Adding Numbers and printing the sum vertically using arrays - java

I have to create a program which adds two integers and prints the sum vertically.
For example, I have.
a=323, b=322.
The output should be:
6
4
5
I've created the code for when the integers are up to two digits, but I want it to work for at least three digits.
Below is the best I could think of.
It may be completely wrong, but the only problem I'm facing is the declaration of array.
It says that the array might not be initialized.
If I set it to null then also it won't assign values to it later.
I know maybe I'm making a big mistake here, but I'll really appreciate if anyone could help me out.
Please keep in mind that I must not use any other functions for this code.
Hope I'm clear.
public class Vert
{
public static void main(String args[])
{
int n,i=0,j,a=323,b=322;
int s[];
n=a+b;
while(n>9)
{
s[i]=n%10;
i++;
s[i]=n/10;
if(s[i]>9)
{
n=s[i];
}
}
j=i;
for(j=i;j>=0;j--)
{
System.out.println(+s[j]);
}
}
}

String conversion seems like cheating, so here's a Stack.
int a = 323, b = 322;
java.util.Stack<Integer> stack = new java.util.Stack<>();
int n = a + b;
while (n > 0) {
stack.push(n % 10);
n = n / 10;
}
while (!stack.isEmpty())
System.out.println(stack.pop());
If an array is required, you need two passes over the sum
int a = 323, b = 322;
// Get the size of the array
int n = a + b;
int size = 0;
while (n > 0) {
size++;
n = n / 10;
}
// Build the output
int s[] = new int[size];
n = a + b;
for (int i = size - 1; n > 0; i--) {
s[i] = n % 10;
n = n / 10;
}
// Print
for (int x : s) {
System.out.println(x);
}

To initialize an array, you need to specify the size of your array as next:
int s[] = new int[mySize];
If you don't know the size of your array, you should consider using a List of Integer instead as next:
List<Integer> s = new ArrayList<Integer>();
Here is how it could be done:
// Convert the sum into a String
String result = String.valueOf(a + b);
for (int i=0; i <result.length();i++) {
// Print one character corresponding to a digit here per line
System.out.println(result.charAt(i));
}

I'd do it like this:
int a = 322;
int b = 322;
int sum = a + b;
String s = Integer.toString(sum);
for(int i = 0; i < s.length(); i++) {
System.out.println(s.charAt(i));
}
But your problem looks like an array is required.
The steps are same as in my solution:
Use int values
Sum the int values (operation)
Convert the int value in an array/string
Output the array/string

Related

Adding elements to the beginning of an array of ints

Working on an addBefore() method that adds a new element to the beginning of an array of ints and then causes the existing elements to increase their index by one.
This is what is showing in the console when trying to run --
java.lang.RuntimeException: Index 1 should have value 11 but instead has 0
at IntArrayListTest.main(IntArrayListTest.java:67)
Below is the code I have so far.
public class IntArrayList {
private int[] a;
private int length;
private int index;
private int count;
public IntArrayList() {
length = 0;
a = new int[4];
}
public int get(int i) {
if (i < 0 || i >= length) {
throw new ArrayIndexOutOfBoundsException(i);
}
return a[i];
}
public int size() {
return length;
}
public void set(int i, int x) {
if (i < 0 || i >= a.length) {
throw new ArrayIndexOutOfBoundsException(i);
}
a[i] = x;
}
public void add(int x) {
if (length >= a.length) {
int[] b = new int[a.length * 2];
for (int i = 0; i < a.length; i++) {
b[i] = a[i];
}
a = b;
//count += 1;
}
a[length] = x;
count++;
length = length + 1;
}
public void addBefore(int x) {
int[] b = new int[a.length*2];
for (int i = 0; i < a.length; i++) {
b[i+a.length] = a[i];
}
a = b;
a[index] = x;
length ++;
}
}
Whether you add first or last, you need to only grow the array size if it is already full.
The count field seems to be exactly the same as length, and index seems unused and meaningless as a field, so remove them both.
To rearrange values in an array, use this method:
System.arraycopy(Object src, int srcPos, Object dest, int destPos, int length)
You two "add" methods should then be:
public class IntArrayList {
private int[] a; // Underlying array
private int length; // Number of added elements in a
// other code
public void add(int x) {
if (length == a.length) {
int[] b = new int[a.length * 2];
System.arraycopy(a, 0, b, 0, length);
a = b;
}
a[length++] = x;
}
public void addBefore(int x) {
if (length < a.length) {
System.arraycopy(a, 0, a, 1, length);
} else {
int[] b = new int[a.length * 2];
System.arraycopy(a, 0, b, 1, length);
a = b;
}
a[0] = x;
length++;
}
}
If the answer requires you to do the looping yourself then something like this should work fine (one of a few ways to do this, but is O(n)) :
public void addBefore(int x) {
if(length + 1 >= a.length){
int[] b = new int[a.length*2];
b[0] = x;
for (int i = 0; i < length; i++) {
b[i + 1] = a[i];
}
a = b;
} else {
for (int i = length; i >= 0 ; i--) {
a[i + 1] = a[i];
}
a[0] = x;
}
length++;
}
I noticed this started running a "speed test" - not sure how useful a test like that is, as it would be based on cpu performance, rather than testing complexity of the algorithm ..
you had three problems with your solution:
you increased the length of a every time the method was called. this would quickly create an OutOfMemoryException
when you copied values from a to b, you did b[i+a.length] = a[i]; which means the values would be copied to the middle of b instead of shift just one place
at the end, you put the new value in the end of the array instead of at the beginning.
all this I was able to see because I used a debugger on your code. You need to start using this tool if you want to be able to detect and fix problems in your code.
so fixed solution would do this:
check if a is full (just like it is done with add() method) and if so, create b, and copy everything to it and so on)
move all values one place ahead. the easiest way to do it is to loop backwards from length to 0
assign new value at the beginning of the array
here is a working solution:
public void addBefore(int x) {
// increase length if a is full
if (length >= a.length) {
int[] b = new int[a.length * 2];
for (int i = 0; i < a.length; i++) {
b[i] = a[i];
}
a = b;
}
// shift all values one cell ahead
for (int i = length; i > 0; i--) {
a[i] = a[i-1];
}
// add new value as first cell
a[0] = x;
length ++;
}
}
You can use the existing Java methods from the Colt library. Here is a small example that uses a Python syntax (to make the example code small I use Jython):
from cern.colt.list import IntArrayList
a=IntArrayList()
a.add(1); a.add(2) # add two integer numbers
print "size=",a.size(),a
a.beforeInsert(0, 10) # add 10 before index 0
print "size=",a.size(),a
You can use DataMelt program to run this code. The output of the above code is:
size= 2 [1, 2]
size= 3 [10, 1, 2]
As you can see, 10 is inserted before 1 (and the size is increased)
Feel free to change the codding to Java, i.e. importing this class as
import cern.colt.list.IntArrayList
IntArrayList a= new IntArrayList()
You could use an ArrayList instead and then covert it to an Integer[] Array which could simplify your code. Here is an example below:
First create the ArrayList:
ArrayList<Integer> myNums = new ArrayList<Integer>();
Next you can add the values that you want to it, but I chose to just add the numbers 2-5, to illustrate that we can make the number 1 the first index and automatically increment each value by one index. That can simplify your addBefore() method to something such as this:
public static void addBefore(ArrayList<Integer> aList) {
int myInt = 1;
aList.add(0, myInt);
}
Since your ArrayList has ONE memory location in Java, altering the Array within a method will work (this would also work for a regular Array). We can then add any value to the beginning of the ArrayList. You can pass an Integer to this method as the second argument (int x), if you want, but I simply created the myInt primitive to simplify the code. I know that in your code you had the (int x) parameter, and you can add that to this method. You can use the ArrayList.add() method to add the int to index 0 of the Array which will increment each Array element by 1 position. Next you will need to call the method:
addBefore(myNums);//You can add the int x parameter and pass that as an arg if you want here
Next we can use the ArrayList.toArray() method in order to covert the ArrayList to an Integer Array. Here is an example below:
Integer[] integerHolder = new Integer[myNums.size()];
Integer[] numsArray = (Integer[])myNums.toArray(integerHolder);
System.out.println(Arrays.toString(numsArray));
First we create an ArrayHolder that will be the same size as your ArrayList, and then we create the Array that will store the elements of the ArrayList. We cast the myNums.toArray() to an Integer Array. The results will be as follows. The number 1 will be at index 0 and the rest of your elements will have incremented by 1 index:
[1, 2, 3, 4, 5]
You could do the entire process within the addBefore() method by converting the Array to an ArrayList within the method and adding (int x) to the 0 index of the ArrayList before converting it back into an Array. Since an ArrayList can only take a wrapper class object you'll simply need to convert the int primitive Array into the type Integer for this to work, but it simplifies your addBefore() method.

How to use create 2 Array with User input with Std.readAllInts()

This is how it should work, i Put in put for the 1 st array like: 2 3 1 2 3 4 5 6 then 2 and 3 are row and colum and the rest are the values. Problem is the 1st array work, but when i reach EOF ( ctrl+z) then there is out of bound exception. Which mean i cant input value for the 2nd Array like the 1st one. I know there is anotherway where that i can declare array size first then value. But how could i fix this f i still want to usr StdIn.readAllInts() ?
public class MatrixMult {
public static void main(String[] args){
System.out.println("First Matrix Config");
int[] einGabeMatrix1= StdIn.readAllInts();
int zeileM1 = einGabeMatrix1[0];
int spalteM1 = einGabeMatrix1[1];
int[][] ersteMatrix= new int [zeileM1][spalteM1];
int k=2;
int sum;
for(int i=0;i<zeileM1-1;i++){
for(int j=0;j<spalteM1-1;j++){
ersteMatrix[i][j]=einGabeMatrix1[k];
k++;
}
}
System.out.println("Second Matrix Config");
int[] einGabeMatrix2 = StdIn.readAllInts();
int zeileM2 = einGabeMatrix2[0];
int spalteM2 = einGabeMatrix2[1];
int h=2;
int[][] zweiteMatrix= new int [zeileM2][spalteM2];
for(int m=0;m<zeileM2-1;m++){
for(int n=0;n<spalteM2-1;n++){
zweiteMatrix[m][n]=einGabeMatrix2[h];
h++;
}
}
int[][] ergebnisMatrix= new int [zeileM1][spalteM2];
for (int t = 0; t < zeileM1; t++) {
for (int c = 0; c < spalteM2; c++) {
sum = 0;
for (int d = 0; d < spalteM1; d++) {
sum = sum + ersteMatrix[t][d] * zweiteMatrix[d][c];
}
ergebnisMatrix[t][c] = sum;
}
}
for(int i=0;i<zeileM1;i++){
for(int j=0;j<spalteM1;j++){
System.out.println(ergebnisMatrix[i][j]);
}
}
}
}
// This is StdIn.readAllInts(), standard method by java.
public static int[] readAllInts() {
String[] fields = readAllStrings();
int[] vals = new int[fields.length];
for (int i = 0; i < fields.length; i++)
vals[i] = Integer.parseInt(fields[i]);
return vals;
}
It looks like the issue is coming from the fact that StdIn.readAllInts() reads until the EOF. There will be no values left to read by the time your code gets to the second call.
I would suggest instead using the StdIn.readInt() call to read each integer one at a time and then you can use it within your loop to read the exact number of values your need.
Here is an example of how you could get the first 2 integers to find your matrix size:
int zeileM1 = StdIn.readInt();
int spalteM1 = StdIn.readInt();
You will also need to apply this method in your for loops to read the data into the matrix.

What is wrong with this simple application of binary addition algorithm in Java?

I am trying to apply a simple binary addition algorithm in Java, but I am getting errors repeatedly. Could anyone please help me. Thanks!
public class BasicBinaryAddition {
public static void main(String[] args)
{
int []sum = new int [3];
int []a = {0,1,0};
int []b = {1,1,0};
int carry =0;
int bitIndex;
for (bitIndex =0; bitIndex < a.length; bitIndex++)
{
int bitSum = a[bitIndex] + b[bitIndex] + carry;
sum[bitIndex] = bitSum%2;
double d = bitSum/2;
carry = (int) Math.floor(d);
}
sum[bitIndex] = carry;
for (int i = 0 ; i <= sum.length-1; i++)
System.out.print(sum[I]+"");
}
}
Your sum array is too short to hold the full result. Make it length 4, this will avoid the ArrayIndexOutOfBoundsException.
The way you calculate the carry could also be simplified to
carry = bitSum / 2;
This is an integer division and yields an integer result.
Adding two 3-bit numbers could result in 4 bits. So, you should declare
int []sum = new int [4];
Also, you should use small i in the last for-loop
System.out.print(sum[i]+"");
You should put (sum[bitIndex] = carry) inside the loop;
public class BasicBinaryAddition {
public static void main(String[] args)
{
int []sum = new int [3];
int []a = {0,1,0};
int []b = {1,1,0};
int carry =0;
int bitIndex;
for (bitIndex =0; bitIndex < a.length-1; bitIndex++)
{
int bitSum = a[bitIndex] + b[bitIndex] + carry;
sum[bitIndex] = bitSum%2;
double d = bitSum/2;
carry = (int) Math.floor(d);
sum[bitIndex] = carry;
}
for (int i = 0 ; i <= sum.length-1; i++)
System.out.print(sum[I]+"");
}
}
This looks like homework so I will just point you in the right direction.
The first problem is in your line:
System.out.print(sum[I]+"");
You are using "I" instead of "i" so the compiler is upset that it can't find a variable "I".
The second problem is that you have an overflow problem. If you add 2 binary numbers of length 3 it is possible to have an answer that is 4 bits long.
101 + 100 = 1001
When you perform the line: sum[bitIndex] = carry, bit index equals 3, which means the 4th item of the array sum (remember Java arrays start from index 0). However sum is declared to be of length 3, so Java's head explodes in rage that you tried to access an element of an array that is out of bounds. The answer to this problem is to declare sum to be of length 4.
Another thing that can be cleaned up are the lines:
double d = bitSum/2;
carry = (int) Math.floor(d);
This can be simplified to:
carry = bitSum/2;
This works because carry and bitSum are both integers, so Java will perform integer division, which just ignore anything that would happen after the decimal in regular division.

How to add to array without ArrayIndexOutOfBoundsException?

Can anyone answer this question?
public class AddingArray {
public static void main(String[] args){
int arry1[] = {2,3,4,5,6,7,9};
int arry2[] = {4,3,7,9,3,5};
for(int i = 0; i <arry1.length; i++){
int result = arry1[i] + arry2[i];
System.out.println("Result "+result);
}
}
}
Whenever I try executing the above code I get the error Exception in
thread "main" java.lang.ArrayIndexOutOfBoundsException: 6 at
basics.AddingArray.main(AddingArray.java:9)
But,my output should be like this 6,6,11,14,9,12,9
As people have mentioned, one of yours arrays is literally shorter than the other. Take two blocks and overlay them over only one block. The second (in this case index 1 block) would fall into the abyss, because the block that was supposed to catch it never existed.
I would make sure both of them are of the same size. If you do want to leave em as they are, I would do this:
int result = 0;
try
{
for(int i = 0, length = array2.length; i < length; i++)
{
result = array1[i] + array2[i];
System.out.println("Result is: " + result);
}
catch(Exception e)
{
System.out.println("You tried to do something that resulted in an error");
System.out.println("Your previous result was: " + result);
}
}
SO, assuming that I still recall how to do basic arrays, what this code will do is that it will catch any errors thrown by your code.
Let's make this as simple and understandable as possible, with no fancy annotations:
You have two int arrays, of not equal lengths, and you wish to add the index-paired numbers to an array of the sums. However, if one of the arrays does not have any more numbers, the value of the longest array will be the result.
public class AddingArray {
public static void main(String[] args){
int arry1[]={2,3,4,5,6,7,9};
int arry2[]={4,3,7,9,3,5};
You need to determine the length of the longest array. You can do this with a Math.max()-method, where you give the length of each array as parameters:
int biggestArrayLength = Math.max(arry1.length, arry2.length);
Then, instead of for(int i=0;i<arry1.length;i++){, you write:
for(int i=0;i<biggestArrayLength;i++){
Now it doesn't matter which of the two arrays is the biggest one.
Inside the loop, I would define two ints, representing a value from each of the two arrays:
int value1 = arry1[i];
int value2 = arry2[i];
however, this will give an error when the smallest array does not have any more elements. We need to check if the array actually has an element with index i. index numbers in arrays start with 0. so if the length is 7, the 7 elements will have index numbers from 0-6. In other words, only index numbers that are lower (and not equal) to length, is valid numbers:
int value1 = 0;
int value2 = 0;
if(arry1.length > i){
value1 = arry1[i];
}
if(arry2.length > i){
value2 = arry2[i];
}
int result = value1 + value2;
System.out.println("Result "+result);
}
}
}
Now, if you need to put these in a third array, say named sumArray, this would be the complete code:
public class AddingArray {
public static void main(String[] args){
int arry1[]={2,3,4,5,6,7,9};
int arry2[]={4,3,7,9,3,5};
int biggestArrayLength = Math.max(arry1.length, arry2.length);
int[] sumArray = new int[biggestArrayLength];
for(int i=0;i<biggestArrayLength;i++){
int value1 = 0;
int value2 = 0;
if(arry1.length > i){
value1 = arry1[i];
}
if(arry2.length > i){
value2 = arry2[i];
}
int result = value1 + value2;
sumArray[i] = result;
System.out.println("Result "+result);
}
}
}
It is because your loop will go from 0 to 6 (which is the array1.length - 1) and your array2 only has 6 elements (so from 0 to 5).
So when you are accessing arry2[6]; It will give you the java.lang.ArrayIndexOutOfBoundsException.
You could change your for loop to go to the length of the smallest array:
for(int i = 0; i < arry2.length; i++){ /*Do what you want */ }
Or add an element in array2, but that is yours to decide since I do not know your requirements.
Because arry1 is longer than arry2 when you make the last iteration through the loop arry2[i] returns null because there is no element to return.
either do:
if(arry2[i] != null) {
//run your adding code
}
or change your arrays to be the same size
Edit: The reason it is not working properly is because you are using the length of the largest array as the conditional within the for loop. This condition allows you to attempt to access the 2nd array at a location that does not exist, which is why you are getting an ArrayIndexOutOfBoundsException.
Can we stop the downvoting?
End edit----
If you want to add up all of the elements in the array use this code.
public class AddingArray {
public static void main(String[] args){
int arry1[]={2,3,4,5,6,7,9};
int arry2[]={4,3,7,9,3,5};
int result = 0;
for(int i=0;i<arry1.length;i++){
result+=arry1[i];
}
for(int j=0; j < array2.length; j++){
result += array2[j];
}
System.out.println("Result: "+ result);
}
}
if you are trying to sum individual elements as you loop you can do the following. This will properly handle 2 arrays of different length regardless of which one is longer.
public class AddingArray {
public static void main(String[] args){
int arry1[]={2,3,4,5,6,7,9};
int arry2[]={4,3,7,9,3,5};
int result = 0;
for(int i=0;i<arry1.length;i++){
result=arry1[i];
if(i < array2.length){
result += array2[i];
}
System.out.println("Result: "+ result);
}
for(int j = i; j < array2.length; j++){
System.out.println("Result: "+ array2[j]);
}
}
}

Creating multiple nested loops to generate two numbers that move through the length of a Array

As the title reads, I have been thinking about creating multiple nested loops that aim to achieve one purpose. Move two generated random numbers between 0-9 through each possible possition of an array.
For example, App generates first number (fNum) 1 and second number (sNum) 6. It then moves these numbers in the array which containts ABC. However firstNum and secondNum will need to also try all the possible combinations, so each one will need to be different with each loop.
-1ABC6
-A1BC6
-AB1C6
-ABC16
-ABC61
-AB6C1
-A6BC1
-6ABC1
-A6B1C
-A61BC
-A16BC
-A1B6C
-A1BC6
and so on...
I beleive the best way will be to create a method for generating a counter, which increments the numbers which I can call.
private int getNextNumber(int num) {
if (num == 0) {
return num;
} else {
num++;
}
if (num < 10) {
return num;
} else {
return -1;
}
}
Then I will need multiple nested loops... I have decided to go for several loops which will go infinitly.
while (j < maxlen) {
//J = 0 and maxlen = length of text so in this case 3 as it is ABC
//Add two numbers and check against answer
while (fNum != -1 || sNum != -1) {
//incrememnt numbers
fNum = getNextNumber(fNum);
System.out.println(fNum);
sNum = getNextNumber(sNum);
System.out.println(fNum);
}
String textIni = "ABC";
int lenOfText = textIni.length();
char[] split = textIni.toCharArray();
for (int i = 0; i < lenOfText; i++) {
//here it will look at the length of the Text and
//try the possible positions it could be at....
//maybe wiser to do a longer loop but I am not too sure
}
}
Since you don't need to store all possible combinations, we will save some memory using only O(n) storage with an iterative solution. I propose you a basic implementation but don't expect to use it on large arrays since it has a O(n³) complexity.
public static void generateCombinationsIterative(List<Integer> original, int fnum, int snum) {
int size = original.size();
for (int i=0 ; i<=size ; i++) {
List<Integer> tmp = new ArrayList<>(original);
tmp.add(i,fnum);
for (int j=0 ; j<=size + 1 ; j++) {
tmp.add(j,snum);
System.out.print(tmp + (i == size && j == size + 1 ? "" : ", "));
tmp.remove(j);
}
}
}
For your culture, here is an example of a recursive solution, which takes a lot of memory so don't use it if you don't need to generate the lists of results. Nevertheless, this is a more general solution that can deal with any number of elements to insert.
public static List<List<Integer>> generateCombinations(List<Integer> original, Deque<Integer> toAdd) {
if (toAdd.isEmpty()) {
List<List<Integer>> res = new ArrayList<>();
res.add(original);
return res;
}
int element = toAdd.pop();
List<List<Integer>> res = new LinkedList<>();
for (int i=0 ; i<=original.size() ; i++)
// you must make a copy of toAdd, otherwise each recursive call will perform
// a pop() on it and the result will be wrong
res.addAll(generateCombinations(insertAt(original,element,i),new LinkedList<>(toAdd)));
return res;
}
// a helper function for a clear code
public static List<Integer> insertAt(List<Integer> input, int element, int index) {
List<Integer> result = new ArrayList<>(input);
result.add(index,element);
return result;
}
Note that I did not use any array in order to benefit from dynamic data structures, however you can call the methods like this :
int[] arr = { 1,2,3 };
int fnum = 4, snum = 5;
generateCombinationsIterative(Arrays.asList(arr),fnum,snum);
generateCombinations(Arrays.asList(arr),new LinkedList<>(Arrays.asList(fnum,snum));
Note that both methods generate the combinations in the same order.

Categories