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How can I handle branching calculations in Java?

I am trying to find the longest possible path based on how many connections a variable number has, without repeating connections. The way I thought of doing this was creating a list that holds all points that have already been gone through, but when a path ends, and I need to check a new path, all of those old connections remain in the list. How can I restart my list from the initial point?
Putting it in the recursive function itself would just clear the list each time. Is there a better option than using a list?
Relevant code:
package testapp;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.io.IOException;
import java.util.List;
class TestApp {
// Store list of objects we have already matched with
static List<NumberObject> holdingList = new ArrayList<NumberObject>();
//Test objects
static int[] array1 = {2,2};
static int[] array2 = {3,1};
static int[] array3 = {2,1};
static int[] array4 = {1,1};
static NumberObject eight = new NumberObject(array1, 8);
static NumberObject two = new NumberObject(array2, 2);
static NumberObject three = new NumberObject(array3, 3);
static NumberObject four = new NumberObject(array4, 4);
// Test objects ^^
public static int longestSequence(int[][] grid) {
// TODO: implement this function
// Code exists here not relevant to the problem
//Setting up a new numberList array for testing
NumberObject[] newNumberList = {eight, two, three, four};
NumberObject[] connections1 = {two, four};
NumberObject[] connections2 = {two, three};
//Adding connections
eight.connections = connections1;
four.connections = connections2;
for (NumberObject s: newNumberList){
recursive(s);
}
return 0;
}
public static void recursive(NumberObject object){
for (NumberObject x: holdingList){
System.out.println(x);
}
if (!holdingList.contains(object)){
holdingList.add(object);
if (object.hasConnections()){
NumberObject[] newobject = object.getConnections();
for(NumberObject y: newobject){
recursive(y);
}
}
else {
System.out.println(holdingList.size());
return;
}
}
else {
System.out.println(holdingList.size());
return;
}
}
public static void main(String[] args) throws IOException {
BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
int numRows = 0;
int numCols = 0;
String[] firstLine = reader.readLine().split("\\s+");
numRows = Integer.parseInt(firstLine[0]);
numCols = Integer.parseInt(firstLine[1]);
int[][] grid = new int[numRows][numCols];
for (int row = 0; row < numRows; row++) {
String[] inputRow = reader.readLine().split("\\s+");
for (int col = 0; col < numCols; col++) {
grid[row][col] = Integer.parseInt(inputRow[col]);
}
}
int length = longestSequence(grid);
System.out.println(length);
}
}
class NumberObject {
int[] id;
int value;
NumberObject[] connections;
//Constructor
public NumberObject(int[] id, int value){
this.id = id;
this.value = value;
}
//print statement
public String toString(){
return ("NumberOject: Id = " + id + "\nValue = " + value);
}
//Check if it has connections
public boolean hasConnections(){
if (connections == null){
return false;
}
else if (connections.length != 0){
return true;
}
else
return false;
}
//Return the connections it has
public NumberObject[] getConnections(){
return connections;
}
}
Ideally, the image displays what I want to happen.
Instead, all the old branching connections remain on holdingList.
it should be noted paths can branch off to more than two other objects.

Instead of storing the list in a field, you could just pass an instance of a copy of your list to the function as an argument. So the signature of your function recursive would look like:
public static void recursive(NumberObject object, List<NumberObject> visited)
To hide this implementation detail, I recommend writing two functions, whereby the second function just passes an empty list to the other one.
However, I'd choose a different approach since yours acquires as many new lists as entries are in your tree. In the following implementation, you only have one list per "tree end". Moreover, just like in the previous suggestion, this keeps your class stateless.
static List<NumberObject> findLongestPath(NumberObject currentNode) {
if (currentNode.getConnectedNodes().isEmpty()) {
List<NumberObject> result = new ArrayList<>();
result.add(currentNode);
return result;
}
List<NumberObject> longestPath = currentNode.getConnectedNodes().stream()
.map(PathFinder::findLongestPath)
.max(Comparator.comparing(List::size))
.get();
longestPath.add(currentNode);
return longestPath;
}

Initializing 2D ArrayList for for-loop summation (Java)

I am trying to sum N pairs of ints--an Nx2 ArrayList--and return the N summations as an ArrayList. While I understand it is not necessary to set up a class to accomplish this, I would like to do so as practice for future projects.
import java.util.ArrayList;
public class SumsInLoop {
public SumsInLoop(int numberOfPairs, ArrayList<ArrayList<Integer>> numbersList) {}
public ArrayList<Integer> getSums(int numberOfPairs, ArrayList<ArrayList<Integer>> numbersList) {
ArrayList<Integer> pairsOfSums = new ArrayList<Integer>();
for (ArrayList<Integer> Pair : numbersList) {
int x = Pair.get(0);
int y = Pair.get(1);
int sum = x + y;
pairsOfSums.add(sum);
}
System.out.println(pairsOfSums);
return pairsOfSums;
}
The data that I am given is a random assortment of N pairs (numbersOfPairs) of integers, e.g. 612673 108695. I would like to add these pairs of integers to a 2D ArrayList (numbersList) that will be called by getSums.
However, I am having difficulties initializing numbersList. My main function is as follows:
public static void main(String[] args) {
int myNumberOfPairs = 13;
ArrayList[][] myNumbersList = new ArrayList[13][2];
myNumbersList[0][0] = new ArrayList<>();
myNumbersList[0][0].add(612673);
myNumbersList[0][1].add(108695);
myNumbersList[1][0] = new ArrayList<>();
myNumbersList[1][0].add(756875);
myNumbersList[1][1].add(496058);
SumsInLoop mySum = new SumsInLoop(myNumberOfPairs,myNumbersList);
mySum.getSums(myNumberOfPairs, myNumbersList);
The last two lines of code throw errors, asking me to change myNumbersList to type ArrayList<List<Integer>> which throws even more errors, even after changing all 2D ArrayLists to type ArrayList<List<Integer>>.
So, my two questions are as follows:
How can I initialize an NxM ArrayList and populate it correctly?
Is there a faster way of accomplishing this task while still using a class method?
P.S. I'm used to coding in Python and am self-teaching myself Java, so any other information or resources you can provide me with are much appreciated.

You may want to simplify your input by using 2D array of int : int[][] myNumbersList = new int[13][2];
The expected output in that case is a 1D array of int[13] that can be obtained as follows (demonstrated with 2 pairs. See mcve ) :
public class SumsInLoop {
//pairsOfInts should be an [n][2] array
private static int[] sumOfPairs(int[][] pairsOfInts) {
int[] sums = new int[pairsOfInts.length];
for(int pairIndex = 0; pairIndex < pairsOfInts.length; pairIndex++) {
sums[pairIndex]= pairsOfInts[pairIndex][0]+pairsOfInts[pairIndex][1];
}
return sums;
}
public static void main(String[] args) {
int numberOfPairs = 2;
int[][] pairsOfInts = new int[numberOfPairs][2];
pairsOfInts[0] = new int[] {612673,108695 };
pairsOfInts[1] = new int[] {756875,496058 };
int[] sumOfPairs = sumOfPairs(pairsOfInts);
System.out.println(Arrays.toString(sumOfPairs));
}
}
If you want a solution implemented with List you can make use of javafx Pair (or make your own pair class.
The input can be defined as List<Pair<Integer,Integer>> pairsOfInts = new ArrayList<>();
The out put can be an array as above, or a List<Integer>:
import java.util.ArrayList;
import java.util.List;
import javafx.util.Pair;
public class SumsInLoop {
private static List<Integer> sumOfPairs(List<Pair<Integer, Integer>> pairsOfInts) {
List<Integer> sums = new ArrayList<>();
for(Pair<Integer,Integer> pair : pairsOfInts) {
sums.add(pair.getKey()+ pair.getValue());
}
return sums;
}
public static void main(String[] args) {
List<Pair<Integer,Integer>> pairsOfInts = new ArrayList<>();
pairsOfInts.add (new Pair<>(612673,108695 ));
pairsOfInts.add (new Pair<>(756875,496058));
List<Integer> sumOfPairs = sumOfPairs(pairsOfInts);
System.out.println(sumOfPairs);
}
}

The (compile) exception you are getting is due to the fact that you expect a ArrayList<ArrayList<Integer>>, but pass an ArrayList[][]. (which is not the same in Java)
In your case you'd need (in the main method):
ArrayList<ArrayList<Integer>> myNumbersList = new ArrayList</* when java > 6 ;)*/>(13);
this only sets the capacity of the (parent) list (..and the underlying/internal backing array)
to initialize the child lists, you'd not come around looping (somehow...even not in python :):
for (int i = 0; i < 13; i++) {
myNumbersList.add(new ArrayList<Integer>(2));
}
Depends on what means "correctly" ...but I assume with "random data", ideally you would again inner loop:
java.util.Random rnd = new Random(/*seed default current timestamp*/);
//...
for (int i = 0; i < 13; i++) {
ArrayList<Integer> innerList = new ArrayList<>(2);
for (int j = 0; j < 2; j++) {
innerList.add(rnd.netxInt(/*bound default Integer.MAX_VALUE*/) /*+/-/% offset*/);
}
myNumberList.add(innerList);
}
Sorry I am not aware of one (faster way), but much depends on the "input format".

Since you already know the amount of values in a pair, an ArrayList is unnecessary. You can create your own, simpler implementation of a pair.
class Pair {
public final int left;
public final int right;
public Pair(int left, int right){
this.left = left;
this.right = right;
}
}
You can then access the values by creating a pair object and accessing its fields.
Pair p = new Pair(10, 7);
System.out.println(p.left); // 10
System.out.println(p.right); // 7
You can then more easily redefine your getSums method.
public static List<Integer> getSums(List<Pair> pairs){
List<Integer> pairsOfSums = new ArrayList<>();
for(Pair pair : pairs){
int sum = pair.left + pair.right;
pairsOfSums.add(sum);
}
return pairsOfSums;
}
Please also notice the function can be static and you don't need to pass the number of pairs. The for-each loop will cycle through all of them regardless.
Initializing the array is then easier than the method you have described in the question.
List<Pair> pairs = new ArrayList<>();
pairs.add(new Pair(7, 10));
pairs.add(new Pair(18, 3));
pairs.add(new Pair(-6, 0));
pairs.add(new Pair(4, 2));
System.out.println(SumsInLoop.getSums(pairs));

All Unique Permutations - Stack overflow issue

I am trying to solve the following problem:
Given a collection of numbers that might contain duplicates, return
all possible unique permutations.
Here is my code:
public class Solution {
public ArrayList<ArrayList<Integer>> permute(ArrayList<Integer> a) {
HashMap<ArrayList<Integer>, Boolean> unique = new HashMap<>();
ArrayList<ArrayList<Integer>> results = new ArrayList<>();
permu(results, a, 0, unique);
return results;
}
private void permu(ArrayList<ArrayList<Integer>> results, final ArrayList<Integer> a, int item, HashMap<ArrayList<Integer>, Boolean> unique) {
for(int i = 0; i < a.size(); i++) {
ArrayList<Integer> aClone = new ArrayList<>(a);
// swap
int backup = aClone.get(i);
aClone.set(i, aClone.get(item));
aClone.set(item, backup);
if(!unique.containsKey(aClone)) {
results.add(aClone);
unique.put(aClone, true);
permu(results, aClone, i, unique); //<--- Stack overflow error
}
}
}
}
I have a stack overflow error on this the call to the recurrence, line (19)

Well, I don't like to ruin good job interview questions, but this question was fun to think about, so...
Here's a super-L337 answer for generating all unique permutations very quickly and without using much memory or stack. If you use this in a job interview, the interviewer will ask you to explain how it works. If you can do that, then you deserve it :)
Note that I permuted the chars in a string instead of the instead of integers in a list, just because it's easier to pretty-print the result.
import java.util.Arrays;
import java.util.function.Consumer;
public class UniquePermutations
{
static void generateUniquePermutations(String s, Consumer<String> consumer)
{
char[] array = s.toCharArray();
Arrays.sort(array);
for (;;)
{
consumer.accept(String.valueOf(array));
int changePos=array.length-2;
while (changePos>=0 && array[changePos]>=array[changePos+1])
--changePos;
if (changePos<0)
break; //all done
int swapPos=changePos+1;
while(swapPos+1 < array.length && array[swapPos+1]>array[changePos])
++swapPos;
char t = array[changePos];
array[changePos] = array[swapPos];
array[swapPos] = t;
for (int i=changePos+1, j = array.length-1; i < j; ++i,--j)
{
t = array[i];
array[i] = array[j];
array[j] = t;
}
}
}
public static void main (String[] args) throws java.lang.Exception
{
StringBuilder line = new StringBuilder();
generateUniquePermutations("banana", s->{
if (line.length() > 0)
{
if (line.length() + s.length() >= 75)
{
System.out.println(line.toString());
line.setLength(0);
}
else
line.append(" ");
}
line.append(s);
});
System.out.println(line);
}
}
Here is the output:
aaabnn aaanbn aaannb aabann aabnan aabnna aanabn aananb aanban aanbna
aannab aannba abaann abanan abanna abnaan abnana abnnaa anaabn anaanb
anaban anabna ananab ananba anbaan anbana anbnaa annaab annaba annbaa
baaann baanan baanna banaan banana bannaa bnaaan bnaana bnanaa bnnaaa
naaabn naaanb naaban naabna naanab naanba nabaan nabana nabnaa nanaab
nanaba nanbaa nbaaan nbaana nbanaa nbnaaa nnaaab nnaaba nnabaa nnbaaa

How to create combinations of values in Java?

I have the following map: Map<Integer,String[]> map = new HashMap<Integer,String[]>();
The keys are integers and the values are arrays (could also be replaced by lists).
Now, I would like to get all possible combinations of the values among the keys. For example, let's say the map contains the following entries:
key 1: "test1", "stackoverflow"
key 2: "test2", "wow"
key 3: "new"
The combinations consists of
("test1","test2","new")
("test1","wow","new")
("stackoverflow", "test2", "new")
("stackoverflow", "wow", "new")
For this I imagine a method boolean hasNext() which returns true if there is a next pair and a second method which just returns the next set of values (if any).
How can this be done? The map could also be replaced by an other data structure.

The algorithm is essentially almost the same as the increment algorithm for decimal numbers ("x -> x+1").
Here the iterator class:
import java.util.Iterator;
import java.util.Map;
import java.util.NoSuchElementException;
import java.util.TreeSet;
public class CombinationsIterator implements Iterator<String[]> {
// Immutable fields
private final int combinationLength;
private final String[][] values;
private final int[] maxIndexes;
// Mutable fields
private final int[] currentIndexes;
private boolean hasNext;
public CombinationsIterator(final Map<Integer,String[]> map) {
combinationLength = map.size();
values = new String[combinationLength][];
maxIndexes = new int[combinationLength];
currentIndexes = new int[combinationLength];
if (combinationLength == 0) {
hasNext = false;
return;
}
hasNext = true;
// Reorganize the map to array.
// Map is not actually needed and would unnecessarily complicate the algorithm.
int valuesIndex = 0;
for (final int key : new TreeSet<>(map.keySet())) {
values[valuesIndex++] = map.get(key);
}
// Fill in the arrays of max indexes and current indexes.
for (int i = 0; i < combinationLength; ++i) {
if (values[i].length == 0) {
// Set hasNext to false if at least one of the value-arrays is empty.
// Stop the loop as the behavior of the iterator is already defined in this case:
// the iterator will just return no combinations.
hasNext = false;
return;
}
maxIndexes[i] = values[i].length - 1;
currentIndexes[i] = 0;
}
}
#Override
public boolean hasNext() {
return hasNext;
}
#Override
public String[] next() {
if (!hasNext) {
throw new NoSuchElementException("No more combinations are available");
}
final String[] combination = getCombinationByCurrentIndexes();
nextIndexesCombination();
return combination;
}
private String[] getCombinationByCurrentIndexes() {
final String[] combination = new String[combinationLength];
for (int i = 0; i < combinationLength; ++i) {
combination[i] = values[i][currentIndexes[i]];
}
return combination;
}
private void nextIndexesCombination() {
// A slightly modified "increment number by one" algorithm.
// This loop seems more natural, but it would return combinations in a different order than in your example:
// for (int i = 0; i < combinationLength; ++i) {
// This loop returns combinations in the order which matches your example:
for (int i = combinationLength - 1; i >= 0; --i) {
if (currentIndexes[i] < maxIndexes[i]) {
// Increment the current index
++currentIndexes[i];
return;
} else {
// Current index at max:
// reset it to zero and "carry" to the next index
currentIndexes[i] = 0;
}
}
// If we are here, then all current indexes are at max, and there are no more combinations
hasNext = false;
}
#Override
public void remove() {
throw new UnsupportedOperationException("Remove operation is not supported");
}
}
Here the sample usage:
final Map<Integer,String[]> map = new HashMap<Integer,String[]>();
map.put(1, new String[]{"test1", "stackoverflow"});
map.put(2, new String[]{"test2", "wow"});
map.put(3, new String[]{"new"});
final CombinationsIterator iterator = new CombinationsIterator(map);
while (iterator.hasNext()) {
System.out.println(
org.apache.commons.lang3.ArrayUtils.toString(iterator.next())
);
}
It prints exactly what's specified in your example.
P.S. The map is actually not needed; it could be replaced by a simple array of arrays (or list of lists). The constructor would then get a bit simpler:
public CombinationsIterator(final String[][] array) {
combinationLength = array.length;
values = array;
// ...
// Reorganize the map to array - THIS CAN BE REMOVED.

I took this as a challenge to see whether the new Java 8 APIs help with these kind of problems. So here's my solution for the problem:
public class CombinatorIterator implements Iterator<Collection<String>> {
private final String[][] arrays;
private final int[] indices;
private final int total;
private int counter;
public CombinatorIterator(Collection<String[]> input) {
arrays = input.toArray(new String[input.size()][]);
indices = new int[arrays.length];
total = Arrays.stream(arrays).mapToInt(arr -> arr.length)
.reduce((x, y) -> x * y).orElse(0);
counter = 0;
}
#Override
public boolean hasNext() {
return counter < total;
}
#Override
public Collection<String> next() {
List<String> nextValue = IntStream.range(0, arrays.length)
.mapToObj(i -> arrays[i][indices[i]]).collect(Collectors.toList());
//rolling carry over the indices
for (int j = 0;
j < arrays.length && ++indices[j] == arrays[j].length; j++) {
indices[j] = 0;
}
counter++;
return nextValue;
}
}
Note that I don't use a map as an input as the map keys actually don't play any role here. You can use map.values() though to pass in the input for the iterator. With the following test code:
List<String[]> input = Arrays.asList(
new String[] {"such", "nice", "question"},
new String[] {"much", "iterator"},
new String[] {"very", "wow"}
);
Iterator<Collection<String>> it = new CombinatorIterator(input);
it.forEachRemaining(System.out::println);
the output will be:
[such, much, very]
[nice, much, very]
[question, much, very]
[such, iterator, very]
[nice, iterator, very]
[question, iterator, very]
[such, much, wow]
[nice, much, wow]
[question, much, wow]
[such, iterator, wow]
[nice, iterator, wow]
[question, iterator, wow]

Finding the number of Disjoint Sets

I'm trying to find the number of disjoint sets for given N sets and M relations. For example given a relation "i j" ,I have to merge the sets containing these two elements. M and N can be as large as 100000.
I tried using ArrayList of Hashsets. But couldn't implement it efficiently. This was my code:
import java.io.InputStreamReader;
import java.io.BufferedReader;
import java.util.*;
import java.lang.Object;
class fire
{
public static void main(String[] args)throws Exception
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n,m;
int t=Integer.parseInt(br.readLine());
String st[];
while(t-->0)
{
st=br.readLine().split(" ");
n=Integer.parseInt(st[0]);
m=Integer.parseInt(st[1]);
ArrayList<HashSet<Integer>> list = new ArrayList<HashSet<Integer>>(n+1);
for(int i=0;i<n+1;i++)
{
list.add(i, new HashSet<Integer>());
list.get(i).add(i);
}
int a,b;
while(m-->0)
{
st=br.readLine().split(" ");
a=Integer.parseInt(st[0]);
b=Integer.parseInt(st[1]);
if(list.get(a).contains(a))
{
if(list.get(b).contains(b))
{
Iterator<Integer> it = list.get(b).iterator();
while(it.hasNext())
{
list.get(a).add(new Integer((int)it.next()));
}
list.get(b).clear();
}
else
{
for(int i=1;i<n+1;i++)
if(list.get(i).contains(b))
{
if(i!=a)
{
Iterator<Integer> it = list.get(i).iterator();
while(it.hasNext())
list.get(a).add(new Integer((int)it.next()));
list.get(i).clear();
}
break;
}
}
}
else
{
for(int i=1;i<n+1;i++)
if(list.get(i).contains(a))
{
if(list.get(b).contains(b))
{
Iterator<Integer> it = list.get(b).iterator();
while(it.hasNext())
list.get(a).add(new Integer((int)it.next()));
list.get(b).clear();
}
else
{
for(int j=1;j<n+1;j++)
if(list.get(j).contains(b))
{
if(i!=j)
{
Iterator<Integer> it = list.get(j).iterator();
while(it.hasNext())
list.get(a).add(new Integer((int)it.next()));
list.get(j).clear();
}
break;
}
}
break;
}
}
}
int size=0,prod=1;
int num=0;
for(int i=1;i<n+1;i++)
{
num=list.get(i).size();
if(num!=0)
{
prod*=num;
size++;
}
}
System.out.println(size+" "+prod);
}
}
};
This is a problem from Codechef. The solution is right but I'm getting TimeLimitExceeded for this problem. Should I work on improving this code or would I have to use a different data structure? Any ideas would be really grateful :). Thank you.

You should use disjoint set forest data structure for this problem. Very easy to implement and extremely effiecient.