Why do I get wrong result - java

I am doing Leetcode 118 Pascal's Triangle and my code is as below. The testcase is five. I should get
[[1],[1,1],[1,2,1],[1,3,3,1],[1,4,6,4,1]].
However, my output is
[[1,4,6,4,1],[1,4,6,4,1],[1,4,6,4,1],[1,4,6,4,1],[1,4,6,4,1]]
I know that the line with comments was not right. But can anyone let me know why I can't just add row to the result? Thanks!
class Solution {
public List<List<Integer>> generate(int numRows) {
List<List<Integer>> result = new ArrayList();
List<Integer> row = new ArrayList();
for(int i = 0; i < numRows; i++ ){
row.add(0,1);
for(int j = 1; j < row.size()-1; j++){
row.set(j, row.get(j) + row.get(j+1));
}
result.add(row); // this line should be result.add(new ArrayList(row));
}
return result;
}
}

Your result list is a list of references to row. The first time you iterate, row refers to an array like [1]. As you keep going through the iterations, you're continuously dereferencing and referencing the variable row.
Since result is essentially [row ,row , row, row ... and so on], you're seeing the result you see here. You'll get the same object as the same block of memory is being referenced and you're overwriting it each time.
On the other hand, new ArrayList(row) actually creates a new block of memory with duplicated data from row - hence this works.

Related

Error with adding element to list - repeating n-time last element

Could anyone help me and point out what is wrong here? All the fields are non-static. Function shuffle returns each time different array.
List<Point[]> someArray = new ArrayList<Point[]>();
for(int i = 0; i < 4; i++) {
Point[] temporary = new Point[50];
temporary = shuffle(pointsArray.getPoints());
someArray.add(temporary);
print(someArray.get(i));
}
The result of print in the first loop is ok since they're different. Right after getting out of the loop, I want to print elements of list someArray again.
for(int i = 0; i < 4; i++) {
print(someArray.get(i));
}
The result here is 4 time of the last element.

Inputting values to a full 2d array

I'm currently trying to figure out how to add values into a full 2d array. Any help would be appreciated.
This is what i currently have.
public static ObjectA[][] addValue(ObjectA value, ObjectA[][] oldArray)
{
//Creates a new array with an extra row
ObjectA[][] newArray = new ObjectA[oldArray.length +1][oldArray[0].length]
for (int i= 0 ; i < newArray.length; i++)
{
for (int ii = 0; ii <= newArray[0].length; ii++)
{
// when the index exceeds the oldArray
// it will add the value to the newArray
if (i <= oldArray.length)
{
newArray[i][ii] = oldArray[i][ii];//copies all values into newArray
}
else
{
newArray[i][ii] = value; //adds value to the last row
}
}
}
return newArray;
}
What I currently have done is input a value to the new row however the method is going to be called multiple times to add more than one value. Which mean it's going to create multiple rows rather than adding to the next available column.
EDIT:
mistyped the data type the array and value are suppoed to be objects.
First, your code throws an IndexOutOfBoundsException. Here is why:
Consider the if (i <= oldArray.length) clause. Say, oldArray.length is 3. When i = 3, newArray[i][ii] = oldArray[i][ii] line seeks the oldArray[3][ii] elements but there are no such elements. All the possible elements of oldArray is oldArray[0][ii], oldArray[1][ii] and oldArray[2][ii], since counting starts with 0 in programming.
Second, I didn't get the point of adding another row for each next value. If you're not going to add a set of values to each row, then, why do you consider expanding number of rows?
This is a typical situation when you need to make a tradeoff between element access complexity and complexity of adding new column
If you need fast column adding without new structure allocation you should use LinkedList as a storage of rows and call list.add(row) every time you need to add a new column so your code will look like:
public static void addValue(int value, LinkedList<int[]> list) {
int[] row_you_need_to_add = new int[list.get(0).length];
for (int i = 0; i < list.get(0).length; i++) {
row_you_need_to_add[i] = value;
}
list.add(row_you_need_to_add);
}
As 2D Array is an Array which consist of an array within an Array. So at every index of 2D array there is another array is present and has a specific size.
public static ObjectA[][] addValue(ObjectA value, ObjectA[][] oldArray) {
ObjectA[][] newArray = new ObjectA[oldArray.length +1][oldArray[0].length]
for (int i= 0 ; i < newArray.length; i++) {
for (int ii = 0; ii <= newArray[0].length; ii++) {
// when the index exceeds the oldArray
// it will add the value to the newArray
if (i <= oldArray.length) {
newArray[i][ii] = oldArray[i][ii];//copies all values into newArray
} else {
newArray[i][ii] = value; //adds value to the last row
}
}
}
return newArray;
}

Java mapping data from two arrays and insert into third array

I am having difficulty mapping the data in two lists into a third list. My sample data is as follows:
Categorylist ID: 1,2,3,4,5,6,7,8,9,10,42,46,49,50
CurrentMonthByCat ID: 1,2,3,4,5,6,7,8,9,10,42,49,50
(the transaction amount value for CurrentMonthByCat: 92,46,40,180,60,10,1510,200,500,10,234,12)
There is a missing 46 in the currentMonthByCat. I am trying to do it in a way such that if the currentMonthByCat ID does not exist from the categorylist ID, I will insert 0 into the third list rather than getting the transaction amount from CurrentMonthByCat and push it into the third list.
ArrayList<Double> total = new ArrayList<Double>();
for(int i = 0; i < categorylist.size(); i++){
for(int j = 0; j < currentMonthByCat.size(); j++){
if(categorylist.get(i).getCategoryID().equals(currentMonthByCat.get(j).getCategory().getCategoryID())){
Log.d("IIIII", categorylist.get(i).getCategoryID());
Log.d("JJJJJ", currentMonthByCat.get(j).getCategory().getCategoryID());
total.add((double)currentMonthByCat.get(j).getTransactionAmt());
}else{
total.add(0.0);
}
}
}
for(int k = 0; k < total.size(); k++){
Log.d("KKKKK", String.valueOf(total.get(k)));
}
But the printed out result of total list is:
92,0,0,0,0,0,0,0,0,0,0,0,0,0,46,0,0,0...
What I expected is:
92,46,40,180,60,10,1510,200,500,10,0,234,12
I wanted to insert 0 only if the ID in currentMonthByCat does not match the ID in categorylist. For instance, the ID 46 which is the 3rd position from the right.
I realized the reason is because firstly I inserted 92 into third array, then the categorylist ID is still at 1, then it will compare with all the rest in the currentMonthByCat before moving to ID 2. That is why the unnecessary zeros. But I not sure how to actually sort it to achieve what I wanted.
Any ideas?
Thanks in advance.
It's easy one. you can't take a decision of adding zero or value in the total array unless the inner loop finish. so probably you add element existAtIndex and initialize it with -1 and in the loop if you find the element then assign the index to existAtIndex and break the loop or if it's not exist then you add zero. so the code will be something like :
ArrayList<Double> total = new ArrayList<Double>();
int existAtIndex;
for(int i = 0; i < categorylist.size(); i++){
// search for the element index
existAtIndex = -1;
for(int j = 0; j < currentMonthByCat.size(); j++){
if(categorylist.get(i).getCategoryID().equals(currentMonthByCat.get(j).getCategory().getCategoryID())){
existAtIndex = j;
break;
}
}
// add the value in the element index or add zero if the element not exist
if (existAtIndex != -1) {
total.add((double)currentMonthByCat.get(existAtIndex).getTransactionAmt());
}
else {
total.add(0.0);
}
}
for(int k = 0; k < total.size(); k++){
Log.d(String.valueOf(total.get(k)));
}
For better code you could use contains method to check for the item if exist or not in arrayList instead of using the basic loop. Good luck
You have a lot of code for what you are trying to do here. I think the following snippet does what you want in a very readable and maintainable way.
//First of all we are interested in getting a transaction amount for each value in currentMonthByCat
//so loop around using your object (not sure what it's called)
for(CurrentMonth value : currentMonthByCat){
//check if it's present.
//We create a new method here that gets you your category list as a list of integers.
//This is key to making the whole method much more readable.
if(categorylist.getIdsAsList().contains(value.getCategory().getCategoryID())){
//it is so add it
total.add(value.getTransactionAmt());
} else {
//it's not so add a 0
total.add(0.0);
}
}
The getIdsAsList method could look like this:
public List<Integer> getIdsAsList(){
List<Integer> result = new ArrayList<>();
for (CategoryListItem item : categorylist) {
result.add(item.getCategoryId());
}
return result;
}
Put your values into Map<Integer, Double>
Map<Integer, Double> map = new HashMap<Integer, Double>();
for (int i = 0; i < currentMonthByCat.size(); ++i) {
//... categoryId = currentMonthByCat.get(i).categoryId
//... amount = currentMonthByCat.get(i).amount
map.put(categoryId, amount);
}
Then traverse the map using values from Categorylist ID:
// create result arraylist
ArrayList<Double> total = new ArrayList<Double>();
for (int i = 0; i < categorylist.size(); ++i) {
Double amount = map.get(categorylist.get(i));
if (amount == null) {
total.add(0.0);
} else {
total.add(amount);
}
}
The result list total will contain amounts for existing mappings, or zeros for non-existent.
Other way
If it is guaranteed categorylist is sorted and CurrentMonthByCat is sorted
you can then traverse one of the lists while keeping index/cursor to the other list and not iterating the other list from the beginning, but from previously remembered cursor value, resulting in better average performance than n^2

Copy part of two-dimensional array

So I have a matrix M1 that I build at the beginning of my method each method call (this is all done in a non-static environment). The issue is that, depending on an integer index, say n I want the matrix M2 (in method call #2 for example) to contain rows 1, 2... n of M1 (the number of rows that I need for M changes each method call, but the number of columns persist).
My method in pseudocode;
int myMethod(numRows, numColumns, n) {
Initialize M as matrix with dimensions(numRows, numColumns)
if (n > 0) {
**copy rows 1,2...,n from previous M matrix**
}
**do stuff with M**
return M[numRows][numColumns];
What is a smart way to accomplish this? I hope it's clear what I'm asking for. Something worth noting is that Mi can be 'taller' or 'shorter' (it's always exactly as 'wide') as Mj for i > j
Maybe return the array and pass the array as a parameter.
int[][] mymethod(int[][]oldArray, int noOfRows){
int[][] newArr;
if(noOfRows > 0){
//here copy from oldArray to new;
//sorry this is where I can't remember, you will have
//to do something to get column size maybe oldArray[0].length()
// but that may not work if each row has different length.
newArr = new int[noOfRows][];
for(int i=0; i < noOfRows; i++){
for(int j = 0; j < noOfRows; j++){
newArr[i][j] = oldArray[i][j];
}
}
}
return newArr;
}
after this just call the method.
int [][] a = mymethod(dArray,5);
int [][] b = mymethod(a, 2);
You could use a two dimensional vector, and then perhaps the clone() method. The awesome thing about vector is that it can grow and shrink as you need it to.

Finding Duplicate Array Elements

I've been struggling to create a function to essentially find all the indices of duplicate elements in a multi-dimensional array(unsorted), in this case a 5x5 array, and then using the indices found changing the parallel elements in a score array. But only find duplicates within columns and not comparatively to the other columns in the array Here is what I've done so far, with research online. The main problem with this code is that it will find all the duplicate elements but not the originals. For example: if the array holds the elements:
{{"a","a","a"},{"b","b","b"},{"a","c","a"}}, then it should change the parallel score array to: {{0,1,0},{1,1,1},{0,1,0}}. But instead it only recognizes the last row and top the top row's duplicates.
Code:
public static void findDuplicates(String a[][])
{
System.out.println("*Duplicates*");
Set set = new HashSet();
for(int j = 0; j<a.length; j++)
{
for(int i=0; i < a[0].length; i++)
{
if(!set.contains(a[i][j]))
{
set.add(a[i][j]);
}
else
{
System.out.println("Duplicate string found at index " + i + "," + j);
scores[i][j] -= scores[i][j];
}
}
set = new HashSet();
}
}
I know my explanation is a bit complicated, but hopefully it is understandable enough. Thanks,
Jake.
Your logic is incorrect. Your outer loop is j and inner loop is i but you're doing:
set.add(a[i][j]);
It should be the other way around:
set.add(a[j][i]);
Technically you could get an out of bounds exception if the array isn't NxN. But you can state that as a precondition.
For some reason you're also setting to 0 with:
scores[i][j] -= scores[i][j];
Why not just:
scores[i][j] = 0;
But to find duplicates within columns:
public static void findDuplicates(String a[][]) {
for (int col=0; col<a[0].length; col++) {
Map<String, Integer> values = new HashMap<String, Integer>();
for (int row=0; row<a.length; row++) {
Integer current = values.put(a[row][col], row);
if (current != null) {
scores[row][col] = 0;
scores[current][col] = 0;
}
}
}
}
How does this work?
I've renamed the loop variables to row and col. There's no reason to use i and j when row and col are far more descriptive;
Like you I assume the input array is correct as a precondition. It can be NxM (rather than just NxN) however;
I use a Map to store the index of each value. Map.put() returns the old value if key is already in the Map. If that's the case you've found a duplicate;
The current (row,col) and (current,col) are set to 0. Why subtract the score from itself rather than simply setting to 0?
if the value "a" is found 3+ times in a column then scores[current][col] will be set to 0 more than once, which is unnecessary but not harmful and makes for simpler code.
I've declared the Map using generics. This is useful and advisable. It says the Map has String keys and Integer values, which saves some casting;
It also uses auto-boxing and auto-unboxing to convert an int (the loop variable) to and from the wrapper class Integer.

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