Can someone explain to me how this code works?
It lets the user input numbers up until 1000, then it prints the original inputted numbers, the even and the odd, all in a separate array. But I just don't understand the parts where there is gem++ and gem1++ when it outputs the even and odd not the number of the even and odd numbers.
And after putting this
double even[] = new double[gem];
double odd[] = new double [gem1];
why does it need to repeat gem=0 and gem1=0 again? I'm so sorry if I ask too many question, I'm just confused, I just learned java last week.
public class wutt {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
System.out.print("Enter no. of elements you want in array : ");
int n = s.nextInt();
if (1 <= n && n <= 1000) {
double a[] = new double[n];
int gem = 0, gem1 = 0;
System.out.println("Enter all the elements : ");
for (int i = 0; i < n; i++) {
a[i] = s.nextInt();
if (a[i] % 2 == 0)
gem++;
else
gem1++;
}
double even[] = new double[gem];
double odd[] = new double[gem1];
gem = 0;
gem1 = 0;
for (int i = 0; i < a.length; i++) {
if (a[i] % 2 == 0) {
even[gem] = a[i];
gem++;
} else {
odd[gem1] = a[i];
gem1++;
}
}
System.out.println("Original: " + Arrays.toString(a));
System.out.println("Odd: " + Arrays.toString(odd));
System.out.println("Even: " + Arrays.toString(even));
} else
System.out.print("Invalid input");
}
}
If you want the program stops after the user enters a number greater than 1000 o less than 0 you need to add the break statement in your if condition.
if (size < 0 || size > 1000) {
System.out.println("Size must be between 0 and 1000");
break;
}
the code before double even[] = new double[gem];
double odd[] = new double [gem1]; is trying to get the number of odds occurred and the number of even occurred and put all inputted elements in array a.
after all that ,now what we got is a array of numbers called a containing all the inputted elements. and two numbers called gem and gem1, which contains the number of odds occurred and the number of even occurred.
so
we get gem(numberOfEvens), gem1(numberOfOdds) and list a
next, we need to put all odds from a to a new array called odd[] with size gem1, and
put all evens from a to a new array called even[] with size gem. at this point, the duty of variable gem1 and gem is done. they become useless.
now we need to go through the list and pick the odd and even out and put them in the array one by one in a sequential way. that's why we need two new variables with 0 initialized.
in this case, because gem and gem1 are useless aready, they are reassigned to help manipulate the tree arrays a , odd[] and even[]
So the user inputs the number of elements he/she wants in the array (n)
double a[] = new double[n]; // a is the array that is initialised to accommodate n elements
int gem = 0, gem1 = 0; // gem is the counter for "even" numbers and "gem1" the counter for odd numbers, and like every good counter, they start at 0
System.out.println("Enter all the elements : ");
for (int i = 0; i < n; i++) { // so we ask the user to input n elements
a[i] = s.nextInt(); // here we read every input and put it in the a array
if (a[i] % 2 == 0) // if the new number is even
gem++; // we increase the even counter "gem"
else // otherwise, when it is an odd number
gem1++; // we increase the odd counter
}
double even[] = new double[gem]; // now we create a new array where we want to hold all the even numbers, we do that by telling it how many even numbers we have counted before (gem)
double odd[] = new double[gem1]; // and a new array for all odd numbers (gem1 was our counter)
gem = 0; // now we reinitialise the counters, because we want to start from the beginning
gem1 = 0;
for (int i = 0; i < a.length; i++) { // in order to copy all numbers from the a array into the two other arrays for even and odd numbers, we iterate over the whole length of the a array. i is the index for the "a" array
if (a[i] % 2 == 0) { // ever even number we encounter
even[gem] = a[i]; // we put in the even array
gem++; // while gem, the "even numbers counter" is our index for the "even" array
} else {
odd[gem1] = a[i]; // odd numbers are for the odd array
gem1++; // while the former "odd numbers counter" now serves as our "odd" array index
}
}
and that's pretty much it. First the user inputs all numbers in a single array and simply counts how many odd and how many even numbers where inputted,
then two new arrays are created, one for the even and one for the odd numbers and since we counted them, we know how big these two new arrays have to be.
And finally all numbers are again iterated over and put in their according array.
At the end you have 3 array, one that holds all numbers, one that holds the even numbers and one with only the odd numbers.
EDIT
here are a few minor changes you could make without changing the nature of that method:
double allNumbers[] = new double[n]; // "allNumbers" is way more specific than "a"
int oddCounter = 0; // "oddCounter" instead of "gem"
int evenCounter = 0; // numbers in variables like "gem1" is really bad practice, because numbers don't say anything about the nature of the variable
System.out.println("Enter all the elements : ");
for (int i = 0; i < n; i++) {
allNumbers[i] = s.nextInt();
if (allNumbers[i] % 2 == 0) {
evenCounter++;
} else {
oddCounter++;
}
}
// until here nothing changes but the names
double even[] = new double[evenCounter];
double odd[] = new double[oddCounter];
int oddIndex = 0; // and here we create new variables, instead of reusing old ones
int evenIndex = 0; // there is absolutely no performance gain in reusing primitives like this - it's just confusing
for (int i = 0; i < allNumbers.length; i++) {
if (allNumbers[i] % 2 == 0) {
even[evenIndex++] = allNumbers[i]; // the "++" can be done directly in the first expression. that's just to make it shorter.
} else {
odd[oddIndex++] = allNumbers[i]; // it is not more performant nor easier to read - just shorter
}
}
EDIT (again)
This is how the arrays look like, say when you enter 4 numbers:
gem = 0
gem1 = 0
n = 4 // user said 4
a = [ , , , ] // array a is empty but holds the space for 4 numbers
a = [1, , , ] // user enters 1
^
i=0
gem1 = 1 // 1 is an odd number -> gem1++
a = [1,4, , ] // user entered "4"
^
i=1
gem = 1 // 4 is an even number -> gem++
a = [1,4,2, ] // user entered "2"
^
i=2
gem = 2 // 24 is an even number -> gem++
a = [1,4,2,7] // user entered "7"
^
i=3
gem1 = 2 // 7 is an odd number -> gem1++
then we fill the other arrays
even = [ , ] // gem is 2, so we have 2 even numbers
odd = [ , ] // gem1 is 2, so we have 2 odd numbers
a = [1,4,2,7]
^
i=0
odd[1, ] // for i=0, a[i] is 1, which is an odd number
a = [1,4,2,7]
^
i=1
even = [4, ] // for i=1, a[i] is 4, which is an even number
a = [1,4,2,7]
^
i=2
even = [4,2] // for i=2, a[i] is 2, which is an even number
a = [1,4,2,7]
^
i=3
odd = [1,7] // for i=3, a[i] is 7, which is an odd number
and in the end you have
a = [1,4,2,7]
even = [4,2]
odd = [1,7]
Related
The first line of the input contains a single integer T denoting the number of test cases. The description of T test cases follows.
The first line of each test case contains a single integer N.
The second line contains N space-separated integers A1,A2,…,AN.
Output
For each test case, print a single line containing one integer ― the desired number of pairs.
Example Input
2
3
2 4 2
3
0 2 3
Example Output
1
0
My solution looks like:
class Codechef {
public static void main (String[] args) throws java.lang.Exception {
ArrayList<Integer> resultList = new ArrayList<Integer>();
Scanner scanner = new Scanner(System.in);
int T;
T = scanner.nextInt();
for(int i = 0; i < T; i++) {
int N;
N = scanner.nextInt();
int A[] = new int[N];
for(int j = 0; j < N; j++) {
A[j] = scanner.nextInt();
}
quick_sort(A, 0, N-1);
int pos = 0, pairs = 0;
while(pos < A.length - 1) {
if(A[pos] == A[pos + 1]) {
pos += 2;
pairs += 1;
} else {
++pos;
}
}
resultList.add(pairs);
}
for(int pairCount : resultList) {
System.out.println(pairCount);
}
scanner.close();
}
}
It successfully runs the example test cases but fails on submission, My question is, if the input is something like 1 1 2 2 1, then what should be the answer, 3? as there are 2 pairs of 1, and 1 of 2's.
Also, what will be the suggested data structure to be used for this purpose, Java with primitive data types is taking too much longer to execute it with 40,000 input values. What's wrong with my solution
To answer your first question, I'd say yes that each pair of 1's would count separately so you'd get 3.
I think your code is failing since you're only counting touching pairs after you sort.
For example,
1 1 1, you find the first pair at index 0/1, but then advance pos += 2.This means you're missing the two other pairs of 1's.
Your solution seems to be O(nlogn) because of sorting but I can think of a O(n) solution.
int[] backing = new int [10];
for (int j = 0; j < N; j++) {
int x = scanner.nextInt();
backing[x]++;
}
//At this point, you have a backing array with the frequency of each integer
You'll want something similar to this to calculate the number of pairs. It's the frequency of each integer choose 2, since you want to choose each occurrence of a pair.
So for example if you know you have 5 1's, then you'll compute:
5!/(2!*3!) = 10
I want to write a code that asks for three numbers dig(1), dig(2), dig(3) and displays a sequence of numbers dig(1), dig(2), dig(3), ..., dig(n) according to this rule:
a = dig(2) - dig(1)
b = dig(3) - dig(2)
dig(n) = dig(n-1) + a if n is odd
dig(n) = dig(n-1) + b if n is even
Example:
7, 8, 5, 6, 3, 4, 1, 2, -1, 0
It asks an user for three integers dig1, dig2, dig3
It asks a number N ≥ 3 which will be the whole sequence count.
It prints a sequence beginning with
Then prints the sequence, beginning with the three leading integers, followed by N-3 other terms that follow the pattern defined by the first three integers. See examples below for more information.
(The sequence begins with n = 1, but of course the array starts at 0.)
int dig1 = 0;
int dig2 = 0;
int dig3 = 0;
int a;
int b;
int n = 0;
int i;
dig1 = scnr.nextInt();
dig2 = scnr.nextInt();
dig3 = scnr.nextInt();
n = scnr.nextInt();
int[] array = new int[n];
array[0] = dig1;
array[1] = dig2;
array[2] = dig3;
a = dig2 - dig1;
b = dig3 - dig2;
for (i = 3; i < n; i++){
if(i%2 == 0){
array[i] = b + array[i-1];
}
else{
array[i] = a + array[i-1];
}
}
System.out.println(array[i]);
}
}
whenever I try to print this out, I get this error:
java.lang.ArrayIndexOutOfBoundsException
Another example: if I put in the numbers: 0 1 0 9 into my input, I should receive back the sequence 0 1 0 1 0 1 0 1 0
Printing array[n-1] gives me back just the final output. I'm trying to iterate through each number.
Sorry if that's unclear, but any help would be great, thank you.
Your System.out.println(array[i]); seems to be out of the for loop. Then i will be equal to n. And there is no element with index n in array with length n. The elements are from 0 to n-1.
At first I thought that you were choosing an array size less than 3, and that the code was failing here, during the initialization of the array:
array[0] = dig1;
array[1] = dig2;
array[2] = dig3;
But now, I actually think that the last line of code is the problem. Have a closer look at your for loop:
for (i=3; i < n; i++) {
if (i % 2 == 0) {
array[i] = b + array[i-1];
}
else {
array[i] = a + array[i-1];
}
}
System.out.println(array[i]);
Appreciate that at the end of the for loop, i, the loop counter, will always be n, the size of the array. Accessing array[n] will always be out of bounds, because an array in Java has a highest addressable index of one less than the actual number of buckets. So, if you were intending to just print the final element, then use this:
System.out.println(array[i-1]);
or
System.out.println(array[n-1]);
at the line n=scrn.nextInt(), you are assigning to n a integer from scrn but you don't know the value of that integer.
Then with this:
int[] array = new int[n];
Your array is of size of n elements starting from index 0 to index n-1.
So let's suppose the value you affected to n with .nextInt() is inferior to 3: initial value of i, the loop will even not be reached, because you will be out of bounds just here:
array[0] = dig1;
array[1] = dig2;
array[2] = dig3;
since you will be attempting to affect a value on an array at an index that does not exists (n is inferior to three but you are trying to affect three values in an array of size n).
Then if n is superior to 3, since your printing code is out from the loop, the value of i is equal to n and there is no value at index n. remember, the loop will leave only if the condition i<n is false and to reach that condition i must be equal or superior and that will lead to an IndexOutOfBoundsException on printing.
I think you printing line code should be inside the for loop.
I have created a program previously using the BubbleSort method that works to sort numbers in a list that already exists, however, I am having difficulty with trying to manipulate this program in order to allow a user to input the list of numbers to be sorted instead. So far I have:
import java.util.Scanner;
public class MedianValue {
public static void main(String[] args) {
//use scanner to input list of numbers to sort
Scanner scan = new Scanner(System.in);
int[] numbers = new int[] {scan.nextInt()};
//nested for loop
//outer loop just iterating
//inner loop going through and flipping
//checking if out of order (if statement)
int counter = 0;
//outer loop: keep doing this until it's sorted
for(int i = 0; i < numbers.length - 1; i = i + 1)
//put in a inner loop number.length times minus one because we don't want to swap the last element
for(counter = 0; counter < numbers.length - 1; counter = counter + 1)
{
if (numbers [counter] > numbers [counter + 1])
{
int temporary = numbers [counter];
numbers [counter] = numbers [counter + 1];
numbers [counter + 1] = temporary;
}
}
for(int i =0; i < numbers.length; i = i + 1)
{
System.out.print(numbers[i] + " ");
}
}
}
But, in this program, instead of sorting the inputted numbers, the program simply prints the first number that is inputted by the user. I am not sure if I need to move where my scanner function is placed, or add on to it within the loop for it to sort all of the numbers as I want it to do. I am lost on where to change the program if that is the case.
That's because int[] numbers = new int[] {scan.nextInt()}; is a single assigment. scan read a single input and assign to number[0].
You actually need to modify your code for scan to read n numbers and store in n-sized numbers.
something like.
int[] numbers = new int[scan.nextInt()];
for( int i = 0; i < numbers.length; i++)
numbers[i] = scan.nextInt();
The code int[] numbers = new int[] {scan.nextInt()}; will always create an array (not a List) of size 1.
Usually in these kinds of assignments you get n + 1 numbers, for example 5 3 6 2 4 1 would mean "I'm going to give you five numbers. Oh here they are: 3 6 2 4 and 1!"
You probably want something like int[] numbers = new int[scan.nextInt()]; - then loop from 0 to numbers.length to fill the array.
I'm kind of a newbie at Java, and not very good at it. It's a trial and error process for me.
I'm working on a Java program to output the amount of primes in an array. I can get it to output the primes, but I want to also output the quantity of primes. I tried to add each prime to an array list titled "primes" then return "primes.size()" at the end of my program. It doesn't work as intended. The count is actually off. When I create an array of 5 numbers, it outputs 3 primes, 2, 3, and 5. But then it says I have 4 primes. I think it might be counting 1 as a prime. Because when I create an array of 20, the prime numbers output 2,3,5,7,11,13,17 and 19. Then it says the total prime numbers = 9. It should be 8 though.
Here's my code
public class Prime {
public static void main(String[] args) {
int index = 0;
Scanner scan = new Scanner(System. in );
System.out.println("How big would you like the array? ");
int num = scan.nextInt();
int[] array = new int[num];
ArrayList < Integer > primes = new ArrayList < Integer > ();
//System.out.println("How Many threads? ");
//int nThreads = scan.nextInt(); // Create variable 'n' to handle whatever integer the user specifies. nextInt() is used for the scanner to expect and Int.
//Thread[] thread = new Thread[nThreads];
for (int n = 1; n <= array.length; n++) {
boolean prime = true;
for (int j = 2; j < n; j++) {
if (n % j == 0) {
prime = false;
break;
}
}
if (prime) {
primes.add(n);
}
if (prime && n != 1) {
System.out.println(n + "");
}
}
System.out.println("Total Prime numbers = " + primes.size());
System.out.println("Prime Numbers within " + array.length);
}
}
Forgive the sloppiness of it. I actually plan on adding multithreading to it, but I wanted to get this down first.
Any help would be greatly appreciated. Thanks.
You have included 1 in your array of primes, because you started the n for loop at 1. You don't print it because of the final if statement, but it's there in the ArrayList.
Start your n for loop with n = 2. As a consequence, you won't need the final if statement, because n won't be 1 ever. You could print the prime at the same time as you add it to the ArrayList.
I am getting an error and I cant quite understand the cause. I'm trying to get user-input using the Scanner method to assign values to the indices of an array, numbers. The user is allowed to keep - continue - entering values into the array numbers, however, if the user decides to stop, entering 0 will break the assignment of array-index values.
int[] numbers = new int[100];
int[] countIndex = new int[100];
int[] numberOcc = new int[100];
for (int i = 0; i < countIndex.length; i++)
{
countIndex[i] = i;
}
System.out.println("Enter your integer values: ");
for (int i = 0; i < numbers.length; i++)
{
if(input.nextInt() == 0)
{
break;
}
else
{
numbers[i] = input.nextInt();
}
}
This piece of code is a small part of a much larger code, however, it is very important. The error I'm getting is that the code is allowing the user to enter twice before moving to the next index, but the first integer entered it always skipped:
Enter your integer values:
1
2
0
0 occurs 99 times
2 occurs 1 time
UPDATES
System.out.println("Enter your integer values: ");
for (int i = 0; i < numbers.length; i++)
{
numbers[i] = input.nextInt();
if(numbers[i] == 0)
{
break;
}
}
Why is it that the 1 is ignored as input?
Also, is using input.nextInt() as the condition of an if-else statement bad code design?
From the comments
However, a 0 is still being placed in the index,
By default, int values will be assigned 0. So when you don't assign non zero values to int, they will still be 0.
One way around that is to use arraylist. Then once you enter all the numbers, you can convert arraylist to array.
Simple example without any input validation
List<Integer> list = new ArrayList<Integer>();
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int[] defaultArray = new int[n];
// do validation for n
//assume n = 5 with values 1,2,0,3,2
for (int i = 0; i < n; i++) {
int temp = in.nextInt();
if (temp != 0) {
list.add(temp);
defaultArray[i] = temp;
}
}
Integer[] positiveArray = new Integer[list.size()];
positiveArray = list.toArray(positiveArray);
System.out.println(Arrays.toString(defaultArray)); //will print [1, 2, 0, 3, 2]
System.out.println(Arrays.toString(positiveArray)); //will print [1, 2, 3, 2]
If you want to convert Integer arraylist to int array, just search on google and there are many similar questions.
Demo