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How to count inputs in an array in java

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.

Java. Division of input so that number will be divided into 3 equal parts but trick is in description of question

Let's say I have number x=6 I want to divide this number in such a way that I can run 3 loops based on the 1st, 2nd and 3rd part.
For example: x=6 then 1st loop (1-2), 2nd loop (3-4), 3rd loop (5-6).
Example 2: x=3000 then 1st loop (1-1000), 2nd loop (1001-2000), 3rd loop (2001-3000). I don't want to put manually because x can be any "even" number.
x will be number that can be equally divided like 3,6,9,12,15,18 .....

You can do this by the following code. There is no validation as you said the number would be divisible
int x; //your value
int step = x/3;
for(int i=0;i<3;i++){
for(int j=(step*i)+1;j<=step*(i+1);j++){
//do something with j
}
}

This will run the the inner loop 3 times and the inner loop will run x/3 times.
public static void executeLoop(int multiple) {
int interval = multiple / 3;
int start = 0;
int end = 0;
for (int i = 0; i < 3; i++) {
start = i * interval;
end = start + interval;
for (int j = start; j < end; j++) {
System.out.printf("i: %d, j: %d\n", i, j);
}
}
}

What do I need to do to my current code to make the program run a multiplication table (using a two dimensional array)?

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.

2D Array Scores

This is the original prompt:
Write program that declares a 2-dimensional array of doubles called scores with three rows and three columns. Use a nested while loop to get the nine (3 x 3) doubles from the user at the command line. Finally, use a nested for loop to compute the average of the doubles in each row and output these three averages to the command line.
Here is my code:
import java.util.Scanner;
public class Scorer {
public static void main(String[] args) {
Scanner scnr = new Scanner(System.in);
double [][] scores = new double[3][3];
double value = 0;
int i = 0;
int j;
while (i < 3) {
j = 0;
while (j < 3) {
System.out.print("Enter a number: ");
value = scnr.nextDouble();
scores[i][j] = value;
j++;
}
i++;
}
int average = 0;
for (i = 0; i < scores.length; i++) {
for (j = 0; j < scores[i].length; j++) {
average += value;
value = value / scores[i][j];
System.out.println(value);
}
}
}
}
The part that I need help on is the nested for loop at the bottom of the code. This code is supposed to compute the average of the numbers that are entered; however, I am confused on how to do that with the nested for loop.

you're almost there!
Here are the things you need to do:
1)you've to initialize the variable 'average' after the first for loop.
because average needs to be 0 i.e., reset after second for loop ends each time.
2)you've defined "value = value / scores[i][j]" . I don't know why you did that, but "value = scores[i][j]" must solve your problem.
3) you should print the average only thrice i.e., after calculating average of each row. so, print average at the end of second for loop.
Hope this makes it clear.
here's the code for your reference:
for (i = 0; i < 3; i++) {
int average = 0;
for (j = 0; j < 3; j++) {
value = scores[i][j];
average += value;
}
System.out.println(average/3);
}

Ever i represents a row, every j represents a column.
You need the average of every row, meaning that for every same i and every different j for that i you need to store the values and calculate the average.
Looks like homework code. We can give you hints but not write it for you :(

Finding repeats in a 2D array

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);