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How to find the largest number in a set of random numbers

Scanner s = new Scanner(System.in);
System.out.println("Type a number");
int n = s.nextInt();
int start = 0;
int largest = 0;
int occurrence = 0;
while(n > start){
int number = (int)(Math.random()*100);
if(number > largest){
largest = number;
}
n--;
System.out.print(number+" ");
}
System.out.println("max is "+largest);
System.out.println("Occurrence is "+occurrence);
I would like to be able to find the occurrence of the largest int, and I am not sure how to go about doing so, in a rather simple way. I tried adding occurrence++; under largest = number;, but that did not work.
For example, I would type 6 as input, and I would get random numbers 54, 74, 61, 89, 13, 89.
The desired output would be max is 89. Occurrence is 2.
Also, I am trying to only get the code to print only 10 numbers per line, then it would skip to the next line and continue.

In case of reoccurrence, increase occurrence variable;
set occurrence to 1 in case of new larger number
You can use
if(number > largest){
largest = number; // found larger value
occurrence = 1; // reset occurrence back to initial
}else if(number == largest){
occurrence++; // keep track of same large value
}

You need to use an else if condition to check equality like this
Scanner s = new Scanner(System.in);
System.out.println("Type a number");
int n = s.nextInt();
int start = 0;
int largest = 0;
int occurrence = 1;
while(n > start){
int number = (int)(Math.random()*100);
if(number > largest){
largest = number;
}else if(number == largest){
occurrence++;
}
n--;
System.out.print(number+" ");
}
System.out.println("max is "+largest);
System.out.println("Occurrence is "+occurrence);
P.S - Initialize occurrence = 1 as there will be at least one time the largest number will be present

I think you are almost there. Basically, you are missing a check on whether the new random number is equal to the current largest number. Something like:
if (number == largest) {
occurrences++;
}
Also, remember to reset occurrences when a new largest number has been found:
if (number > largest){
largest = number;
occurrences = 1;
}

How to Pass This Test Case?

I'm creating the program that determines the largest and smallest number is a series of numbers entered by the user. I've created several tests cases for my code and they all work out, but it fails the most simple test case. When the user inputs a single number. For instance, if the user sets the terminating value to be 25, then enters -1, and finally enters the terminating the value, the output should be
Largest: -1 and Smallest: -1. However, my code will output Largest: 0 and Smallest: -1 -- I why this happens (because I initialized the max value to be 0 before running the loop), but how can I fix this?
Here's my code...
Scanner scan = new Scanner(System.in);
// Declaration variables
double min;
double max = 0;
System.out.println("Enter terminating number: ");
double terminator = scan.nextDouble();
System.out.println("Enter a number: ");
double num = scan.nextDouble();
min = num;
if (num == terminator) {
System.out.println("There must be one number in the list.");
// break;
} else {
while (num != terminator) {
System.out.println("");
num = scan.nextDouble();
if ((num < min) && (num != terminator)) {
double temp = min;
min = num;
max = temp;
} else if ((num > min) && (num != terminator)) {
max = num;
} else {
max = min;
}
}
System.out.println("Largest: " + max);
System.out.println("Smallest: " + min);
}

Instead of initializing max = 0, do max = num just like you already do with min.
It's not clear why you're initializing max differently from min; when a single number has been entered, it's both the minimum and the maximum. Right now, the only code that modifies max is within the loop that reads numbers beyond the first, so the first number has no effect on it.

Printing the two highest values from user input

I have an assignment where I have to write a code which lets the user decide an amount of int values to be written in, and then decides what these values should be. There has to be atleast 2 inputs from the user. The program will then compare the values from the input and then print out the two highest values. So far I managed to print out the highest value, but I'm not sure whats wrong with the way I've done it since the output just becomes 0 if I choose to print out 2 numbers and the highest one is entered in first. And I'm also not sure how to keep track of the second highest number either. Would appreciate some help.
import java.util.Scanner;
public class ToStoersteTall{
public static void main(String[] args){
System.out.println("How many numbers? (minimum 2)?:");
Scanner reader = new Scanner(System.in);
if (reader.hasNextInt()) {
int numbers = reader.nextInt();
if (numbers >= 2) {
System.out.println("Enter value #1");
if (reader.hasNextInt()) {
int num1 = reader.nextInt();
System.out.println("Enter value #2");
if (reader.hasNextInt()) {
int num2 = reader.nextInt();
int biggest = 0;
for (int i = 3; i <= tall; i++) {
System.out.println("Enter value #" + i);
int num3 = reader.nextInt();
biggest = num1;
if(biggest < num3){
biggest = num3;
}
}
System.out.println(biggest);
} else {
System.out.println("Please enter an integer");
}
} else {
System.out.println("Please enter an integer");
}
} else {
System.out.println("Please enter an integer equal or higher than 2.");
}
} else {
System.out.print("Vennligst oppgi et heltall større eller lik 2.");
}
}
}

I have an assignment where I have to write a code which lets the user decide an amount of int values to be written in, and then decides what these values should be. There has to be atleast 2 inputs from the user. The program will then compare the values from the input and then print out the two highest values. So far I managed to print out the highest value, but I'm not sure whats wrong with the way I've done it since the output just becomes 0 if I choose to print out 2 numbers and the highest one is entered in first. And I'm also not sure how to keep track of the second highest number either. Would appreciate some help.
A couple things:
good practice to close scanner (and IO-related resources in general)
reduced if-statement blocks bloat for easier readability
you specify 2 guaranteed numbers, so attempt to parse those before looping
can remove system.exit calls or replace system.exit and move bulk of code back into the larger if-else blocks as originally state in OP (but I refer back to the sake of readability)
added check for the first and second numbers input to make sure high1 is highest value, and high2 is second highest value.
keep order while looping and checking values (note: does not use array), if the number is a new high, replace high1 and move high1's value down to high2, or if the number is a second (new) high, replace high2. If values are equal, this logic is excluded and you may want to specify based on your own constraints
import java.io.IOException;
import java.util.Scanner;
public class ToStoersteTall {
public static void main(String[] args) throws IOException {
System.out.println("How many numbers? (minimum 2)?:");
Scanner reader = new Scanner(System.in);
int n = 0;
if (reader.hasNextInt()) {
n = reader.nextInt();
} else {
System.out.println("Vennligst oppgi et heltall større eller lik 2.");
System.exit(-1); // quits execution
}
if (n < 2) {
System.out.println("Please enter an integer equal or higher than 2.");
System.exit(-2);
}
// Since guaranteed 2 numbers, parse and assign now
int high1 = 0, high2 = 0;
System.out.println("Enter value # 1");
if (reader.hasNextInt())
high1 = reader.nextInt();
System.out.println("Enter value # 2");
if (reader.hasNextInt())
high2 = reader.nextInt();
// check to see if a switch to keep correct highest order, swap values if so
if (high1 < high2) {
int t = high2;
high2 = high1;
high1 = t;
}
// loop won't execute if only 2 numbers input, but will if 3 or more specified at start
for (int i = 2; i < n; ++i) {
System.out.println("Enter value #" + (i + 1));
if (reader.hasNextInt()) {
int t = reader.nextInt();
if (t > high1) {
high2 = high1; // throw away high2 value and replace with high1
high1 = t; // replace high1 value with new highest value
} else if (t > high2) {
high2 = t;
}
} else {
System.out.println("Please enter an interger");
}
}
reader.close();
System.out.println("The two highest numbers are: " + high1 + ", " + high2);
}
}

You're already keeping track of the biggest, so why not keep track of the second biggest? Another easy way of solving this problem is to keep all the numbers in a list, sort the list by number size, and grab the two highest entries.

I tried your code and used an array to solve the problem.
import java.util.Scanner;
public class Main {
static int secondHighest(int... nums) {
int high1 = Integer.MIN_VALUE;
int high2 = Integer.MIN_VALUE;
for (int num : nums) {
if (num > high1) {
high2 = high1;
high1 = num;
} else if (num > high2) {
high2 = num;
}
}
return high2;
}
public static void main(String[] args) {
System.out.println("How many numbers? (minimum 2)?:");
Scanner reader = new Scanner(System.in);
if (reader.hasNextInt()) {
int numbers = reader.nextInt();
int[] array = new int[numbers];
if (numbers >= 2) {
System.out.println("Enter value #1");
if (reader.hasNextInt()) {
int num1 = reader.nextInt();
array[0] = num1;
System.out.println("Enter value #2");
if (reader.hasNextInt()) {
int num2 = reader.nextInt();
array[1] = num2;
int biggest = 0;
for (int i = 3; i <= numbers; i++) {
System.out.println("Enter value #" + i);
int num3 = reader.nextInt();
array[i-1] = num3;
}
System.out.println("second largest number is" + secondHighest(array));
int largest = 0;
for(int i =0;i<array.length;i++) {
if(array[i] > largest) {
largest = array[i];
}
}
System.out.println("Largest number in array is : " +largest);
} else {
System.out.println("Please enter an integer");
}
} else {
System.out.println("Please enter an integer");
}
} else {
System.out.println("Please enter an integer equal or higher than 2.");
}
} else {
System.out.print("Vennligst oppgi et heltall større eller lik 2.");
}
}
}
Test
How many numbers? (minimum 2)?:
6
Enter value #1
3
Enter value #2
4
Enter value #3
5
Enter value #4
6
Enter value #5
7
Enter value #6
8
second largest number is7
Largest number in array is : 8

There is a logic error in your program. If numbers is 2, then the for loop never gets executed, and the value of biggest remains zero because it is never updated. Change your declaration of biggest to reflect the current maximum value found so far.
int biggest = num1 > num2 ? num1 : num2;
That way if the for loop never executes then biggest will be the maximum value of the first two numbers.
As for keeping track of the second highest value, you could introduce another variable secondBiggest, initialised in a similar manner to biggest, and then write logic to update this value in your for loop. However, in my opinion, it would be much easier to change your strategy to hold the entered values into an array, then when all inputs have been entered, calculate whichever values you desire from the array. This would lead to a much cleaner solution IMO.
(I have assumed that tall in the for loop is actually meant to be numbers...)
import java.util.Scanner;
public class Foo{
public static void main(String[] args){
System.out.println("How many numbers? (minimum 2)?:");
Scanner reader = new Scanner(System.in);
if(reader.hasNextInt()){
int numbers = reader.nextInt();
if(numbers >= 2){
int[] list = new int[numbers];
for(int i = 0; i < numbers; i++){
System.out.println("Enter value #" + (i + 1));
if(reader.hasNextInt())
list[i] = reader.nextInt();
}//for
int biggest = 0;
int secondBiggest = 0;
// find the values you want
for(int i = 0; i < numbers; i++){
if(list[i] > biggest){
secondBiggest = biggest;
biggest = list[i];
}//if
else if(list[i] > secondBiggest)
secondBiggest = list[i];
}//for
// print your results
System.out.println("The biggest integer is: " + biggest);
System.out.println("The second biggest integer is: " + secondBiggest);
}//if
}//if
}//main
}//class

java program outputting even/odd numbers

My task is to write a java program that first asks the user how many numbers will be inputted, then outputs how many odd and even numbers that were entered. It is restricted to ints 0-100. My question is: What am I missing in my code?
import java.util.Scanner;
public class Clancy_Lab_06_03 {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int n;
System.out.println("How many numbers will be entered?");
n = input.nextInt();
while (n < 0 || n > 100) {
System.out.println("ERROR! Valid range 0-100. RE-Enter:");
n = input.nextInt();
n++;
}
int odd = 0;
int even = 0;
while (n >= 0 || n <= 100) {
n = input.nextInt();
if (n % 2 == 0) {
even++;
} else {
odd++;
}
}
System.out.println(even + "even" + odd + "odd");
}
}

Second while loop is infinite. Relplace it with something like this:
for (int i = 0; i < n; i++) {
int b = input.nextInt();
if (b % 2 == 0) {
even++;
} else {
odd++;
}
}
Also I don't understand why are you incrementing n in first loop. For example when you will first give -5, you will be asked to re-enter the number. Then you type -1, but it gets incremented and in fact program processes 0, altough user typed -1. In my opinion it is not how it suppose to work and you should just remove this n++.
As you asked in comment - the same using while loop:
while(n > 0) {
n--;
int b = input.nextInt();
if (b % 2 == 0) {
even++;
} else {
odd++;
}
}
Also it is good idea to close input when you no longer need it (for example at the end of main method)
input.close();

You had two issues - first you were incrementing n in the first loop, rather than waiting for the user to enter a valid number.
In the second loop, you weren't comparing the number of entries the user WANTED to make with the number they HAD made - you were over-writing the former with the new number.
This version should work, although I've not tested it as I don't have java on this machine.
Note that we now sit and wait for both inputs, and use different variable names for the "how many numbers will you enter" (n) and "what is the next number you wish to enter" (num) variables? Along with a new variable i to keep track of how many numbers the user has entered.
import java.util.Scanner;
public class Clancy_Lab_06_03
{
public static void main (String[] args)
{
Scanner input = new Scanner (System.in);
int n;
System.out.println ("How many numbers will be entered?");
n = input.nextInt();
//Wait for a valid input
while (n < 0 || n > 100)
{
System.out.println ("ERROR! Valid range 0-100. RE-Enter:");
n = input.nextInt();
}
//Setup variables for the loop
int odd = 0;
int even = 0;
int num;
//Keep counting up until we hit n (where n is the number of entries the user just said they want to make)
for(int i = 0; i < n; i++)
{
//Changed this, because you were over-writing n (remember, n is the number of entries the user wants to make)
//Get a new input
while (num < 0 || num > 100)
{
System.out.println ("ERROR! Valid range 0-100. RE-Enter:");
num = input.nextInt();
}
//Check whether the user's input is even or odd
if (num % 2 == 0)
{
even++;
}
else
{
odd++;
}
}
System.out.println(even + " even. " + odd + " odd.");
}
}

import java.util.Scanner;
public class Test2 {
public static void main(String[] args) {
System.out.println("Enter an Integer number:");
Scanner input = new Scanner(System.in);
int num = input.nextInt();
if ( num % 2 == 0 )
System.out.println("Entered number is even");
else
System.out.println("Entered number is odd");
}
}

My suggestion to you is to have a clear separation of your requirements. From your post, you indicate you need to prompt the user for two distinct data items:
How many numbers will be entered (count)
The values to be analyzed
It is a good practice, especially when you are learning, to use meaningful names for your variables. You are using 'n' for a variable name, then reusing it for different purposes during execution. For you, it is obvious it was difficult to figure out what was 'n' at a particular part of the program.
Scanner input = new Scanner (System.in);
int count;
System.out.println ("How many numbers will be entered?");
count = input.nextInt();
//Wait for a valid input
while (count < 1 || count > 100)
{
System.out.println ("ERROR! Valid range 1-100. RE-Enter:");
count = input.nextInt();
}
Additionally, a count of zero should not be valid. It does not make sense to run a program to evaluate zero values (don't bother a program that does nothing). I believe the lowest count should be one instead.
int odd = 0;
int even = 0;
int value;
do
{
System.out.print("Enter a number between 0 and 100: ");
value = input.nextInt();
while (value < 0 || value > 100)
{
System.out.println ("ERROR! Valid range 0-100. RE-Enter:");
value = input.nextInt();
}
if (value % 2 == 0)
{
even++;
}
else
{
odd++;
}
count--; // decrement count to escape loop
} while (count > 0);
System.out.println(even + " even. " + odd + " odd.");
This example uses a do/while loop because in this case, it is OK to enter the loop at least once. This is because you do not allow the user to enter an invalid number of iterations in the first part of the program. I use that count variable directly for loop control (by decrementing its value down to 0), rather than creating another variable for loop control (for instance , 'i').
Another thing, slightly off topic, is that your requirements were not clear. You only indicated that the value was bounded to (inclusive) values between 0 and 100. However, how many times you needed to repeat the evaluation was not really clear. Most people assume 100 was also the upper bound for your counter variable. Because the requirement is not clear, checking a value greater or equal to 1 for the count might be valid, although highly improbable (you don't really want to repeat a million times).
Lastly, you have to pay attention to AND and OR logic in your code. As it was indicated, your second while loop:
while (n >= 0 || n <= 100) {}
Is infinite. Because an OR evaluation only needs one part to evaluate to TRUE, any number entered will allow the loop to continue. Obviously, the intent was not allow values greater than 100. However, entering 150 allows the loop to continue because 150 >= 0. Likewise, -90 also allows the loop to continue because -90 <= 100. This is when pseudocode helps when you are learning. You wanted to express "a VALUE between lower_limit AND upper_limit." If you reverse the logic to evaluate values outside the limit, then you can say " value below lower_limit OR above upper_limit." These pseudocode expressions are very helpful determining which logical operator you need.
I also took the liberty to add a message to prompt the user for a value. Your program expects the user to enter two numbers (count and value) but only one prompt message is provided; unless they enter an out of range value.

extract even numbers from arrayList
ArrayList numberList = new ArrayList<>(Arrays.asList(1,2,3,4,5,6));
numberList.stream().filter(i -> i % 2 == 0).forEach(System.out::println);

Why doesn't my "While Loop" print the computation of finding the average "score"?

I am writing a program that reads a sequence of positive integers input by the user. User will only enter one integer at a time.Then it will compute the average of those integers. The program will end when user enters 0. (0 is not counted in the average).The program will print out the average once the program ends.
Question: My code stops working when I gets to the while loop hence it doesn't compute the input by user, hence prints out nothing. Why doesn't my while loop compute the average from the user's inputs? Appreciate your guidance :)
import java.util.Scanner;
public class AverageOfIntegers {
public static void main(String[] args) {
int integer;
double sum;
sum = 0;
double average;
Scanner input = new Scanner(System.in);
int count; count = 0;
average = 0;
System.out.println("Please enter an integer: ");
integer = input.nextInt();
while (integer != 0) {
count = count + 1;
sum = sum + integer;
average = sum / count;
}
System.out.println("Average = " + average);
}
}

This is because you are never actually summing over more than one integer. The user only ever enters one number. As a result your loop is essentially acting on just the one number. You need to put the input inside the while loop and save a running sum and count there. Something more like this
while (integer != 0) {
count += 1;
sum += integer;
average = sum / count;
integer = input.nextInt();
}

Explanation
First of all, when you define data types, you can set their default value in the definition. Ex:
double sum = 0;
vs
double sum;
sum = 0;
Secondly, sum = sum + integer; is the same as: sum += integer;
Thirdly, count = count + 1; is the same as: count += 1 OR (and better yet), count++;
As for your actual algorithm, there is one problem and one suggestion:
you are not changing integer's value after each loop. So, you can
either do that in the while condition: while ((integer =
input.nextInt()) != 0) { or, at the end of each loop:
while (integer != 0) {
count ++;
sum += integer;
average = sum / count;
integer = input.nextInt();
}
This is a suggestion for technically better code (in my opinion), but it looks better, is more intuitive and requires less calculations to calculate the average after the while loop is done instead of during. That way, you only calculate it once, where needed, vs. every loop, which is not needed.
________________________________________________________________________________
The Code (complete class)
public class AverageOfIntegers {
public static void main(String[] args) {
int integer;
double sum = 0;
double average = 0;
Scanner input = new Scanner(System.in);
int count = 0;
System.out.println("Please enter an integer: ");
// set integer = to the nextInt() while looping so it calculates properly
while ((integer = input.nextInt()) != 0) {
count ++;
sum += integer;
}
average = sum / count; // calculate the average after the while-loop
System.out.println("Average = " + average);
}
}
________________________________________________________________________________
Example input/output:
Please enter an integer:
5
10
15
0
Average = 10.0
So it did 5 + 10 + 15 = 30 (which is the sum), and then the average is 30 / 3 (30 is the sum, 3 is the count), and that gave you Average = 10.0.

You need to move integer = input.nextInt(); inside the loop, so your program will collect inputs in a loop. See the corrected version:
import java.util.Scanner;
public class AverageOfIntegers {
public static void main(String[] args) {
int integer = 0, count = 0;
double sum = 0.0, average = 0.0;
Scanner input = new Scanner(System.in);
System.out.println("Please enter an integer: ");
integer = input.nextInt();
while (integer != 0) {
count = count + 1;
sum = sum + integer;
System.out.println("Please enter an integer: ");
integer = input.nextInt();
}
average = sum / count;
System.out.println("Average = " + average);
}
}

The problem is that the input.nextInt() should be part of the loop. The way you wrote it, the code gooes into an infinite loop whenever the first input is non-zero. Instead, do:
while ((integer = input.nextInt()) != 0) {
count = count + 1;
sum = sum + integer;
average = sum / count;
}

In the loop:
while (integer != 0) {
count = count + 1;
sum = sum + integer;
average = sum / count;
}
This will only stops when integer is 0, but this variable is not changing in the loop, so it will never be 0 if it wasn't already in the first place.
According to what you said you want to do, you should probably repeat the call to integer = input.nextInt(); inside your loop, lke this:
System.out.println("Please enter an integer: ");
integer = input.nextInt();
while (integer != 0) {
count = count + 1;
sum = sum + integer;
System.out.println("Please enter an integer: ");
integer = input.nextInt();
}
average = sum / count;
Also, as others have said, you only need to compute the average once after the loop, so I moved it too.