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how to add elements in each row seperately and print output in 2D array

Lets assume i have a 2x3 matrix where row denote student and column denote marks.
eg:[[67,80,56],
[32,26,31]]
need to find the average of each row and assign a grade based on avg. if avg>40 then return "p" else return "F".
import java.util.*;
public class Solution
{
public static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
int n=2;
int m=5;
int mark[][]=new int[n][m];
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
mark[i][j]=sc.nextInt();
}
}
String result=grade(mark);
System.out.println("RESULT:"+result);
}
public static String grade(int mark[][])
{
int n=2,m=5,avg=0;
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
int sum=0;
sum=sum+ mark[i][j];
if(j==m)
{
avg=sum/m;
}
}
if(avg>=90)
{
return "A+";
}
else if(avg<40)
{
return "F";
}
}
return null;
}
}
IN the above code my my average value is initialised to 0.scope of average in for loop is not reflected in outside loop.how to correct itenter image description here

You are declaring the sum variable inside your 2nd loop it should be declared before your 2nd loop starts. If you declare it inside it will not have the calculated value from the previous iteration.
Also you can easily calculate avg outside the loop and then gor for your cheking.
EG:
int n=2;
int m=3;
int[][] mark = { { 67,80,56 }, { 32,26,31} };
// grade function
for (int i = 0; i < n; i++) {
int sum=0;
for (int j = 0; j < m ; j++) {
sum=sum+ mark[i][j];
}
int avg = sum / m;
System.out.println(avg);
//your code to check pass/fail
}
Also Don't return if you want to print values for each row. Instead of return use print there.
if(avg>=90)
{
System.out.println("GRADE")
}

First of all, if you need to print the grade of every student separately you will need to call your grade method for every student's marks. So put your method call in a for loop.
for(int i = 0; i < n; i++) {
System.out.println(grade(mark[i]))
}
Consequently, the definition of your method changes to
public static String grade(int mark[]) {
...
}
And at last, your if(j==m) check will always be false, cause in your code, j will never reach m in your for loop. It comes to a simple calculation of array element's sum.
public static String grade(int mark[][])
{
int m = 5, avg = 0;
int sum = 0;
for(int j = 0;j < m; j++) {
sum = sum + mark[i][j];
}
avg = sum / m;
if(avg >= 90)
{
return "A+";
}
else if(avg < 40)
{
return "F";
}
}

Need to find out the duplicate element in array without using Hashmaps

I am a newbie here. I wanted to print out the duplicate elements in an array.
This code will print out the duplicate elements.
Suppose I'm taking an array of size 5 with elements [1,2,5,5,5]
This code will print:
Duplicate elements: 5,5,5 //(since 5 is being repeated thrice.)
But I want the output something like this
Duplicate Elements: 5 //( instead of printing 5 thrice)
import java.util.*;
import java.util.Scanner;
public class duplicateArray{
public static void main(String args[]){
Scanner sc=new Scanner(System.in);
System.out.print("Enter the size of the array: ");
int x =sc.nextInt();
int arr[]=new int[x];
int i,count=0;
for(i=0;i<x;i++){
arr[i]=sc.nextInt();
}
System.out.print("Array: ");
for(i=0;i<x;i++){
System.out.print(arr[i]+" ");
}
System.out.println(" ");
System.out.print("Duplicate elements: ");
for(i=0;i<arr.length;i++){
for(int j=i+1;j<arr.length;j++){
if(arr[i]==arr[j]){
System.out.print(arr[j]+" ");
}
}
}
}
}

The following code does it without creating any additional data structure. For each element, it counts the number of duplicates previously encountered and only prints the first duplicate.
If I were doing this in the real world, I would use a Set but I'm assuming you haven't learnt about them yet, so I'm only using the array that you've already created.
import java.util.Scanner;
public class DuplicateArray {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.print("Enter the size of the array: ");
int x = sc.nextInt();
int[] arr = new int[x];
System.out.print("Enter " + x + " values: ");
for (int i = 0; i < x; i++) {
arr[i] = sc.nextInt();
}
System.out.print("Array: ");
for (int i = 0; i < x; i++) {
System.out.print(arr[i]+" ");
}
System.out.println();
System.out.print("Duplicate elements:");
for (int i = 0; i < arr.length; i++) {
int numDups = 0;
for (int j = 0; j < i; j++) {
if (arr[i] == arr[j]) {
numDups++;
}
}
if (numDups == 1) {
System.out.print(" " + arr[i]);
}
}
System.out.println();
}
}

One solution is to create a separate List to store any duplicates found.
That, in addition to using the .contains() method of the List, you can ensure only one entry per int is made.
public static void main(String[] args) {
// Sample array of ints
int[] ints = {1, 1, 4, 5, 2, 34, 7, 5, 3};
// Create a separate List to hold duplicated values
List<Integer> duplicates = new ArrayList<>();
// Find duplicates
for (int i = 0; i < ints.length; i++) {
for (int j = 0; j < ints.length; j++) {
if (ints[i] == ints[j] && // Are the ints the same value?
i != j && // Ignore if we're looking at the same index
!duplicates.contains(ints[i])) { // Check if our List of duplicates already has this entry
duplicates.add(ints[i]); // Add to list of duplicates
}
}
}
System.out.println("Duplicates: " + duplicates);
}
Output:
Duplicates: [1, 5]

This is pretty simple however you need to sort array before. All you need to know if duplicate for an element exists or not and do the printing in the outside for loop. Rest is described in comments
Arrays.sort(arr); // Sort Array
for (int i = 0; i < arr.length; i++) {
boolean hasDuplicate = false; // Assume that arr[i] is not repeating
for (int j = i + 1; j < arr.length; j++) {
// Check if it is repeating
if (arr[i] == arr[j]) {
// If it repeats
hasDuplicate = true;
}
// Since array is sorted we know that there is no value of arr[i] after this
if (arr[i] != arr[j]) {
// Set i to the last occurrence of arr[i] value
i = j - 1;
break; // Since there no occurrence of arr[i] value there is no need to continue
}
}
// Print the element at i
if (hasDuplicate)
System.out.print(arr[i] + " ");
// In next iteration loop will start from the index next to the last occurrence of value of arr[i]
}

System.out.println("Duplicate Elements : ");
for(int i = 0; i<arr.length; i++){
boolean isDuplicate = false;
for(int k=0;k<i;k++){
if(arr[i]== arr[k]){
isDuplicate = true;
break;
}
}
if(isDuplicate){
continue;
}
int count = 0;
for(int j=0; j<arr.length; j++){
if(arr[i] == arr[j]){
count++;
}
if(count >1){
System.out.println(arr[i]);
break;
}
}
}

Without using Hashmaps, I think your best option would be to first sort the array and then count the duplicates. Since the array is now in order you can print the duplicates after each number switch!
If this is for an assignment go ahead and google bubble sort and implement it as a method.

randomly generate 100 unique numbers using Math.random [duplicate]

my intend is to use simplest java (array and loops) to generate random numbers without duplicate...but the output turns out to be 10 repeating numbers, and I cannot figure out why.
Here is my code:
int[] number = new int[10];
int count = 0;
int num;
while (count < number.length) {
num = r.nextInt(21);
boolean repeat = false;
do {
for (int i=0; i<number.length; i++) {
if (num == number[i]) {
repeat = true;
} else if (num != number[i] && i == count) {
number[count] = num;
count++;
repeat = true;
}
}
} while (!repeat);
}
for (int j = 0; j < number.length; j++) {
System.out.print(number[j] + " ");
}

How about you use a Set instead? If you also want to keep track of the order of insertion you can use a LinkedHashSet.
Random r = new Random();
Set<Integer> uniqueNumbers = new HashSet<>();
while (uniqueNumbers.size()<10){
uniqueNumbers.add(r.nextInt(21));
}
for (Integer i : uniqueNumbers){
System.out.print(i+" ");
}
A Set in java is like an Array or an ArrayList except it handles duplicates for you. It will only add the Integer to the set if it doesn't already exist in the set. The class Set has similar methods to the Array that you can utilize. For example Set.size() is equivalent to the Array.length and Set.add(Integer) is semi-equivalent to Array[index] = value. Sets do not keep track of insertion order so they do not have an index. It is a very powerful tool in Java once you learn about it. ;)
Hope this helps!

You need to break out of the for loop if either of the conditions are met.
int[] number = new int[10];
int count=0;
int num;
Random r = new Random();
while(count<number.length){
num = r.nextInt(21);
boolean repeat=false;
do{
for(int i=0; i<number.length; i++){
if(num==number[i]){
repeat=true;
break;
}
else if(i==count){
number[count]=num;
count++;
repeat=true;
break;
}
}
}while(!repeat);
}
for(int j=0;j<number.length;j++){
System.out.print(number[j]+" ");
}
This will make YOUR code work but #gonzo proposed a better solution.

Your code will break the while loop under the condition: num == number[i].
This means that if the pseudo-generated number is equal to that positions value (the default int in java is 0), then the code will end execution.
On the second conditional, the expression num != number[i] is always true (otherwise the code would have entered the previous if), but, on the first run, when i == count (or i=0, and count=0) the repeat=true breaks the loop, and nothing else would happen, rendering the output something such as
0 0 0 0 0 0...

Try this:
int[] number = new int[10];
java.util.Random r = new java.util.Random();
for(int i=0; i<number.length; i++){
boolean repeat=false;
do{
repeat=false;
int num = r.nextInt(21);
for(int j=0; j<number.length; j++){
if(number[j]==num){
repeat=true;
}
}
if(!repeat) number[i]=num;
}while(repeat);
}
for (int k = 0; k < number.length; k++) {
System.out.print(number[k] + " ");
}
System.out.println();
Test it here.

I believe the problem is much easier to solve. You could use a List to check if the number has been generated or not (uniqueness). Here is a working block of code.
int count=0;
int num;
Random r = new Random();
List<Integer> numbers = new ArrayList<Integer>();
while (count<10) {
num = r.nextInt(21);
if(!numbers.contains(num) ) {
numbers.add(num);
count++;
}
}
for(int j=0;j<10;j++){
System.out.print(numbers.get(j)+" ");
}
}

Let's start with the most simple approach, putting 10 random - potentially duplicated - numbers into an array:
public class NonUniqueRandoms
{
public static void main(String[] args)
{
int[] number = new int[10];
int count = 0;
while (count < number.length) {
// Use ThreadLocalRandom so this is a contained compilable unit
number[count++] = ThreadLocalRandom.current().nextInt(21);
}
for (int j = 0; j < number.length; j++) {
System.out.println(number[j]);
}
}
}
So that gets you most of the way there, the only thing you know have to do is pick a number and check your array:
public class UniqueRandoms
{
public static void main(String[] args)
{
int[] number = new int[10];
int count = 0;
while (count < number.length) {
// Use ThreadLocalRandom so this is a contained compilable unit
int candidate = ThreadLocalRandom.current().nextInt(21);
// Is candidate in our array already?
boolean exists = false;
for (int i = 0; i < count; i++) {
if (number[i] == candidate) {
exists = true;
break;
}
}
// We didn't find it, so we're good to add it to the array
if (!exists) {
number[count++] = candidate;
}
}
for (int j = 0; j < number.length; j++) {
System.out.println(number[j]);
}
}
}

The problem is with your inner 'for' loop. Once the program finds a unique integer, it adds the integer to the array and then increments the count. On the next loop iteration, the new integer will be added again because (num != number[i] && i == count), eventually filling up the array with the same integer. The for loop needs to exit after adding the unique integer the first time.
But if we look at the construction more deeply, we see that the inner for loop is entirely unnecessary.
See the code below.
import java.util.*;
public class RandomDemo {
public static void main( String args[] ){
// create random object
Random r = new Random();
int[] number = new int[10];
int count = 0;
int num;
while (count < number.length) {
num = r.nextInt(21);
boolean repeat = false;
int i=0;
do {
if (num == number[i]) {
repeat = true;
} else if (num != number[i] && i == count) {
number[count] = num;
count++;
repeat = true;
}
i++;
} while (!repeat && i < number.length);
}
for (int j = 0; j < number.length; j++) {
System.out.print(number[j] + " ");
}
}
}

This would be my approach.
import java.util.Random;
public class uniquerandom {
public static void main(String[] args) {
Random rnd = new Random();
int qask[]=new int[10];
int it,i,t=0,in,flag;
for(it=0;;it++)
{
i=rnd.nextInt(11);
flag=0;
for(in=0;in<qask.length;in++)
{
if(i==qask[in])
{
flag=1;
break;
}
}
if(flag!=1)
{
qask[t++]=i;
}
if(t==10)
break;
}
for(it=0;it<qask.length;it++)
System.out.println(qask[it]);
}}

public String pickStringElement(ArrayList list, int... howMany) {
int counter = howMany.length > 0 ? howMany[0] : 1;
String returnString = "";
ArrayList previousVal = new ArrayList()
for (int i = 1; i <= counter; i++) {
Random rand = new Random()
for(int j=1; j <=list.size(); j++){
int newRand = rand.nextInt(list.size())
if (!previousVal.contains(newRand)){
previousVal.add(newRand)
returnString = returnString + (i>1 ? ", " + list.get(newRand) :list.get(newRand))
break
}
}
}
return returnString;
}

Create simple method and call it where you require-
private List<Integer> q_list = new ArrayList<>(); //declare list integer type
private void checkList(int size)
{
position = getRandom(list.size()); //generating random value less than size
if(q_list.contains(position)) { // check if list contains position
checkList(size); /// if it contains call checkList method again
}
else
{
q_list.add(position); // else add the position in the list
playAnimation(tv_questions, 0, list.get(position).getQuestion()); // task you want to perform after getting value
}
}
for getting random value this method is being called-
public static int getRandom(int max){
return (int) (Math.random()*max);
}

Java set/setElementAt not setting the right value

I need to find all the permutations for a given n(user input) without backtracking.
What i tried is:
import java.util.Scanner;
import java.util.Vector;
class Main {
private static int n;
private static Vector<Vector<Integer>> permutations = new Vector<>();
private static void get_n() {
Scanner user = new Scanner(System.in);
System.out.print("n = ");
n = user.nextInt();
}
private static void display(Vector<Vector<Integer>> permutations) {
for (int i = 0; i < factorial(n) - 1; ++i) {
for (int j = 0; j < n; ++j) {
System.out.print(permutations.elementAt(i).elementAt(j) + " ");
}
System.out.println();
}
}
private static int factorial(int n) {
int result = 1;
for (int i = 1; i <= n; ++i) {
result *= i;
}
return result;
}
private static int max(Vector<Integer> permutation) {
int max = permutation.elementAt(0);
for (int i = 1; i < permutation.size(); ++i)
if (permutation.elementAt(i) > max)
max = permutation.elementAt(i);
return max;
}
// CHECKS FOR ELEMENT COUNT AND 0 - (n-1) APPARITION
public static int validate_permutation(Vector<Integer> permutation) {
// GOOD NUMBER OF ELEMENTS
if (max(permutation) != permutation.size() - 1)
return 0;
// PROPER ELEMENTS APPEAR
for (int i = 0; i < permutation.size(); ++i)
if (!permutation.contains(i))
return 0;
return 1;
}
private static Vector<Integer> next_permutation(Vector<Integer> permutation) {
int i;
do {
i = 1;
// INCREMENT LAST ELEMENT
permutation.set(permutation.size() - i, permutation.elementAt(permutation.size() - i) + 1);
// IN A P(n-1) PERMUTATION FOUND n. "OVERFLOW"
while (permutation.elementAt(permutation.size() - i) == permutation.size()) {
// RESET CURRENT POSITION
permutation.set(permutation.size() - i, 0);
// INCREMENT THE NEXT ONE
++i;
permutation.set(permutation.size() - i, permutation.elementAt(permutation.size() - i) + 1);
}
} while (validate_permutation(permutation) == 0);
// OUTPUT
System.out.print("output of next_permutation:\t\t");
for (int j = 0; j < permutation.size(); ++j)
System.out.print(permutation.elementAt(j) + " ");
System.out.println();
return permutation;
}
private static Vector<Vector<Integer>> permutations_of(int n) {
Vector<Vector<Integer>> permutations = new Vector<>();
// INITIALIZE PERMUTATION SET WITH 0
for (int i = 0; i < factorial(n); ++i) {
permutations.addElement(new Vector<>());
for(int j = 0; j < n; ++j)
permutations.elementAt(i).addElement(0);
}
for (int i = 0; i < n; ++i)
permutations.elementAt(0).set(i, i);
for (int i = 1; i < factorial(n); ++i) {
// ADD THE NEXT PERMUTATION TO THE SET
permutations.setElementAt(next_permutation(permutations.elementAt(i - 1)), i);
System.out.print("values set by permutations_of:\t");
for (int j = 0; j < permutations.elementAt(i).size(); ++j)
System.out.print(permutations.elementAt(i).elementAt(j) + " ");
System.out.println("\n");
}
System.out.print("\nFinal output of permutations_of:\n\n");
display(permutations);
return permutations;
}
public static void main(String[] args) {
get_n();
permutations.addAll(permutations_of(n));
}
}
Now, the problem is obvious when running the code. next_permutation outputs the correct permutations when called, the values are set correctly to the corresponding the vector of permutations, but the end result is a mass copy of the last permutation, which leads me to believe that every time a new permutation is outputted by next_permutation and set into the permutations vector, somehow that permutation is also copied over all of the other permutations. And I can't figure out why for the life of me.
I tried both set, setElementAt, and an implementation where I don't initialize the permutations vector fist, but add the permutations as they are outputted by next_permutation with add() and I hit the exact same problem. Is there some weird way in which Java handles memory? Or what would be the cause of this?
Thank you in advance!

permutations.setElementAt(next_permutation(permutations.elementAt(i - 1)), i);
This is literally setting the vector at permutations(i) to be the same object as permutations[i-1]. Not the same value - the exact same object. I think this the source of your problems. You instead need to copy the values in the vector.

Returning wrong percentage

I am trying to calculate how many times two people in a group have the same birthday when given a size of the group. I am also given how many times the simulation is ran. I am trying to return the correct percentage for how many times we have two people share the same birthday out of the given amount of simulations.
I created an array first and then called a method to put the elements in a hashMap which would then show when there are two of the same values in the hashMap. However, I am not getting the correct percentage when running on Android Studio. In fact I am getting a percentage way off. I also declared a global static match variable of type int above this block.
/**
* sameBday: Create a word count mapping from an array
*/
public void sameBday(int[] valueHolder) {
Map<Integer, Integer> myMap = new HashMap<Integer, Integer>();
for(int number: valueHolder){
if(!myMap.containsKey(number)){
myMap.put(number, 1);
}
else if(myMap.containsKey(number)){
myMap.put(number, myMap.get(number) + 1);
match++;
break;
}
}
}
public double calculate(int size, int count) {
double percentage = 0.0;
int[] myArray = new int[size];
for(int i = 1; i <= count; i++){
Random r = new Random(i);
for(int j = 0; j < size; j++){
myArray[j] = r.nextInt(365) + 1;
}
sameBday(myArray);
if(i == count){
percentage = (match * (100.0/i));
}
}
return percentage;
}

Well your code is full of weird things, but that's OK we all did that. The first thing is Map, you don't need it. You can create just good old for loop and by additional check you will not compare the same person (it is i != j condition), but if you really want to do this by map you need to at the end of adding number (as key) to map check if some value of key is higher than 1, if true it's a match.
How to do something at the end of loop?
if(i == count){
percentage = (match * (100.0/i));
}
No, just do this after loop :)
//At the beginning there is int match = 0;
public void sameDayBirthday(int[] birthdays) {
for(int i = 0; i < birthdays.length; i++) {
for(int j = 0; j < birthdays.length; j++) {
if(birthdays[i] == birthdays[j] && i != j) {
match++;
return;
}
}
}
}
public double calculate(int size, int count) {
int[] birthdays = new int[size];
Random r = new Random();
for(int i = 1; i <= count; i++){ //looping through i counts (or 20 counts in this case
for(int j = 0; j < size; j++){ //looping through j times every i iteration
birthdays[j] = r.nextInt(365) + 1;
}
sameDayBirthday(birthdays);
}
return (match * (100.0/(double) count));
}
This code by calling calculate(23, 1000000) got me 50.7685% chance, for 22 persons 47.48690%
I am sorry if I offend you I didn't mean it. Leave a comment if you have questions.

I would use a HashSet and skip the sameBday function:
public double calculate(int size, int count) {
int match = 0;
Random r = new Random();
for(int i = 1; i <= count; i++){ //looping through i counts (or 20 counts in this case
Set<Integer> birthdays = new HashSet<Integer>(size);
for(int j = 0; j < size; j++){ //looping through j times every i iteration
Integer birthday = r.nextInt(365) + 1;
if (birthdays.contains(birthday)) {
match++;
break;
} else {
birthdays.add(birthday);
}
}
}
return (match * (100.0/count));
}