I have made a program that outputs the number of repeats in a 2D array. The problem is that it outputs the same number twice.
For example: I input the numbers in the 2D array through Scanner: 10 10 9 28 29 9 1 28.
The output I get is:
Number 10 repeats 2 times.
Number 10 repeats 2 times.
Number 9 repeats 2 times.
Number 28 repeats 2 times.
Number 29 repeats 1 times.
Number 9 repeats 2 times.
Number 1 repeats 1 times.
Number 28 repeats 2 times.
I want it so it skips the number if it has already found the number of repeats for it. The output should be:
Number 10 repeats 2 times.
Number 9 repeats 2 times.
Number 28 repeats 2 times.
Number 29 repeats 1 times.
Number 1 repeats 1 times.
Here is my code:
import java.util.Scanner;
public class Repeat
{
static Scanner leopard = new Scanner(System.in);
public static void main(String [] args)
{
final int ROW = 10; //Row size
final int COL = 10; //Column size
int [][] num = new int[ROW][COL];
int size;
//Get input
size = getData(num);
//Find repeat
findRepeats(num, size);
}
public static int getData(int [][] num)
{
int input = 0, actualSize = 0; //Hold input and actualSize of array
System.out.print("Enter positive integers (-999 to stop): ");
//Ask for input
for(int i = 0; i < num.length && input != -999; i++)
{
for(int j = 0; j < num[i].length && input != -999; j++)
{
input = leopard.nextInt();
//Check if end
if(input != -999)
{
num[i][j] = input;
actualSize++;
}
}
}
System.out.println();
return actualSize;
}
public static void findRepeats(int [][] num, int size)
{
int findNum;
int total = 0, row = 0, col = 0;
for(int x = 0; x < size; x++)
{
//Set to number
findNum = num[row][col];
//Loop through whole array to find repeats
for(int i = 0; i < num.length; i++)
{
for(int j = 0; j < num[i].length; j++)
{
if(num[i][j] == findNum)
total++;
}
}
//Cycle array to set next number
if(col < num[0].length-1)
col++;
else
{
row++; //Go to next row if no more columns
col = 0; //Reset column number
}
//Display total repeats
System.out.println("Number " + findNum + " appears " + total + " times.");
total = 0;
}
}
}
I know why it is doing it, but I cannot figure out how to check if the number has already been checked for it to skip that number and go to the next number. I cannot use any classes or code that is not used in the code.
Since you cannot use anything other than this, lets say, basic elements of Java consider this:
Make another temporary 2D array with two columns (or just two separate arrays, personally I prefer this one). On the start of the algorithm the new arrays are empty.
When you take a number (any number) from the source 2D structure, first check if it is present in the first temporary array. If it is, just increment the value (count) in the second temporary array for one (+1). If it is not present in the first tmp array, add it to it and increase the count (+1) in the second at the same index as the newly added number in the first (which should be the last item of the array, basically).
This way you are building pairs of numbers in two arrays. The first array holds all your distinct values found in the 2D array, and the second one the number of appearances of the respective number from the first.
At the and of the algorithm just iterate the both arrays in parallel and you should have your school task finished. I could (and anyone) code this out but we are not really doing you a favor since this is a very typical school assignment.
It's counting the number two times, first time it appears in the code and second time when it appears in the code.
To avoid that keep a system to check if you have already checked for that number. I see you use check int array but you haven't used it anywhere in the code.
Do this,
Put the number in the check list if you have already found the count of it.
int count = 0;
check[count] = findNum;
count++;
Note: You can prefill you array with negative numbers at first in order to avoid for having numbers that user already gave you in input.
Next time in your for loop skip checking that number which you have already found a count for
for(int x = 0; x < size; x++) {
findNum = num[row][col];
if(check.containsNumber(findNUm)) { //sorry there is no such thing as contains for array, write another function here which checks if a number exists in the array
//skip the your code till the end of the first for loop, or in other words then don't run the code inside the for loop at all.
}
}
Frankly speaking I think you have just started to learn coding. Good luck! with that but this code can be improved a lot better. A piece of advice never create a situation where you have to use 3 nested for loops.
I hope that you understood my solution and you know how to do it.
All answers gives you some insight about the problem. I try to stick to your code, and add a little trick of swap. With this code you don't need to check if the number is already outputted or not. I appreciate your comments, structured approach of coding, and ask a question as clear as possible.
public static void findRepeats(int [][] num, int size)
{
int findNum;
int total = 1, row = 0, col = 0;
int [] check = new int[size];
while(row < num.length && col < num[0].length)
{
//Set to number
findNum = num[row][col];
//Cycle array to set next number
if(col < num[0].length-1)
col++;
else
{
row++; //Go to next row if no more columns
col = 0; //Reset column number
}
//Loop through whole array to find repeats
for(int i = row; i < num.length; i++)
{
for(int j = col; j < num[i].length; j++)
{
if(num[i][j] == findNum) {
total++;
//Cycle array to set next number
if(col < num[0].length-1)
col++;
else
{
row++; //Go to next row if no more columns
col = 0; //Reset column number
}
if(row < num.length - 1 && col < num[0].length -1)
num[i][j] = num[row][col];
}
}
}
//Display total repeats
System.out.println("Number " + findNum + " appears " + total + " times.");
total = 1;
}
}
you can use a HashMap to store the result. It Goes like this:
// Create a hash map
HashMap arrayRepeat = new HashMap();
// Put elements to the map
arrayRepeat.put(Number, Repeated);
Related
I have n inputs.
these inputs are numbers from 1 to 100.
I want to output the number that appears less than the other ones; also if there are two numbers with the same amount of appearance, I want to output the number that is less than the other one.
I wrote this code but it doesn't work!
Scanner scanner = new Scanner(System.in);
int n=scanner.nextInt(), max=0 , ans=-1;
int[] counter = new int[n];
for(int i=0; i<n; i++)
counter[scanner.nextInt()]+=1;
for(int j=1; j<=100; j++){
if(counter[j]>max)
max=counter[j];
}
for (int i=1; i<=max; i++){
if(counter[i]>0)
if(ans==-1 || counter[ans]>counter[i] || (counter[ans] == counter[i] && i<ans))
ans=i;
}
System.out.print(ans);
There’s a couple of problems with your code, but the main one is the last for loop: You are trying to find the first (ie lowest) number whose counter is equal to max, so your loop should be from 1 to n, not 1 to max.
Another problem is if you are using the number, which is in the range 1-n, as your array index, you need an array of size n+1, not n.
I pinched this from another question regarding the title of yours:
i = input.nextInt (); while (i != 0) { counts [i]++; i = input.nextInt (); } That method increments the number at the position of the user input in the counts array, that way the array holds the number of times a number occurs in a specific index, e.g. counts holds how often 3 occurs.
counter array should contain frequency values for the numbers from 1 to 100 inclusive.
That is, either a shift by 1 should be used when counting the frequency:
int[] counter = new int[100];
for (int i = 0; i < n; i++) {
counter[scanner.nextInt() - 1]++;
}
or 101 may be used as the length of counter array thus representing values in the range [0..100], without shifting by 1.
int[] counter = new int[101];
for (int i = 0; i < n; i++) {
counter[scanner.nextInt()]++;
}
The minimal least frequent number can be found in a single loop (assuming that the counter length is 101).
int minFreq = 101, answer = -1;
for(int j = 1; j <= 100; j++) {
if (counter[j] > 0 && counter[j] < minFreq) { // check valid frequency > 0
minFreq = counter[j];
answer = j;
}
}
System.out.println(answer);
For a wider range of input values (e.g. including negative values) of a relatively small count it is better to use a hashmap instead of a large sparse array.
I'm supposed to write a program using for loops that print out the even indexes of my array. For example, if I create an array that has 10 numbers, it will have indexes from 0-9 so in that case I would print out the numbers at index 2, 4, 6 and 8. This is what I wrote so far but it doesn't work. Please note that I am not trying to print out the even numbers of the array. All I want are the even indexes.
Example I enter the following array: 3,7,5,5,5,7,7,9,9,3
Program output:
5 // (the number at index 2)
5 // (the number at index 4)
7 // (the number at index 6)
9 // (the number at index 8)
My Code:
public class Arrayevenindex
{
public static void main(String[] args)
{
int number; // variable that will represent how many elements the user wants the array to have
Scanner key = new Scanner(System.in);
System.out.println(" How many elements would you like your array to have");
number = key.nextInt();
int [] array = new int [number];
// let the user enter the values of the array.
for (int index = 0; index < number; index ++)
{
System.out.print(" Value" + (index+1) + " :");
array[index] = key.nextInt();
}
// Print out the even indexes
System.out.println("/nI am now going to print out the even indexes");
for (int index = 0; index < array.length; index ++)
{
if (array[number+1]%2==0)
System.out.print(array[number]);
}
}
}
You can just change your for loop and get rid of the inner IF...
for( int index = 0; index < array.length; index += 2) {
System.out.println(array[index]);
}
Just absolutely same thing using java 8 Stream API
Integer[] ints = {0,1,2,3,4,5,6,7,8,9};
IntStream.range(0, ints.length).filter(i -> i % 2 == 0).forEach(i -> System.out.println(ints[i]));
I assume this would be sufficient
// For loop to search array
for (int i = 0; i < array.length; i++) {
// If to validate that the index is divisible by 2
if (i % 2 == 0) {
System.out.print(array[i]);
}
}
This is what I did and it works:also I am not printing out index[0] because technically its not even thats why I started the for loop at 2. Your post did help me a lot. I also thank everyone else as well that took the time to post an answer.
import java.util.Scanner;
public class Arrayevenindex
{
public static void main(String[] args)
{
int number; // variable that will represent how many elements the user wants the array to have
Scanner key = new Scanner(System.in);
System.out.println(" How many elements would you like your array to have");
number = key.nextInt();
int [] array = new int [number];
// let the user enter the values of the array.
for ( int index = 0; index < number; index ++)
{
System.out.print(" Value" + (index+1) + " :");
array[index] = key.nextInt();
}
// Print out the even indexes
System.out.println("/nI am now going to print out the even indexes");
for ( int index = 2; index < array.length; index +=2)
{
System.out.print(array[index] + " ");
}
}
}
I'm trying to make a Java program that uses the input value from a user to calculate and list the products of two numbers up to the entered number. Like if a user enters 2, the program should calculate the products between the two numbers (1 *1, 1*2, 2*1, 2*2) stores the products in a two-dimensional array, and list the products. I'm not sure that I totally understand arrays and so I feel as though my code is problem not right in many instances, can someone please tell me what I should do to my current code to make it work properly. Thanks in advance! :)
import java.util.Scanner;
public class ProductTable {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String inputString;
char letter = 'y';
// Prompt the user to enter an integer
while(letter != 'q') {
System.out.print("Enter a positive integer: ");
int integer = input.nextInt();
// Create an two-dimensional array to store products
int[][] m = new int[integer][3];
for (int j = 1; j <= m.length; j++) {
m[j][0] = input.nextInt();
m[j][1] = input.nextInt();
m[j][2] = input.nextInt();
}
// Display the number title
System.out.print(" ");
for (int j = 1; j <= m.length; j++)
System.out.print(" " + j);
System.out.println("\n--- ");
// Display table body
for (int i = 1; i <= m.length + 1; i++) {
System.out.print(i);
for (int j = 1; j <= m.length + 1; i++) {
System.out.printf("%4d", i * j);
}
System.out.println();
}
// Prompt the user to either continue or quit
System.out.print("Enter q to quit or any other key to continue: ");
String character = input.nextLine();
inputString = input.nextLine();
letter = inputString.charAt(0);
}
}
}
You can achieve your multiplication table by iterating over 2 counter variables as you already did in your output
// Display table body
// loops running out of bounds (until m.length + 1 instead of m.length-1)
for (int i = 1; i <= m.length + 1; i++) {
System.out.print(i);
for (int j = 1; j <= m.length + 1; i++) { // missed to increment j here
System.out.printf("%4d", i * j);
}
System.out.println();
}
Arrays should be based on 0, that means an array with 3 fields has the indexes 0, 1, 2. So the last index is length-1. Your condition <=m.length+1 runs out of bounds. index<length will work since 2 is less than 3, but not 3 less than 3.
You have also a typo: In the inner loop you are doing an increment of i instead of j, so the loop will run infinite.
Try to create an outer and inner loop as you did, but with start index 0 and end condition index<inputValue. Calculate multiTable[index1][index2] = (1+index1)*(1+index2).
Then do a similar loop and print the array fields. You don't need to use an array, you could output directly as you did. But you wanted to practice handling arrays.
Since you want to learn and understand, I don't like just to write the code. imagine an array with 3 fields having the indexes 0, 1, 2:
index 0 | 1 | 2
value 1 | 2 | 3
Now you would check the length at first, that's 3. You start with 0 and count while the counter does not reach the length since the zero based index is 1 below our natural order (0,1,2 vs. 1,2,3). That is the case as long the index is below the length, meaning index<length.
Try this logic (pseudo code). To simplify the problem we include the 0 in our product table. So we get 0*0, 0*1, 0*2... Doing so, we do not have to ignore index 0 or to calculate between index 0 should represent value 1.
maxNumber = userInput()
outer loop idx1 from 0 to maxNumber
inner loop idx2 from 0 to maxNumber
array[idx1][idx2] = (idx1) * (idx2)
end loop
end loop
Do the same to dump your generated array to screen.
First get it running. Afterwards you could try to alter the logic, so that it shows you only numbers from 1 to max.
I would like to fill a 3x3 2D array with values 1,2,3.
I need each number to appear for a given times.
For example:
1 to appear 2 times
2 to appear 4 times
3 to appear 3 times
What I need is to store this numbers to array in a random position.
For Example:
1,2,2
3,2,2
1,3,3
I already did this in a simple way using only 2 different numbers controlled by a counter. So I loop through the 2D array and applying random values of number 1 and number 2.
I'm checking if the value is 1 and add it in the counter and the same with number 2. if one of the counter exceeds the number I have set as the maximum appear times then it continues and applies the other value.
Is there any better approach to fill the 3 numbers in random array position?
See code below:
int [][] array = new int [3][3];
int counter1 =0;
int counter2 =0;
for (int i=0; i<3; i++) {
for (int j=0; j<3; j++) {
array[i][j] = (int)random(1, 3); //1,2
if (arrray[i][j]==1) {
counter1++;
} else if (array[i][j] ==2) {
counter2++;
}
//if it is more than 5 times in the array put only the other value
if (counter1>5) {
array[i][j] = 2;
}
//if it is more than 4 times in the array put only the other value
else if (counter2>4) {
array[i][j] = 1;
}
}
}
I finally did this according to this discussion:
How can I generate a random number within a range but exclude some?, with 1D array for tesing, but it does not always works.
Please see attached code:
int []values = new int[28];
int counter1=0;
int counter2=0;
int counter3=0;
for (int i=0; i<values.length; i++) {
if (counter1==14) {
ex = append(ex, 5);
}
if (counter2==4) {
ex =append(ex, 6);
}
if (counter3==10) {
ex =append(ex, 7);
}
values[i] = getRandomWithExclusion(5, 8, ex);
if (values[i]==5) {
counter1++;
} else if (values[i] ==6) {
counter2++;
} else if (values[i] ==7) {
counter3++;
}
}
int getRandomWithExclusion(int start, int end, int []exclude) {
int rand = 0;
do {
rand = (int) random(start, end);
}
while (Arrays.binarySearch (exclude, rand) >= 0);
return rand;
}
I would like to fill the 1D array with values of 5,6 or 7. Each one a specific number. Number 5 can be added 14 times. Number 6 can be added 4 times. Number 7 can be added 10 times.
The above code works most of the times, however somethimes it does not. Please let me know if you have any ideas
This is the Octave/Matlab code for your problem.
n=3;
N=n*n;
count = [1 2; 2 4; 3 3];
if sum(count(:,2)) ~= N
error('invalid input');
end
m = zeros(n, n);
for i = 1:size(count,1)
for j = 1:count(i,2)
r = randi(N);
while m(r) ~= 0
r = randi(N);
end
m(r) = count(i,1);
end
end
disp(m);
Please note that when you address a 2D array using only one index, Matlab/Octave would use Column-major order.
There are a ton of ways to do this. Since you're using processing, one way is to create an IntList from all of the numbers you want to add to your array, shuffle it, and then add them to your array. Something like this:
IntList list = new IntList();
for(int i = 1; i <= 3; i++){ //add numbers 1 through 3
for(int j = 0; j < 3; j++){ add each 3 times
list.append(i);
}
}
list.shuffle();
for (int i=0; i<3; i++) {
for (int j=0; j<3; j++) {
array[i][j] = list.remove(0);
}
}
You could also go the other way: create an ArrayList of locations in your array, shuffle them, and then add your ints to those locations.
I'm trying to create a method that will search through a 2d array of numbers. If the numbers add up to a certain sum, those numbers should remain and all of the other numbers should be changed to a 0. For example, if the desired sum is 7 and a row contains 2 5 1 2, the result should be 2 5 0 0 after the method is implemented. I have everything functioning but instead of keeping all of the numbers that add up to the sum, only the last number is retained. So, I am left with 0 5 0 0 . I think I need another array somewhere but not sure exactly how to go about implementing it. Any ideas?
public static int[][] horizontalSums(int[][] a, int sumToFind) {
int[][] b = new int[a.length][a[0].length];
int columnStart = 0;
while (columnStart < a[0].length) {
for (int row = 0; row < a.length; row++) {
int sum = 0;
for (int column = columnStart; column < a[row].length; column++) {
sum += a[row][column];
if (sum == sumToFind) {
b[row][column] = a[row][column];
}
}
}
columnStart++;
}
return b;
}
In your example you use 2 5 1 1, would 0 5 1 1 also be a valid response? Or do you just need to find any combination? A recursive function may be the best solution.
If you just need to scan through the array and add up the numbers until the sum is reached then just add a for loop to copy the previous values from the array to the new array when the sum is found. Something like:
if (sum == sumToFind)
{
for (int i= 0; i<= columnStart; i++)
{
b[row][i] = a[row][i];
}
}
if (sum == sumToFind)
{
for (int i= columnStart; i<= column; i++)
{
b[row][i] = a[row][i];
}
}
A minor tweak was all it needed. If you have columnStart and column like in the other answer, it only finds the first number of the series.