We Keep Coding

Generate all possible combinations of an array of size n having exactly k subarrays

It's a combinatorics problem. We can partition the array at max n-1 times and out of it, we have to choose k-1 partitions.
Ex: arr[] = {1, 2, 3, 4, 5, 6}
k = 3, n = 6.
Total possibilities = (n-1)C(k-1) 5C2 = 10.
Expected Output : (3-dimensional array)
[
[[1], [2], [3, 4, 5, 6]],
[[1], [2, 3], [4, 5, 6]],
[[1], [2, 3, 4], [5, 6]],
[[1], [2, 3, 4, 5], [6]],
[[1, 2], [3], [4, 5, 6]],
[[1, 2], [3, 4], [5, 6]],
[[1, 2], [3, 4, 5], [6]],
[[1, 2, 3], [4, 5], [6]],
[[1, 2, 3, 4], [5], [6]],
]
Constraints:
Subarrays should have at least 1 element.
SubArrays should only consist of consecutive elements.
For ex: [[1, 3], [2, 4], [5, 6]] This is not valid.
So far, I have tried a recursive method but it yields wrong results. I can't wrap my head around how to correct it. I need an answer explaining how to approach this problem properly and the intuition behind it first and then correct my code and point out where I went wrong.
My code:
public static void main(String[] args) {
int n = 6;
int k = 3;
int[] arr = new int[]{1,2,3,4,5,6};
ArrayList<ArrayList<ArrayList<Integer>>> result = new ArrayList<ArrayList<ArrayList<Integer>>>();
ArrayList<ArrayList<Integer>> eachComb = new ArrayList<ArrayList<Integer>>();
ArrayList<Integer> eachSubArray = new ArrayList<Integer>();
int index = 0;
generateCombinations(arr, k, index, eachSubArray, eachComb, result);
for (ArrayList<ArrayList<Integer>> each : result) {
for(ArrayList<Integer> eachh: each) {
System.out.println(eachh);
}
System.out.println("-------------------------------");
}
}
private static void generateCombinations(int[] arr, int k, int index, ArrayList<Integer> eachSubArray,
ArrayList<ArrayList<Integer>> eachComb, ArrayList<ArrayList<ArrayList<Integer>>> result) {
if(k==0 || index==arr.length) {
return;
}
if(!eachSubArray.isEmpty()) {
eachComb.add(new ArrayList<>(eachSubArray));
}
if(k==1) {
ArrayList<Integer> last = new ArrayList<Integer>();
for(int i=index; i<arr.length; i++) {
last.add(arr[i]);
}
if(!last.isEmpty()) {
eachComb.add((last));
result.add(new ArrayList<ArrayList<Integer>>(eachComb));
eachComb.remove(eachComb.size()-1);
}
return;
}
eachSubArray.clear();
for(int i=index; i<arr.length; i++) {
eachSubArray.add((arr[i]));
generateCombinations(arr, k-1, i+1, eachSubArray, eachComb, result);
if(!eachComb.isEmpty()) {
eachComb.remove(eachComb.size()-1);
}
}
}

Reverse subset of ArrayList using multidimensional ArrayList in Java

I am trying to reverse a sublist in a List using the indices provided in a multidimensional List.
I don't have much experience using multidimensional lists/arrays. I don't understand why this doesn't work.
/*
Given a List<Integer> list and List<List<Integer>> operations
reverse the sublist and print out the list after all the operations have been done.
Ex: [5, 3, 2, 1, 3]
[[0,1], [1, 3]]
*/
import java.util.*;
public class ReverseParameters {
public static void main(String[] args) {
List<Integer> list = Arrays.asList(5, 3, 2, 1, 3);
List<List<Integer>> operations = new ArrayList<>(2);
for(int i= 0; i < 3; i++){
operations.add(new ArrayList<>());
}
operations.get(0).add(1);
operations.get(1).add(3);
subList(list, operations);
}
public static void subList (List<Integer> list, List<List<Integer>> operations) {
System.out.println(list);
int vertCount = operations.size();
for (int i = 0; i < vertCount; i++) {
int edgeCount = operations.get(i).size();
for (int j = 0; j < edgeCount; j++) {
int startInd = i;
int endInd = operations.get(i).get(j);
int shift = endInd - startInd;
int right = Math.min(i + shift - 1, list.size() - 1);
int temp = 0;
while (startInd < right) {
temp = list.get(startInd);
list.set(startInd, list.get(right));
list.set(right, temp);
startInd+=1;
right-=1;
}
System.out.println();
System.out.printf(" %d %d%n", startInd, endInd);
System.out.println();
}
}
System.out.println(list);
}
}
The output of this code using [[0,1], [1, 3]] as the indecies is:
[5, 2, 3, 1, 3]
but it should be:
[3, 1, 2, 5, 3]
Can someone please help point me in the right direction?

This can be done as simple as this.
public class Main
{
public static void main(String[] args)
{
//Given a List<Integer> list and List<List<Integer>> operations
//reverse the sublist and print out the list after all the operations have been done.
//Ex: [5, 3, 2, 1, 3]
// [[0,1], [1, 3]]
//Target: [3, 1, 2, 5, 3]
//Steps:
// 0 1 2 3 4 (Indices)
//###############
//[5, 3, 2, 1, 3]
//[3, 5, 2, 1, 3] // Swap index of 0 and 1.
//[3, 1, 2, 5, 3] // Swap index of 1 and 3.
List<Integer> list = Arrays.asList(5, 3, 2, 1, 3);
List<List<Integer>> listOfList = new ArrayList<List<Integer>>(2);
listOfList.add(Arrays.asList(0, 1));
listOfList.add(Arrays.asList(1, 3));
for(int i=0; i < listOfList.size(); ++i) {
final int indexA = listOfList.get(i).get(0); //[0 , [1
final int indexB = listOfList.get(i).get(1); // 1], 3]
//Swap the indices.
final int tmpValue = list.get(indexA);
list.set(indexA, list.get(indexB));
list.set(indexB, tmpValue);
}
System.out.println(list);
//[3, 1, 2, 5, 3]
}
}

You are overcomplicating your code with unnecessary variables that makes it difficult to find the problem. Please se a more simple code with explanation:
public static void main(String[] args) {
List<Integer> list = Arrays.asList(5, 3, 2, 1, 3);
List<List<Integer>> operations = new ArrayList<>(2);
// Initialize your operations
operations.add(Arrays.asList(0,1));
operations.add(Arrays.asList(1,3));
subList(list, operations);
}
public static void subList (List<Integer> list, List<List<Integer>> operations) {
// You just iterate over the operations
for (List<Integer> operation : operations) {
// For each operation, store left and right indexes.
int left = operation.get(0);
int right = operation.get(1);
// Iterate until both indexes find each other
while (left < right) {
// Swap left and right elements in input list
int aux = list.get(left);
list.set(left, list.get(right));
list.set(right, aux);
// Now you move your indexes
++left;
--right;
}
}
System.out.println(list);
}
Please note that, depending on what the question asks, you may also need to verify if the operations indexes are within the list boundaries so you won't end up getting an ArrayIndexOutOfBoundsException. So be always careful with edge cases.

You can get subset by List.subList() , and reverse by Collections.reverse().
static void reverseSubsets(List<Integer> list, List<List<Integer>> subsets) {
for (List<Integer> subset : subsets)
Collections.reverse(list.subList(subset.get(0), subset.get(1) + 1));
}
public static void main(String[] args) {
List<Integer> list = Arrays.asList(5, 3, 2, 1, 3);
List<List<Integer>> subsets = List.of(List.of(0, 1), List.of(1, 3));
reverseSubsets(list, subsets);
System.out.println(list);
}
output:
[3, 1, 2, 5, 3]
Note:
toIndex in List.subList(int fromIndex, int toIndex) is exclusive high endpoint of the subList. So you must add 1 to subset.get(1).

Combination of subarrays

Not sure if this has been asked already but I'm having trouble printing combinations of subarrays.
Given say
int[][] x= {{1,2},{3,4,5},{6,7}};
print all valid combinations by choosing one int from each subarray.
some valid outputs are:
1,3,6
1,3,7
1,4,6
1,4,7
..
... etc
SO far my code looks like
public static void main(String args[]) {
int[][] x= {{1,2},{3,4,5},{6,7}};
for(int i = 0; i < x.length; i++){
for (int j = 0; j < 3; j++){
for (int k = 0; k < 2; k++){
System.out.println(x[0][i]+" " + x[1][j] +" "+ x[2][k]);
}
}
My code is throwing an exeption indexoutofbounds. I'm not sure where I'm going out of bounds. I know its the naive solution and brute force and theres prob a better way to solve it but this solution is the first that came across my mind.

Instead of doing i < x.length, try i < x[0].length The length of x is 3 and that of the first element is 2. Hence, the out of bounds exception. Also, it might be a good idea to get x[1].length and x[2].length instead of hard coding 3 and 2.
Try:
public class MyClass {
public static void main(String args[]) {
int[][] x= {{1,2},{3,4,5},{6,7}};
for (int i = 0; i < x[0].length; i++){
for (int j = 0; j < x[1].length; j++){
for(int k = 0; k < x[2].length; k++){
System.out.println(x[0][i] + " " + x[1][j] + " " + x[2][k]);
}
}
}
}
}

Looking for something complex written in Java 8? Here it is:
package com.test;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.List;
import java.util.Objects;
import java.util.stream.Collectors;
import java.util.stream.Stream;
public class ArrayPrint {
#SuppressWarnings("resource")
public static void main(String args[]) {
int[][] x = { { 1, 2 }, { 3, 4, 5 }, { 6, 7 } };
Stream<List<Integer>> inputs = null;
for (int[] set : x) {
List<Integer> list = new ArrayList<Integer>();
Collections.addAll(list, Arrays.stream(set).boxed().toArray(Integer[]::new));
if (inputs == null) {
inputs = Stream.of(list);
} else {
inputs = Stream.concat(inputs, Stream.of(list));
}
}
Stream<List<List<Integer>>> listified = inputs.filter(Objects::nonNull).filter(input -> !input.isEmpty())
.map(l -> l.stream().map(o -> new ArrayList<>(Arrays.asList(o))).collect(Collectors.toList()));
List<List<Integer>> combinations = listified.reduce((input1, input2) -> {
List<List<Integer>> merged = new ArrayList<>();
input1.forEach(permutation1 -> input2.forEach(permutation2 -> {
List<Integer> combination = new ArrayList<>();
combination.addAll(permutation1);
combination.addAll(permutation2);
merged.add(combination);
}));
return merged;
}).orElse(null);
combinations.forEach(System.out::println);
}
}
Output:
[1, 3, 6]
[1, 3, 7]
[1, 4, 6]
[1, 4, 7]
[1, 5, 6]
[1, 5, 7]
[2, 3, 6]
[2, 3, 7]
[2, 4, 6]
[2, 4, 7]
[2, 5, 6]
[2, 5, 7]
With int[][] x = { { 1, 2 }, { 3, 4, 5 }, { 6, 7 }, { 8, 9, 10 } };, Output:
[1, 3, 6, 8]
[1, 3, 6, 9]
[1, 3, 6, 10]
[1, 3, 7, 8]
[1, 3, 7, 9]
[1, 3, 7, 10]
[1, 4, 6, 8]
[1, 4, 6, 9]
[1, 4, 6, 10]
[1, 4, 7, 8]
[1, 4, 7, 9]
[1, 4, 7, 10]
[1, 5, 6, 8]
[1, 5, 6, 9]
[1, 5, 6, 10]
[1, 5, 7, 8]
[1, 5, 7, 9]
[1, 5, 7, 10]
[2, 3, 6, 8]
[2, 3, 6, 9]
[2, 3, 6, 10]
[2, 3, 7, 8]
[2, 3, 7, 9]
[2, 3, 7, 10]
[2, 4, 6, 8]
[2, 4, 6, 9]
[2, 4, 6, 10]
[2, 4, 7, 8]
[2, 4, 7, 9]
[2, 4, 7, 10]
[2, 5, 6, 8]
[2, 5, 6, 9]
[2, 5, 6, 10]
[2, 5, 7, 8]
[2, 5, 7, 9]
[2, 5, 7, 10]
Thus, this code can handle any two dimensional integer array.

Regarding my suggestion from the comments try this code and look at the changes to each for loop. Instead of iterating over predetermined size like for(int k = 0; k < 2; k++) We instead iterate based on the size of the array as follows for(int k = 0; k < x[1].length; k++)
public static void main(String args[]) {
int[][] x = { { 1, 2, 3}, { 3, 4, 5 }, { 6, 7 } };
for (int j = 0; j < x[0].length; j++) {
for (int k = 0; k < x[1].length; k++) {
for (int l = 0; l < x[2].length; l++) {
System.out.println(x[0][j] + " " + x[1][k] + " " + x[2][l]);
}
}
}
}
As long as your outer array only contains three arrays you can do as shown. Otherwise you will need to be a bit more sophisticated with you loops and not hardcode the specific index like for(int k = 0; k < x[0].length; k++) Here we are directly coding index 0.
Hope this helps.

You could always declare an array of objects and cast to the data-type of your choice like this.
public class PrintArray{
public static void main(String[] args){
Object[][] myArray = {{1, 2, 3},
{4, 5, null},
{7, 8, null}};
for(int x = 0; x < myArray.length; x++){
for(int y = 0; y < 3; y++){
if(myArray[x][y] != null){
System.out.print((int)myArray[x][y]);
}
}
System.out.println();
}
}
}
I hope this helps

How do you split an Array List into sublists everytime there is a duplicate value in the Array List

I have the following array list which contains the following
point ids (1,2,3,4,1,8,5,6,8,9,7,9). I am using Java 7
I was wondering how it could be split into sublists i.e the sublists below
(1,2,3,4,1)
(8,5,6,8)
(9,7,9)
I have had problems trying to use a loop within a loop (i.e check each point
from the outer loop with each of the other points in the inner loop) to get
index positions (starPosIndex and endPosIndex) where there are duplicate point ids and ArrayList.sublist(startPosIndex,endPosIndex) to get the correct sublist
int startPos = 0;
int endPos = 0;
for (int j = 0; j < polygonList3.size(); j++){
Point pointToCheck = polygonList3.get(j);
for (int k = 1; k < polygonList3.size(); k++){
Point pointToCheck2 = polygonList3.get(k);
if (pointToCheck.getID() == pointToCheck2.getID()){
startPos = startPos + endPos;
endPos = endPos + k;
//startPos = startPos + endPos;
//for (int startPos = j; startPos < polygonList3.size(); startPos = (startPos) + endPos) {
//endPos = Math.min(startPos + endPos, polygonList3.size());
finalPolygonLists.add(new ArrayList<Point>(polygonList3.subList(startPos, endPos)));//originalPtsSublist2);
//}
}
}

I would solve it in the following manner:
Allocate a HashSet to contain unique values encountered
Allocate a new list for the first sublist
Iterate over the whole list, adding each value to the set. When we encounter a value that is already in the set, we are done with the first sublist, so clear the set, and allocate a new sublist
After iteration, you will have your list of sublists, obtained in O(n) runtime

You can walk along the list, and create slices of the list (using List#subList) as you go. This can be done efficiently, by always checking whether the first element of the current segment of the list appears somewhere else in the list. If it does, you can store this "slice", and continue with the "tail" of the list. If it doesn't, you are finished (and the tail of the list may or may not be part of the result - that's up to you)
Implemented here as an example:
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
public class ListSlicing
{
public static void main(String[] args)
{
runTest(1,2,3,4,1,8,5,6,8,9,7,9);
runTest(1,2,3,4);
runTest(1,1,1,1);
runTest(1,2,1,2,1,2,1,2,1,2,1,2);
runTest();
}
private static void runTest(Integer ... numbers)
{
List<Integer> list = Arrays.asList(numbers);
System.out.println("Input: "+list);
System.out.println("Output: "+slices(list));
}
private static <T> List<List<T>> slices(List<T> input)
{
List<List<T>> slices = new ArrayList<List<T>>();
List<T> current = input;
while (current.size() > 0)
{
T first = current.get(0);
int appearance = current.subList(1, current.size()).indexOf(first);
if (appearance == -1)
{
slices.add(current);
return slices;
}
List<T> slice = current.subList(0, appearance+2);
slices.add(slice);
current = current.subList(appearance+2, current.size());
}
return slices;
}
}
The output is
Input: [1, 2, 3, 4, 1, 8, 5, 6, 8, 9, 7, 9]
Output: [[1, 2, 3, 4, 1], [8, 5, 6, 8], [9, 7, 9]]
Input: [1, 2, 3, 4]
Output: [[1, 2, 3, 4]]
Input: [1, 1, 1, 1]
Output: [[1, 1], [1, 1]]
Input: [1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2]
Output: [[1, 2, 1], [2, 1, 2], [1, 2, 1], [2, 1, 2]]
Input: []
Output: []

The following code tracks the last position for each number and as soon as it founds a duplicate, it will create the sublist and clears all previously tracked entries.
List<Integer> list = Arrays.asList( 1,2,3,4,1,8,5,6,8,9,7,9);
List<List<Integer>> sublists = new ArrayList<>();
Map<Integer,Integer> lastPos = new HashMap<>();
for(int i = 0; i < list.size(); i++) {
Integer current = list.get(i);
if(lastPos.containsKey(current)){
sublists.add(list.subList(lastPos.get(current), i+1));
lastPos.clear();
} else {
lastPos.put(current, i);
}
}
System.out.println(sublists);

Permutation between two vectors

I´m studying some techniques of algorithms and got stucked in one problem, I need to do a permutation between two groups. For example:
[1,2,3] e [5,6,7]
Need to generate:
[5,2,3] e [1,6,7]
[5,6,3] e [1,2,7]
........
And so on.
From this what I've done so far is do a permutation in one vector between yourself.
Passing one vector [1,2,3]. Generate the answer:
123
132
213
231
321
312
Based on the code below:
public void permutar(int[] num, int idx) {
for (int i = idx; i < num.length; i++) {
swap(num, i, idx);
permutar(num, idx + 1);
swap(num, i, idx);
}
if (idx == num.length - 1) {
for (int i = 0; i < num.length; i++) {
System.out.print(num[i]);
}
System.out.println("");
}
}
public void swap(int[] num, int a, int b) {
int aux = num[a];
num[a] = num[b];
num[b] = aux;
}
How to do a permutation between this two vectors?

Although you did not precisely describe what you are looking for, and attempt to answer: It seems like you are just looking for all 3-element subsets of the input (1,2,3,5,6,7). Each subset is the first vector of one solution, and the respective remaining elements the other vector.
Here is an example how this may be computed, based on a ChoiceIterable utility class that I wrote a while ago:
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Iterator;
import java.util.LinkedHashSet;
import java.util.List;
import java.util.NoSuchElementException;
import java.util.Set;
public class CombinationsOfVectors
{
public static void main(String[] args)
{
List<Integer> input = Arrays.asList(1,2,3,5,6,7);
ChoiceIterable<Integer> c = new ChoiceIterable<Integer>(3, input);
for (List<Integer> v0 : c)
{
Set<Integer> s = new LinkedHashSet<Integer>(input);
s.removeAll(v0);
List<Integer> v1 = new ArrayList<Integer>(s);
System.out.println(v0+" and "+v1);
}
}
}
// From https://github.com/javagl/Combinatorics/blob/master/src/
// main/java/de/javagl/utils/math/combinatorics/ChoiceIterable.java
// See the GitHub repo for a commented version
class ChoiceIterable<T> implements Iterable<List<T>>
{
private final List<T> input;
private final int sampleSize;
private final long numElements;
public ChoiceIterable(int sampleSize, List<T> input)
{
this.sampleSize = sampleSize;
this.input = input;
long nf = factorial(input.size());
long kf = factorial(sampleSize);
long nmkf = factorial(input.size() - sampleSize);
long divisor = kf * nmkf;
long result = nf / divisor;
numElements = result;
}
private static long factorial(int n)
{
long f = 1;
for (long i = 2; i <= n; i++)
{
f = f * i;
}
return f;
}
#Override
public Iterator<List<T>> iterator()
{
return new Iterator<List<T>>()
{
private int current = 0;
private final int chosen[] = new int[sampleSize];
{
for (int i = 0; i < sampleSize; i++)
{
chosen[i] = i;
}
}
#Override
public boolean hasNext()
{
return current < numElements;
}
#Override
public List<T> next()
{
if (!hasNext())
{
throw new NoSuchElementException("No more elements");
}
List<T> result = new ArrayList<T>(sampleSize);
for (int i = 0; i < sampleSize; i++)
{
result.add(input.get(chosen[i]));
}
current++;
if (current < numElements)
{
increase(sampleSize - 1, input.size() - 1);
}
return result;
}
private void increase(int n, int max)
{
if (chosen[n] < max)
{
chosen[n]++;
for (int i = n + 1; i < sampleSize; i++)
{
chosen[i] = chosen[i - 1] + 1;
}
}
else
{
increase(n - 1, max - 1);
}
}
#Override
public void remove()
{
throw new UnsupportedOperationException(
"May not remove elements from a choice");
}
};
}
}
The output in this example will be
[1, 2, 3] and [5, 6, 7]
[1, 2, 5] and [3, 6, 7]
[1, 2, 6] and [3, 5, 7]
[1, 2, 7] and [3, 5, 6]
[1, 3, 5] and [2, 6, 7]
[1, 3, 6] and [2, 5, 7]
[1, 3, 7] and [2, 5, 6]
[1, 5, 6] and [2, 3, 7]
[1, 5, 7] and [2, 3, 6]
[1, 6, 7] and [2, 3, 5]
[2, 3, 5] and [1, 6, 7]
[2, 3, 6] and [1, 5, 7]
[2, 3, 7] and [1, 5, 6]
[2, 5, 6] and [1, 3, 7]
[2, 5, 7] and [1, 3, 6]
[2, 6, 7] and [1, 3, 5]
[3, 5, 6] and [1, 2, 7]
[3, 5, 7] and [1, 2, 6]
[3, 6, 7] and [1, 2, 5]
[5, 6, 7] and [1, 2, 3]
If this is not what you have been looking for, you should describe more clearly and precisely what the intended result is.
(E.g. whether or not
[1, 2, 3] and [5, 6, 7]
and
[5, 6, 7] and [1, 2, 3]
count as different results is up to you, but you may filter the results accordingly)

Your task is equal to list all 3-element subsets of your 6-element set. Since the order inside the 3-Element-set does not matter, you should define an order, like 'the smaller number comes first'.
Then the algorithm gets obvious: list = [1 2 3 5 6 7]
EDIT: Set1 should always be the set with number 1 in it, to avoid symmetrical identical solutions.
For all the numbers i from 2 to 5
for all the numbers j from i+1 to 5
Set1 = {list[1], list[i], list[j]}
Set2 = "the other numbers"
This should give right your ordered list from your 9-element comment.
These are nested loops, obviously.

As mentioned in comments to OP question, you have to generate Combination. The usual usecase is to take some subset of elements from a set. In this problem you take subset which represents elements taken from the first set and the rest will be taken from the second.
To implement combination I recommend a for loop counting from 0 to 2^n (where n is number of elements in one array). Then take this number and represent it in binary. Now each 0 or 1 in binary representation says from which set should be given number (and second set will be exact opposite).
I guess you have this as a homework or mental excercise so code is not included :]