I have an ArrayList<ArrayList<int[]>>, which has an undefined number of int[].
Example input (note that text "= Condition 0/1/2" is not part of the ArrayList).
= Condition 0
[2, 6], [3, 7]
= Condition 1
[1, 3], [2, 4], [2, 3]
= Condition 2
[1, 2]
(PART A) I wish to create all possible permutations of all the int[] pairs (ArrayList<int[]> option = new ArrayList<>();, line 9), provided there is only one pair from each condition, e.g. first from C0, second from
[2,6], [1,3], [1,2]
[2,6], [2,4], [1,2]
[2,6], [2,3], [1,2]
[3,7], [1,3], [1,2]
and so on...
(PART B) Once I have all possible permutations (2^numOfConditions), I combine each value from ArrayList<int[]> pairs and put all integers into a Set to make sure I only get unique numbers. Eventually, I return the shortest set (i.e., the one that has the biggest number of repeating ints).
public static Set<Integer> findOptimalPairs(ArrayList<ArrayList<int[]>> conditions) {
// A
ArrayList<ArrayList<int[]>> listOfConditions = new ArrayList<>();
for (int i = 0; i < conditions.get(0).size(); i ) {
for (int j = 0; j < conditions.get(1).size(); j ) {
for (int k = 0; k < conditions.get(2).size(); k ) {
ArrayList<int[]> option = new ArrayList<>();
option.add(conditions.get(0).get(i));
option.add(conditions.get(1).get(j));
option.add(conditions.get(2).get(k));
listOfConditions.add(option);
}
}
}
//A
//B
Set<Integer> best = new HashSet<>();
for (ArrayList<int[]> pairs : listOfConditions) {
Set<Integer> curr = new HashSet<>();
for (int[] pair : pairs) {
for (int num : pair) curr.add(num);
}
best = (best.size() > 0 && best.size() <= curr.size()) ? best : curr;
}
return best;
//B
}
PART B works just fine, but how can I modify A, so it will be able to go through conditions' sizes that are different from 3? I hardcoded the values (0,1,2) for now, because I have no idea how to modify it for larger collections.
Have spent so much time on this already, and I feel like the answer is not particularly difficult.
Java 11, if that matters. Thank you.
CodePudding user response：
Something like this should work
public static void setListOfConditions(ArrayList<ArrayList<int[]>> conditions, ArrayList<ArrayList<int[]>> listOfConditions, int depth, ArrayList<Integer> indices){
// when we get to the end then this code creates and adds the option to listOfCondition. This would be equivalent to the code you have inside all the for-loops.
if (conditions.size() == depth){
ArrayList<int[]> option = new ArrayList<>();
// this loop does what the lines
// option.add(conditions.get(0).get(i)); etc
// did in your code.
for (int l = 0; l < indices.size(); l ) {
option.add(conditions.get(l).get(indices.get(l)));
}
// add to listOfConditions
listOfConditions.add(option);
// remove the last index such that it can be changed.
indices.remove(indices.size()-1);
return;
}
// recursive call to the function in a for-loop.
for (int k = 0; k < conditions.get(depth).size(); k ) {
// the variable indices contains for example i,j,k values. Here we make sure that i,j and k have the correct values.
// In the first iteration we get a list of the number 0. (You can think of this as the start of your first loop when i = 0.)
// In every iteration after that we keep adding indices.
// Example iteration1: i = 0
// Example iteration2: i = 0, j = 0
// Example iteration3: i = 0, j = 0, k = 0
// In this example we have reached the end so the code inside the if-statement will be run.
try {
indices.set(depth, k);
} catch (IndexOutOfBoundsException e) {
indices.add(depth, k);
}
// recursive call to the function where depth is increased everytime.
setListOfConditions(conditions, listOfConditions, depth 1, indices);
}
}
public static Set<Integer> findOptimalPairs(ArrayList<ArrayList<int[]>> conditions) {
// A
ArrayList<ArrayList<int[]>> listOfConditions = new ArrayList<>();
// this function gives all the conditions by adding them to listOfConditions.
setListOfConditions(conditions,listOfConditions,0,new ArrayList<>());
// A
// B
Set<Integer> best = new HashSet<>();
for (ArrayList<int[]> pairs : listOfConditions) {
Set<Integer> curr = new HashSet<>();
for (int[] pair : pairs) {
for (int num : pair) curr.add(num);
}
best = (best.size() > 0 && best.size() <= curr.size()) ? best : curr;
}
return best;
// B
}
In your code you have loops inside loops. Here, we use recursive calls to do the for loops. I have tried to make comparisons to the code you currently have. So when I refer to i,j,k in the comments, I am referring to how you are currently using them.
CodePudding user response：
how can I modify A, so it will be able to go through conditions' sizes that are different from 3? I hardcoded the values (0,1,2) for now, because I have no idea how to modify it for larger collections.
Basically you create a 1D array that will hold at each index the number of int[] arrays for that set.
Then you create an additional 1D array of the same size and you "count" using the stored sizes to know when to reset each counter. If you've ever counted in other bases you should know exactly what I mean, except we are treating the left side as the least significant digit.
For instance, with your data set of:
[2, 6], [3, 7]
[1, 3], [2, 4], [2, 3]
[1, 2]
There are three rows, so you'd make an array of size three and store the number of items in each set.
Here I'm hard-coding just to show the process. In the actual code it will be dynamic based on the data passed in:
int[] setSize = {2, 3, 1};
Now we start off a new array of the same size, with all values at zero (which is the default anyways):
int[] counters= {0, 0, 0}
Starting from the left, we increment the value in that index until we reach "set size". Whenever we hit "set size" we reset that index back to zero and increment the column to the right (following the same rule for each column of resetting and incrementing to the right).
These would be the sequences generated:
[0, 0, 0]
[1, 0, 0]
[0, 1, 0]
[1, 1, 0]
[0, 2, 0]
[1, 2, 0]
The numbers represent what index from the int[] array you should pick from each set in that combination. So then we just translate those indices to the corresponding int[] array from the input data set.
This approach is ITERATIVE, requiring no recursion.
Here's what that code might look like:
import java.util.*;
import java.util.ArrayList;
class Main {
public static void main(String[] args) {
ArrayList<ArrayList<int[]>> conditions = new ArrayList<ArrayList<int[]>>();
ArrayList<int[]> condition1 = new ArrayList<int[]>();
condition1.add(new int[] {2,6});
condition1.add(new int[] {3,7});
conditions.add(condition1);
ArrayList<int[]> condition2 = new ArrayList<int[]>();
condition2.add(new int[] {1,3});
condition2.add(new int[] {2,4});
condition2.add(new int[] {2,3});
conditions.add(condition2);
ArrayList<int[]> condition3 = new ArrayList<int[]>();
condition3.add(new int[] {1,2});
conditions.add(condition3);
System.out.println("Input Data:");
display(conditions);
System.out.println();
ArrayList<ArrayList<int[]>> combos = findOptimalPairs(conditions);
System.out.println("Output Data:");
display(combos);
}
public static void display(ArrayList<ArrayList<int[]>> data) {
for(ArrayList<int[]> set : data) {
for(int i=0; i<set.size(); i ) {
System.out.print(Arrays.toString(set.get(i)));
if (i<(set.size()-1)) {
System.out.print(", ");
}
}
System.out.println();
}
}
public static ArrayList<ArrayList<int[]>> findOptimalPairs(ArrayList<ArrayList<int[]>> conditions) {
ArrayList<ArrayList<int[]>> combos = new ArrayList<ArrayList<int[]>>();
int combinations = 1;
int[] setSize = new int[conditions.size()];
for(int i=0; i<conditions.size(); i ) {
ArrayList<int[]> set = conditions.get(i);
int size = set.size();
setSize[i] = size;
combinations = combinations * size;
}
int[] counters = new int[setSize.length];
for(int i=0; i<combinations; i ) {
ArrayList<int[]> combo = new ArrayList<int[]>();
for(int j=0; j<counters.length; j ) {
combo.add(conditions.get(j).get(counters[j]));
}
combos.add(combo);
for(int j=0; j<counters.length; j ) {
if (counters[j]<(setSize[j]-1)) {
counters[j] ;
break;
}
else {
counters[j] = 0;
}
}
}
return combos;
}
}
Output generated:
Input Data:
[2, 6], [3, 7]
[1, 3], [2, 4], [2, 3]
[1, 2]
Output Data:
[2, 6], [1, 3], [1, 2]
[3, 7], [1, 3], [1, 2]
[2, 6], [2, 4], [1, 2]
[3, 7], [2, 4], [1, 2]
[2, 6], [2, 3], [1, 2]
[3, 7], [2, 3], [1, 2]