println(t1[i]);
}
}
}
The code compiles, but it does not sort alphabetical. Please help me.

You have three errors in your code.
The first error is in the inner for loop, in the place where you do the check statement, it should be i < t1.length - t -1 not i < t1.length -1. You subtract t because you do not want to loop through the whole array again, only the first part of it.
The second and third errors are in the if statement. You need to turn the greater than symbol into a lesser than symbol, because the way you have the compareTo method set up, it will return a negative number.
The other error in this line is that in the compareTo parameter you put 1 + 1 it actually should be just i, because you want one less than the object it is comparing to.
The fixed working code is below (Comments are what you originally had):
public static void main(String[] args) {
Scanner reader = new Scanner(System.in);
String tempStr;
System.out.print("Enter the strings > ");
String s1 = new String(reader.nextLine());
String[] t1 = s1.split(", ");
for (int t = 0; t < t1.length - 1; t++) {
for (int i= 0; i < t1.length - t -1; i++) {
if(t1[i+1].compareTo(t1[i])<0) {
tempStr = t1[i];
t1[i] = t1[i + 1];
t1[i + 1] = tempStr;
}
}
}
for (int i = 0; i < t1.length; i++) {
System.out.println(t1[i]);
}
}

please change
String[] t1 = s1.split(", ");
to
String[] t1 = s1.split("");
This will solve the issue.