Counting occurrences of integers in an array in Java - java

Note: no mapping, no sorting
Here's my code:
public static void countArray(int[] n){
int[] m = new int[n.length]; //50 elements of integers between values of 10 & 20
int count = 0;
int sum = 0;
for ( int i = 0; i < n.length ; i++){
m[i] = n[i]; //make a copy of array 'n'
System.out.print(m[i]+" ");
}System.out.println();
for ( int j =0; j < n.length ; j++){
count =0;
for(int i = 0; i < n.length ; i++){
if (n[j]%m[i]==0 && n[j] == m[i])
count++;
}if ( n[j]%m[j] == 0)
System.out.println(m[j] + " occurs = " + count);
}
}
So the problem is: I get repeating results like : "25 occurs = 5", on different lines.
What I think: the problem occurs because of if ( n[j]%m[j] == 0)
so I tried if ( n[j]%m[j+1] == 0). Another problem occurs since m[j] will be m[50] so it crashes but sort of give me the results that I want.
Result that I want: something like this: no repetitions and covers all the random integers on a set
17 occurs = 3
23 occurs = 2
19 occurs = 3
15 occurs = 2
12 occurs = 2

With some adaptation your code should work :
public static void countArray(int[] n){
boolean [] alreadyCounted = new boolean[n.length];
for (int i = 0; i < n.length ; i++){
int count = 0;
if (alreadyCounted[i]) {
// skip this one, already counted
continue;
}
for(int j = 0; j < n.length ; j++){
if (n[i] == n[j]) {
// mark as already counted
alreadyCounted[j] = true;
count++;
}
}
System.out.println(n[i] + " occurs = " + count);
}
}
You could definitely use the same logic with better code, I just tried to follow the original "coding style";
This is O(n^2) solution (read "very slow").
If you could use sorting, you could do it in O(n log(n)) - that is fast.
With mapping you could do it in O(n) - that is blazingly fast;

If you exploit the input limit you can lose the nested loop:
public static void main(String[] args)
{
//6 elements of integers between values of 10 & 20
int[] countMe = { 10, 10, 20, 10, 20, 15 };
countArray(countMe);
}
/** Count integers between values of 10 & 20 (inclusive) */
public static void countArray(int[] input)
{
final int LOWEST = 10;
final int HIGHEST = 20;
//Will allow indexes from 0 to 20 but only using 10 to 20
int[] count = new int[HIGHEST + 1];
for(int i = 0; i < input.length; i++)
{
//Complain properly if given bad input
if (input[i] < LOWEST || HIGHEST < input[i])
{
throw new IllegalArgumentException("All integers must be between " +
LOWEST + " and " + HIGHEST + ", inclusive");
}
//count
int numberFound = input[i];
count[numberFound] += 1;
}
for(int i = LOWEST; i <= HIGHEST; i++)
{
if (count[i] != 0)
{
System.out.println(i + " occurs = " + count[i]);
}
}
}

try this :(sort the array and then count the occurence of element)
public static void countArray(int[] n) {
int count = 0;
int i, j, t;
for (i = 0; i < n.length - 1; i++) // sort the array
{
for (j = i + 1; j < n.length; j++) {
if (n[i] > n[j]) {
t = n[i];
n[i] = n[j];
n[j] = t;
}
}
}
for (i = 0; i < n.length;)
{
for (j = i; j < n.length; j++) {
if (n[i] == n[j])
{
count++;
} else
break;
}
System.out.println(n[i] + " occurs " + count);
count = 0;
i = j;
}
}

Here's a nice, efficient way to do it, rather more efficiently than the other solutions posted here. This one runs in O(n) time, where the array is of length n. It assumes that you have some number MAX_VAL, representing the maximum value that you might find in your array, and that the minimum is 0. In your commenting you suggest that MAX_VAL==20.
public static void countOccurrences(int[] arr) {
int[] counts = new int[MAX_VAL+1];
//first we work out the count for each one
for (int i: arr)
counts[i]++;
//now we print the results
for (int i: arr)
if (counts[i]>0) {
System.out.println(i+" occurs "+counts[i]+" times");
//now set this count to zero so we won't get duplicates
counts[i]=0;
}
}
It first loops through the array increasing the relevant counter each time it finds an element. Then it goes back through, and prints out the count for each one. But, crucially, each time it prints the count for an integer, it resets that one's count to 0, so that it won't get printed again.
If you don't like the for (int i: arr) style, this is exactly equivalent:
public static void countOccurrences(int[] arr) {
int[] counts = new int[MAX_VAL+1];
//first we work out the count for each one
for (int i=0; i<arr.length; i++)
counts[arr[i]]++;
//now we print the results
for (int i=0; i<arr.length; i++)
if (counts[arr[i]]>0) {
System.out.println(arr[i]+" occurs "+counts[arr[i]]+" times");
//now set this count to zero so we won't get duplicates
counts[arr[i]]=0;
}
}

Related

Java How do i repeat sort until no swaps are done in bubblesort?

I am taking 10 elements and performing a bubble sort on them. I want to add an algorithm that repeats the sort until no swaps are needed to make this more efficient.
Essentially I want to:
repeat until no swaps done in a pass
For elements 1 to (n-1)
compare contents of element value 1 with the contents of the next value
if value 1 is greater than value 2
then swap the values
This is what I have done so far :
{
//create array
int[] iList = new int[10];
Scanner sc = new Scanner(System.in);
//takes in array input for 10 numbers
System.out.println("Enter a array of numbers ");
for(int i = 0; i< 10; i++ )
{
int num = i + 1;
System.out.println("Enter number " + num);
iList[i] = sc.nextInt();
}
//Bubble sorts the array
System.out.println("The array =");
for(int a = 0; a < iList.length; a++ )
{
for(int b = a+1; b < iList.length; b++)
{
if(iList[a] > iList[b])
{
int iTemp = iList[a];
iList[a] = iList[b];
iList[b] = iTemp;
}
System.out.println("Progress = " + Arrays.toString(iList) );
}
}
} ```
Here is my implementation :
public static void sort(int[] nums) {
boolean isSwapped;
int size = nums.length - 1;
for (int i = 0; i < size; i++) {
isSwapped = false;
for (int j = 0; j < size - i; j++) {
if (nums[j] > nums[j+1]) {
int temp = nums[j];
nums[j] = nums[j + 1];
nums[j + 1] = temp;
isSwapped = true;
}
}
if (!isSwapped) break;
}
System.out.println("Sorted Array: " + Arrays.toString(nums));
}

How to get indices as well as array numbers printed out horizontally?

I have an assignment of which a part is to generate n random numbers between 0-99 inclusive in a 1d array, where the user enters n. Now, I have to print out those numbers formatted like this:
What is your number? 22 //user entered
1 2 3 4 5 6 7 8 9 10
----random numbers here---------
11 12 13 14 15 16 17 18 19 20
-----random numbers here--------
21 22
---two random numbers here---
Using those numbers, I have find lots of other things, (like min, max, median, outliers, etc.) and I was able to do so. However, I wasn't able to actually print it out in the format shown above, with no more than 10 numbers in one row.
Edit: Hello, I managed to figure it out, here's how I did it:
int counter = 0;
int count2 = 0;
int count3 = 0;
int add = 0;
int idx = 1;
int idx2 = 0;
if (nums > 10)
{
count3 = 10;
count2 = 10;
}
else
{
count3 = nums;
count2 = nums;
}
if (nums%10 == 0) add = 0;
else add = 1;
for (int i = 0; i < nums/10 + add; i++)
{
for (int j = 0; j < count3; j++)
{
System.out.print(idx + "\t");
idx++;
}
System.out.println();
for (int k = 0; k < count2; k++)
{
System.out.print(numbers[idx2] + "\t");
idx2++;
counter++;
}
System.out.println("\n");
if (nums-counter > 10)
{
count3 = 10;
count2 = 10;
}
else
{
count3 = nums-counter;
count2 = nums-counter;
}
}
Thank you to everyone who helped! Also, please let me know if you find a way to shorten what I have done above.
*above, nums was the number of numbers the user entered
I'd use a for-loop to make an array of arrays: and then formatting the lines using those values:
var arr_random_n = [1,2,3,4,5,6,7,8,9,0,1,2,3,6,4,6,7,4,7,3,1,5,7,9,5,3,2,54,6,8,5,2];
var organized_arr = [];
var idx = 0
for(var i = 0; i < arr.length; i+=10){
organized_arr[idx] = arr.slice(i, i+10); //getting values in groups of 10
idx+=1 //this variable represents the idx of the larger array
}
Now organized_arr has an array of arrays, where each array in index i contains the values to be printed in line i.
There's probably more concise ways of doing this. but this is very intuitive.
Let me know of any improvements.
Something like this might be what you're looking for.
private static void printLine(String msg)
{
System.out.println("\r\n== " + msg + " ==\r\n");
}
private static void printLine(int numDisplayed)
{
printLine(numDisplayed + " above");
}
public static void test(int total)
{
int[] arr = new int[total];
// Fill our array with random values
for (int i = 0; i < total; i++)
arr[i] = (int)(Math.random() * 100);
for (int i = 0; i < total; i++)
{
System.out.print(arr[i] + " ");
// Check if 10th value on the line, if so, display line break
// ** UNLESS it's also the last value - in that case, don't bother, special handling for that
if (i % 10 == 9 && i != total - 1)
printLine("Random Numbers");
}
// Display number of displayed elements above the last line
if (total < 10 || total % 10 != 0)
printLine(total % 10);
else
printLine(10);
}
To print 10 indexes on a line then those elements of an array, use two String variables to build the lines, then print them in two nested loops:
for (int i = 0; i < array.length; i += 10) {
String indexes = "", elements = "";
for (int j = 0; j < 10 && i * 10 + j < array.length; j++) {
int index = i * 10 + j;
indexes += (index + 1) + " "; // one-based as per example in question
elements += array[index] + " ";
}
System.out.println(indexes);
System.out.println(elements);
}

Java Array: Finding how many numbers are less then the Mean

Okay so im trying to find the amount of numbers that are less then
the mean of the first array. Everything but the last part is working and i
cant figure it out. The code at the bottom is what im having problems with.
for example. if i enter 1 2 3 4 5. the mean is 3 and, 1 and 2 are less then 3. so the answer would be 2 numbers.
public static void main(String[] args) {
Scanner in=new Scanner(System.in);
System.out.println("How many integers should we analyze?" );
int num;
num=in.nextInt();
while ( num <= 2)
{
System.out.println( "Please reenter, integer must be greater than 1" );
num=in.nextInt();
}
int[] arr = new int[num];
System.out.println( "Please enter the "+ num +" integers:" );
for (int i = 0; i < arr.length; i++)
{
arr[i] = in.nextInt();
}
System.out.print("Number of integers input: " + num);
System.out.println();
double total = 0;
for( int element : arr) {
total += element;
}
System.out.print("Total: " + (int) total);
System.out.println();
double mean = 0;
if ( arr.length > 0) {
mean = total / arr.length;
}
System.out.print("Mean: " + mean );
int big = arr[0];
for (int i = 0 ; i < arr.length; i++) {
if (arr[i] > big) {
big = arr[i];
}
}
System.out.println();
System.out.print("Largest: " + big);
System.out.println();
///////////////////////////////////////////////////////////////////////////////////////////////
int less;
for(int i=0;i<mean;i++) {
int num2 = i;
int[] arr2 = new int[num2];
int count = 0;
while ( num2 != 0 )
{
num2/=10;
++count;
System.out.print("Numbers less than the mean: " + count);
}
}
//////////////////////////////////////////////////////////////////////////////////////////////
}
}
you could use this code below
int count = 0;
for(int i =0;i< arr.length;i++) {
if(arr[i] < mean)
count++;
}
System.out.println("numbers less than mean " + count);
what this does is it loops through all of the original integers and if one is less than the mean, the count variable goes up by 1.
You should iterate over the array and check each element if they are under the mean. If yes, then increment an integer.
int less = 0;
for(int i = 0; i < arr.length; i++) {
if(arr[i] < mean) {
less++;
}
}
System.out.print("Numbers less than the mean: " + less);

error: Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException

I need help with this error:
Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 54
at Main.oddSort(Main.java:44)
at Main.main(Main.java:19)
I understand that this error occurs because I am trying either trying to assign too many values in the array correct? I just don't know how to fix it or why what I did was wrong.
The assignment is to generate 100 random numbers, and to call two different functions, one after the other to assign the odd and even numbers into two different arrays and to display them.
Here is the section that is giving me trouble:
public static int[] oddSort ( int input[] )
{
int amountOfOdd = 0;
int j = 0;
for(int i = 0; i < input.length; i++)
{
if (input[i] % 2 != 0)
amountOfOdd++;
}
int[] odd = new int[amountOfOdd];
for(int i = 0; i <= 99; i++)
{
if (input[i] % 2 != 0)
/*it's this line specifically that doesn't work, according to the debugger*/
odd[j] = input[i];
j++;
}
return odd;
}
Here is the full thing:
public class Main
{
public static void main(String[] args)
{
int[] numbers = new int[100];
for(int i = 0; i < numbers.length-1; i++)
numbers[i] = (int)(Math.random() * 26);
int[] odd = oddSort(numbers);
int[] even= evenSort(numbers);
System.out.println("The odd numbers are:");
display( odd );
System.out.println("The even number are:");
display( even );
}
public static int[] oddSort ( int input[] )
{
int amountOfOdd = 0;
int j = 0;
for(int i = 0; i < input.length; i++)
{
if (input[i] % 2 != 0)
amountOfOdd++;
}
int[] odd = new int[amountOfOdd];
for(int i = 0; i <= 99; i++)
{
if (input[i] % 2 != 0)
odd[j] = input[i];
j++;
}
return odd;
}
public static int[] evenSort ( int input[] )
{
int amountOfEven = 0;
int j = 0;
for(int i = 0; i < input.length; i++)
{
if (input[i] % 2 != 0)
amountOfEven++;
}
int[] even = new int[amountOfEven];
for(int i = 0; i < input.length; i++)
{
if (input[i] % 2 != 0)
even[j] = input[i];
j++;
}
return even;
}
public static void display (int input[] )
{
for (int i = 0; i < input.length; i++)
System.out.print(input[i] + " ");
}
}
The cause of the exception is the fact that you did not put the brackets around the body of your if condition. Should be like this :
for(int i = 0; i <= 99; i++)
{
if (input[i] % 2 != 0){
odd[j] = input[i];
j++;
}
}
Right now, what you are actually doing is increasing j every time through the loop, not every time there is an odd number. So of course you will get this exception when at least one value in the input is not odd since the odd array will have a size that is less than the size of input.

Find the prime-->sieve way

I tried it several times but still gives me ArrayOutOfIndex. But i want to save the memory so i use
boolean[]isPrime = new boolean [N/2+1];
instead of
boolean[]isPrime = new boolean [N+1];
This gives me ArrayOutOfIndex for line 23 and 47
line 23:
for (int i = 3; i <= N; i=i+2) {
isPrime[i] = true;
}
line 47:
for (int i = 3; i <= N; i=i+2) {
if (isPrime[i]) primes++;
...
}
Full code:
public class PrimeSieve {
public static void main(String[] args) {
if (args.length < 1) {
System.out.println("Usage: java PrimeSieve N [-s(ilent)]");
System.exit(0);
}
int N = Integer.parseInt(args[0]);
// initially assume all odd integers are prime
boolean[]isPrime = new boolean [N/2+1];
isPrime[2] = true;
for (int i = 3; i <= N; i=i+2) {
isPrime[i] = true;
}
int tripCount = 0;
// mark non-primes <= N using Sieve of Eratosthenes
for (int i = 3; i * i <= N; i=i+2) {
// if i is prime, then mark multiples of i as nonprime
if (isPrime[i]) {
int j = i * i;
while (j <= N){
tripCount++;
isPrime[j] = false;
j = j + 2*i;
}
}
}
System.out.println("Number of times in the inner loop: " + tripCount);
// count and display primes
int primes = 0;
if(N >= 2 ){
primes = 1;
}
for (int i = 3; i <= N; i=i+2) {
if (isPrime[i]) primes++;
if (args.length == 2 && args[1].equals("-s"))
; // do nothing
else
System.out.print(i + " ");
}
System.out.println("The number of primes <= " + N + " is " + primes);
}
}
You should store and access the array using the same indexing function: isPrime[i/2]
When you change the size of your array from [N+1] to [N/2+1], you need to also update the end-conditions of your for-loops. Right now your for-loops run until i=N, so you are trying to do isPrime[i] when i > (N/2+1) ... so you get an ArrayIndexOutOfBoundsException.
Change this:
for (int i = 3; i <= N; i=i+2)
to this:
for (int i = 3; i <= N/2; i=i+2)
Well, for example if N=50 your isPrime only holds 26 elements, and you're trying to access the elements at 3,5..47,49 (which, of course, is out of bounds)
What you probably want is to use i/2 (as the index) inside your loops, that way you are still iterating over the numbers 3,5..47,49, but you use the correct indexes of your vector.

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