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Finding the element with the lowest difference in an ArrayList when compared to a constant

I am trying to have this code determine which element has the closest value to a constant.
In this code the variable boxes = 5, any element that has boxCapacity >= boxes is added to an ArrayList. From that list, the one with the closest boxCapacity to boxes should be used. I am able to select those greater than boxes, but unable to pick that with the closest boxCapacity.
public void deliver(double miles, int boxes) {
for (int i = 0; i < cars.size(); i++){
if (cars.get(i).getBoxCapacity() >= boxes){
deliveryCars = new ArrayList<Car>();
deliveryCars.add(cars.get(i));
smallest = deliveryCars.get(0).getBoxCapacity();
for(j = 0; j < deliveryCars.size(); j++){
if (deliveryCars.get(j).getBoxCapacity() < smallest) {
smallest = deliveryCars.get(j).getBoxCapacity();
k++;
}
}
}
}
System.out.println("Delivering with " + deliveryCars.get(k).getPlate());
}
I tried to make a new list, but it has not been working out.

You can simplify your code to something that looks like that
public void deliver(double miles, int boxes){
// check if there are cars availible
if (!cars.isEmpty()) {
// assume that first car in a list is best for delivery
int smallest = cars.get(0).getBoxCapacity();
Car deliveryCar = cars.get(0);
// iterating over all cars in a list
// but still compares to the first car in a list
for (Car car : cars) {
if (car.getBoxCapacity() >= boxes
&& car.getBoxCapacity() < smallest) {
deliveryCar = car;
}
}
System.out.println("Delivering with " + deliveryCar.getPlate());
}
}

Using Java 8 streams...
Car deliveryVehicle = cars
.stream()
.filter(c -> c.getBoxCapacity() > boxes)
.min(Comparator.comparingInt(Car::getBoxCapacity))
.orElse(null);
Assuming your cars was an iterable/streamable collection, this creates a stream, filters it to extract all instances where the capacity is greater than boxes, finds the element with the smallest capacity, and returns it, or null if there were no cars with more than boxes capacity.
You can then do whatever you want with the returned Car object, like call getPlate(). Remember to check for null for the case where no acceptable car was found.

This answer builds from the clarifications that #DataDino provided, namely that the goal is to find the car with the lowest boxCapacity that is greater than the targetCapacity.
Given the list of cars, you can filter out any car with a boxCapacity smaller than the target, and then select the minimum boxCapacity from what is left.
List<Car> cars = List.of(new Car(8), new Car(3), new Car(5), new Car(6));
int suggestedCapacity = 4;
Optional<Car> bestFit = cars.stream()
.filter(car -> car.getBoxCapacity() >= suggestedCapacity)
.min(Comparator.comparing(Car::getBoxCapacity));
if (bestFit.isPresent()) {
System.out.println("found car with capacity " + bestFit.get().getBoxCapacity());
} else {
System.out.println("No suitable car found");
}
The Streams api takes care of the list manipulation and keeping track of the internal state of the minimum for you.

You do not need a new list. The task you've outlined is a sequential search through an unordered list, and unless I misunderstand your goal, you only need a single for loop -- that is, you only need to iterate through the list one time. Since you are looking for a single item and you don't need to look at more than one item at a time to see if it's the best one so far, you only need one variable to keep track of its location in the list.
Here's a working sample. Notice the variable names describe the purpose (e.g. "mimimumBoxCapacity" instead of the ambiguous "boxes"). This helps me better understand what my code is doing.
// print the plate number of the car with the smallest boxCapacity
// greater than a specified minimum
public void deliver(List<Car> cars, double miles, int minimumBoxCapacity)
{
if ((cars != null) && (cars.size() > 0))
{
int indexOfBestMatch = -1; // negative index means no match yet
for (int i = 0; i < cars.size(); i++)
{
if (cars.get(i).getBoxCapacity() > minimumBoxCapacity)
{
if (indexOfBestMatch < 0)
{
// this is the only match seen so far; remember it
indexOfBestMatch = i;
}
else
{
// found a better match; replace the old best match
if (cars.get(i).getBoxCapacity() < cars.get(indexOfBestMatch).getBoxCapacity())
{
indexOfBestMatch = i;
}
}
}
}
if (indexOfBestMatch >= 0)
{
System.out.println("Delivering with " + cars.get(indexOfBestMatch).getPlate());
}
}
}
This code illustrates how your algorithm would need to change to do what you want. DataDino's answer using a Car variable to keep track of the best fit is even clearer, especially where the method returns a Car result and lets the calling logic decide what to do with that result.
(Your original code didn't compile, because variables like "smallest" and "deliveryCars" weren't defined before they were used. It would be helpful in the future if you post code that compiles, even if it doesn't yet do what you want it to do.)

So I think what you're saying is that you have of list of values say [0, 1, 2, 3, 4, 5, 6] and then you are given another number say 4, and what you want to do is to select the number from the list that is the smallest of all the numbers greater than 4, so in this case you'd want to choose 5, right?
Well, there are a ton of ways to do that. But the fastest way to do it is to go through the list one time and keep track of the smallest number greater than your 'targetNumber':
...
public Integer getNextBiggest(List<Integer> numbers, int targetNumber)
{
// Set the default 'nextInt' to the largest possible value.
int nextInt = Integer.MAX_VALUE;
for (int i = 0; i < numbers.length; i++) {
// If the current number is greater than our targetNumber.
if (sortedList.get(i) > targetNumber) {
// Set the nextInt variable to the MINIMUM of current number
// and the current nextInt value.
nextInt = Math.min(nextInt, sortedList.get(i));
}
}
return nextInt;
}
...
So that would work, but your list is a bit more complicated since you're using Objects and not integers, that said, it's a simple conversion:
First some assumptions:
cars is a List.
Car object has a getBoxCapacity() method that returns an int.
Car object has a getPlate() method that returns a String.
Ok so it might look like this:
...
public Car getBestFitCar(List<Car> cars, int targetBoxCapacity)
{
// Default best fit is null. No best first yet.
Car bestFit = null;
for (int i = 0; i < cars.length; i++) {
Car current = cars.get(i);
// If the current Car box capacity is greater than our target box capacity.
if (current.getBoxCapacity() > targetBoxCapacity) {
// Set the bestFit variable to the Car that has the MINIMUM
// 'box capacity' between the current Car and the bestFit Car.
if (bestFit == null || bestFit.getBoxCapacity() > current.getBoxCapacity()) {
bestFit = current;
}
}
}
return bestFit;
}
public static void main(String[] args) {
List<Car> cars = new ArrayList<>();
// add some cars here...
int targetBoxCapacity = 5;
Car bestFit = getBestFitCar(cars, targetBoxCapacity);
if (bestFit != null) {
System.out.println("Best fit car is: " + bestFit.getPlate());
} else {
System.out.println("No car can fit " + targetBoxCapacity + " boxes.");
}
}
Update:
I've seen some nice responses using streams, but I'd to add some caution. Streams make writing the code faster/more readable but would end up being less efficient in time/space than a solution with simple loops. This solution only uses O(1) extra space, and O(n) time in the worst case.
I'd figure the stream answers would use O(n) extra space, and O(n * n log n) time in the worst case.
So, if you have a tiny list I'd say go with the simple really cool streams solutions, but you have a list of a lot of elements that you'll be better off with a more traditional approach.

Recursive backtracking in Java for solving a crossword

I need to solve a crossword given the initial grid and the words (words can be used more than once or not at all).
The initial grid looks like that:
++_+++
+____+
___+__
+_++_+
+____+
++_+++
Here is an example word list:
pain
nice
pal
id
The task is to fill the placeholders (horizontal or vertical having length > 1) like that:
++p+++
+pain+
pal+id
+i++c+
+nice+
++d+++
Any correct solution is acceptable, and it's guaranteed that there's a solution.
In order to start to solve the problem, I store the grid in 2-dim. char array and I store the words by their length in the list of sets: List<Set<String>> words, so that e.g. the words of length 4 could be accessed by words.get(4)
Then I extract the location of all placeholders from the grid and add them to the list (stack) of placeholders:
class Placeholder {
int x, y; //coordinates
int l; // the length
boolean h; //horizontal or not
public Placeholder(int x, int y, int l, boolean h) {
this.x = x;
this.y = y;
this.l = l;
this.h = h;
}
}
The main part of the algorithm is the solve() method:
char[][] solve (char[][] c, Stack<Placeholder> placeholders) {
if (placeholders.isEmpty())
return c;
Placeholder pl = placeholders.pop();
for (String word : words.get(pl.l)) {
char[][] possibleC = fill(c, word, pl); // description below
if (possibleC != null) {
char[][] ret = solve(possibleC, placeholders);
if (ret != null)
return ret;
}
}
return null;
}
Function fill(c, word, pl) just returns a new crossword with the current word written on the current placeholder pl. If word is incompatible with pl, then function returns null.
char[][] fill (char[][] c, String word, Placeholder pl) {
if (pl.h) {
for (int i = pl.x; i < pl.x + pl.l; i++)
if (c[pl.y][i] != '_' && c[pl.y][i] != word.charAt(i - pl.x))
return null;
for (int i = pl.x; i < pl.x + pl.l; i++)
c[pl.y][i] = word.charAt(i - pl.x);
return c;
} else {
for (int i = pl.y; i < pl.y + pl.l; i++)
if (c[i][pl.x] != '_' && c[i][pl.x] != word.charAt(i - pl.y))
return null;
for (int i = pl.y; i < pl.y + pl.l; i++)
c[i][pl.x] = word.charAt(i - pl.y);
return c;
}
}
Here is the full code on Rextester.
The problem is that my backtracking algorithm doesn't work well. Let's say this is my initial grid:
++++++
+____+
++++_+
++++_+
++++_+
++++++
And this is the list of words:
pain
nice
My algorithm will put the word pain vertically, but then when realizing that it was a wrong choice it will backtrack, but by that time the initial grid will be already changed and the number of placeholders will be reduced. How do you think the algorithm can be fixed?

This can be solved in 2 ways:
Create a deep copy of the matrix at the start of fill, modify and return that (leaving the original intact).
Given that you already pass around the matrix, this wouldn't require any other changes.
This is simple but fairly inefficient as it requires copying the matrix every time you try to fill in a word.
Create an unfill method, which reverts the changes made in fill, to be called at the end of each for loop iteration.
for (String word : words.get(pl.l)) {
if (fill(c, word, pl)) {
...
unfill(c, word, pl);
}
}
Note: I changed fill a bit as per my note below.
Of course just trying to erase all letter may erase letters of other placed words. To fix this, we can keep a count of how many words each letter is a part of.
More specifically, have a int[][] counts (which will also need to be passed around or be otherwise accessible) and whenever you update c[x][y], also increment counts[x][y]. To revert a placement, decrease the count of each letter in that placement by 1 and only remove letters with a count of 0.
This is somewhat more complex, but much more efficient than the above approach.
In terms of code, you might put something like this in fill:
(in the first part, the second is similar)
for (int i = pl.x; i < pl.x + pl.l; i++)
counts[pl.y][i]++;
And unfill would look something like this: (again for just the first part)
for (int i = pl.x; i < pl.x + pl.l; i++)
counts[pl.y][i]--;
for (int i = pl.x; i < pl.x + pl.l; i++)
if (counts[pl.y][i] == 0)
c[pl.y][i] = '_';
// can also just use a single loop with "if (--counts[pl.y][i] == 0)"
Note that, if going for the second approach above, it might make more sense to simply have fill return a boolean (true if successful) and just pass c down to the recursive call of solve. unfill can return void, since it can't fail, unless you have a bug.
There is only a single array that you're passing around in your code, all you're doing is changing its name.
See also Is Java "pass-by-reference" or "pass-by-value"?

You identified it yourself:
it will backtrack, but by that time the initial grid will be already
changed
That grid should be a local matrix, not a global one. That way, when you back up with a return of null, the grid from the parent call is still intact, ready to try the next word in the for loop.
Your termination logic is correct: when you find a solution, immediately pass that grid back up the stack.

Java - Return random index of specific character in string

So given a string such as: 0100101, I want to return a random single index of one of the positions of a 1 (1, 5, 6).
So far I'm using:
protected int getRandomBirthIndex(String s) {
ArrayList<Integer> birthIndicies = new ArrayList<Integer>();
for (int i = 0; i < s.length(); i++) {
if ((s.charAt(i) == '1')) {
birthIndicies.add(i);
}
}
return birthIndicies.get(Randomizer.nextInt(birthIndicies.size()));
}
However, it's causing a bottle-neck on my code (45% of CPU time is in this method), as the strings are over 4000 characters long. Can anyone think of a more efficient way to do this?

If you're interested in a single index of one of the positions with 1, and assuming there is at least one 1 in your input, you can just do this:
String input = "0100101";
final int n=input.length();
Random generator = new Random();
char c=0;
int i=0;
do{
i = generator.nextInt(n);
c=input.charAt(i);
}while(c!='1');
System.out.println(i);
This solution is fast and does not consume much memory, for example when 1 and 0 are distributed uniformly. As highlighted by #paxdiablo it can perform poorly in some cases, for example when 1 are scarce.

You could use String.indexOf(int) to find each 1 (instead of iterating every character). I would also prefer to program to the List interface and to use the diamond operator <>. Something like,
private static Random rand = new Random();
protected int getRandomBirthIndex(String s) {
List<Integer> birthIndicies = new ArrayList<>();
int index = s.indexOf('1');
while (index > -1) {
birthIndicies.add(index);
index = s.indexOf('1', index + 1);
}
return birthIndicies.get(rand.nextInt(birthIndicies.size()));
}
Finally, if you need to do this many times, save the List as a field and re-use it (instead of calculating the indices every time). For example with memoization,
private static Random rand = new Random();
private static Map<String, List<Integer>> memo = new HashMap<>();
protected int getRandomBirthIndex(String s) {
List<Integer> birthIndicies;
if (!memo.containsKey(s)) {
birthIndicies = new ArrayList<>();
int index = s.indexOf('1');
while (index > -1) {
birthIndicies.add(index);
index = s.indexOf('1', index + 1);
}
memo.put(s, birthIndicies);
} else {
birthIndicies = memo.get(s);
}
return birthIndicies.get(rand.nextInt(birthIndicies.size()));
}

Well, one way would be to remove the creation of the list each time, by caching the list based on the string itself, assuming the strings are used more often than they're changed. If they're not, then caching methods won't help.
The caching method involves, rather than having just a string, have an object consisting of:
current string;
cached string; and
list based on the cached string.
You can provide a function to the clients to create such an object from a given string and it would set the string and the cached string to whatever was passed in, then calculate the list. Another function would be used to change the current string to something else.
The getRandomBirthIndex() function then receives this structure (rather than the string) and follows the rule set:
if the current and cached strings are different, set the cached string to be the same as the current string, then recalculate the list based on that.
in any case, return a random element from the list.
That way, if the list changes rarely, you avoid the expensive recalculation where it's not necessary.
In pseudo-code, something like this should suffice:
# Constructs fastie from string.
# Sets cached string to something other than
# that passed in (lazy list creation).
def fastie.constructor(string s):
me.current = s
me.cached = s + "!"
# Changes current string in fastie. No list update in
# case you change it again before needing an element.
def fastie.changeString(string s):
me.current = s
# Get a random index, will recalculate list first but
# only if necessary. Empty list returns index of -1.
def fastie.getRandomBirthIndex()
me.recalcListFromCached()
if me.list.size() == 0:
return -1
return me.list[random(me.list.size())]
# Recalculates the list from the current string.
# Done on an as-needed basis.
def fastie.recalcListFromCached():
if me.current != me.cached:
me.cached = me.current
me.list = empty
for idx = 0 to me.cached.length() - 1 inclusive:
if me.cached[idx] == '1':
me.list.append(idx)
You also have the option of speeding up the actual searching for the 1 character by, for example, useing indexOf() to locate them using the underlying Java libraries rather than checking each character individually in your own code (again, pseudo-code):
def fastie.recalcListFromCached():
if me.current != me.cached:
me.cached = me.current
me.list = empty
idx = me.cached.indexOf('1')
while idx != -1:
me.list.append(idx)
idx = me.cached.indexOf('1', idx + 1)
This method can be used even if you don't cache the values. It's likely to be faster using Java's probably-optimised string search code than doing it yourself.
However, you should keep in mind that your supposed problem of spending 45% of time in that code may not be an issue at all. It's not so much the proportion of time spent there as it is the absolute amount of time.
By that, I mean it probably makes no difference what percentage of the time being spent in that function if it finishes in 0.001 seconds (and you're not wanting to process thousands of strings per second). You should only really become concerned if the effects become noticeable to the user of your software somehow. Otherwise, optimisation is pretty much wasted effort.

You can even try this with best case complexity O(1) and in worst case it might go to O(n) or purely worst case can be infinity as it purely depends on Randomizer function that you are using.
private static Random rand = new Random();
protected int getRandomBirthIndex(String s) {
List<Integer> birthIndicies = new ArrayList<>();
int index = s.indexOf('1');
while (index > -1) {
birthIndicies.add(index);
index = s.indexOf('1', index + 1);
}
return birthIndicies.get(rand.nextInt(birthIndicies.size()));
}

If your Strings are very long and you're sure it contains a lot of 1s (or the String you're looking for), its probably faster to randomly "poke around" in the String until you find what you are looking for. So you save the time iterating the String:
String s = "0100101";
int index = ThreadLocalRandom.current().nextInt(s.length());
while(s.charAt(index) != '1') {
System.out.println("got not a 1, trying again");
index = ThreadLocalRandom.current().nextInt(s.length());
}
System.out.println("found: " + index + " - " + s.charAt(index));
I'm not sure about the statistics, but it rare cases might happen that this Solution take much longer that the iterating solution. On case is a long String with only a very few occurrences of the search string.
If the Source-String doesn't contain the search String at all, this code will run forever!

One possibility is to use a short-circuited Fisher-Yates style shuffle. Create an array of the indices and start shuffling it. As soon as the next shuffled element points to a one, return that index. If you find you've iterated through indices without finding a one, then this string contains only zeros so return -1.
If the length of the strings is always the same, the array indices can be static as shown below, and doesn't need reinitializing on new invocations. If not, you'll have to move the declaration of indices into the method and initialize it each time with the correct index set. The code below was written for strings of length 7, such as your example of 0100101.
// delete this and uncomment below if string lengths vary
private static int[] indices = { 0, 1, 2, 3, 4, 5, 6 };
protected int getRandomBirthIndex(String s) {
int tmp;
/*
* int[] indices = new int[s.length()];
* for (int i = 0; i < s.length(); ++i) indices[i] = i;
*/
for (int i = 0; i < s.length(); i++) {
int j = randomizer.nextInt(indices.length - i) + i;
if (j != i) { // swap to shuffle
tmp = indices[i];
indices[i] = indices[j];
indices[j] = tmp;
}
if ((s.charAt(indices[i]) == '1')) {
return indices[i];
}
}
return -1;
}
This approach terminates quickly if 1's are dense, guarantees termination after s.length() iterations even if there aren't any 1's, and the locations returned are uniform across the set of 1's.

checking if my array elements meet requirements

I need to create a method which checks each element in my array to see if it is true or false, each element holds several values such as mass, formula, area etc for one compound, and in total there are 30 compounds (so the array has 30 elements). I need an algorithm to ask if mass < 50 and area > 5 = true .
My properties class looks like:
public void addProperty (Properties pro )
{
if (listSize >=listlength)
{
listlength = 2 * listlength;
TheProperties [] newList = new TheProperties [listlength];
System.arraycopy (proList, 0, newList, 0, proList.length);
proList = newList;
}
//add new property object in the next position
proList[listSize] = pro;
listSize++;
}
public int getSize()
{
return listSize;
}
//returns properties at a paticular position in list numbered from 0
public TheProperties getProperties (int pos)
{
return proList[pos];
}
}
and after using my getters/setters from TheProperties I put all the information in the array using the following;
TheProperties tp = new properties();
string i = tp.getMass();
String y = tp.getArea();
//etc
theList.addProperty(tp);
I then used the following to save an output of the file;
StringBuilder builder = new StringBuilder();
for (int i=0; i<theList.getSize(); i++)
{
if(theList.getProperties(i).getFormatted() != null)
{
builder.append(theList.getProperties(i).getFormatted());
builder.append("\n");
}
}
SaveFile sf = new SaveFile(this, builder.toString());
I just cant work out how to interrogate each compound individually for whether they reach the value or not, reading a file in and having a value for each one which then gets saved has worked, and I can write an if statement for the requirements to check against, but how to actually check the elements for each compound match the requirements? I am trying to word this best I can, I am still working on my fairly poor java skills.

Not entirely sure what you are after, I found your description quite hard to understand, but if you want to see if the mass is less than 50 and the area is greater than 5, a simple if statement, like so, will do.
if (tp.getMass() < 50 && tp.getArea() > 5) {}
Although, you will again, have to instantiate tp and ensure it has been given its attributes through some sort of constructor.

Lots of ways to do this, which makes it hard to answer.
You could check at creation time, and just not even add the invalid ones to the list. That would mean you only have to loop once.
If you just want to save the output to the file, and not do anything else, I suggest you combine the reading and writing into one function.
Open up the read and the write file
while(read from file){
check value is ok
write to file
}
close both files
The advantage of doing it this way are:
You only loop through once, not three times, so it is faster
You never have to store the whole list in memory, so you can handle really large files, with thousands of elements.

In case the requirements changes, you can write method that uses Predicate<T>, which is a FunctionalInterface designed for such cases (functionalInterfaces was introduced in Java 8):
// check each element of the list by custom condition (predicate)
public static void checkProperties(TheList list, Predicate<TheProperties> criteria) {
for (int i=0; i < list.getSize(); i++) {
TheProperties tp = list.get(i);
if (!criteria.apply(tp)) {
throw new IllegalArgumentException(
"TheProperty at index " + i + " does not meet the specified criteria");
}
}
}
If you want to check if mass < 50 and area > 5, you would write:
checkProperties(theList, new Predicate<TheProperties> () {
#Override
public boolean apply(TheProperties tp) {
return tp.getMass() < 50 && tp.getArea() > 5;
}
}
This can be shortened by using lambda expression:
checkProperties(theList, (TheProperties tp) -> {
return tp.getMass() < 50 && tp.getArea() > 5;
});

Putting elements into a sorted list

Folks, my method needs to add a new element into already a sorted list, i.e. in a proper position. The point is the method has to add the objects in a diagonal sort. For example,
board.set(1,1,11);
board.set(2,4,33);
board.set(3,4,66);
board.set(3,2,44);
board.set(3,3,55);
board.set(1,4,88);
board.set(0,2,77);
board.set(0,5,99);
board.set(2,1,22);
The result should be:
[(2,1,22), (3,2,44), (1,1,11), (3,3,55), (3,4,66), (0,2,77), (2,4,33), (1,4,88), (0,5,99)]
But my program prints this:
[(3,4,66), (3,3,55), (3,2,44), (2,4,33), (2,1,22), (1,4,88), (1,1,11), (0,5,99), (0,2,77)]
i.e. it does not put the objects into the proper positions.
I have a LinkedList<RowColElem<T>>leftDiagSeq where the objects are added and put into a proper position "on the go". What is my code missing?
NOTE: Im not allowed to use comparators, comparable interface!
Code
LinkedList<RowColElem<T>> rowColSeq;
private void sortedLeftDiagSeq(int row, int col, T x){
RowColElem<T> object = new RowColElem<T>(row, col, x);
ListIterator<RowColElem<T>> iter = leftDiagSeq.listIterator();
RowColElem<T> inListObject;
boolean added = false;
while(iter.hasNext()){
inListObject = iter.next();
if(object.getRow()-1 < inListObject.getRow() ||
object.getRow()-1 == inListObject.getRow() &&
object.getCol()-1 < inListObject.getCol()){
if( iter.hasPrevious() ){
iter.add(object);
}
}
}
}

Primary criterion is an elements "distance" from the main diagonal, negative distances indicating the lower triangle matrix.
if( object.getCol() - object.getRow() < inListObject.getCol() - inListObject.getRow()
||
object.getCol() - object.getRow() == inListObject.getCol() - inListObject.getRow() &&
object.getCol() < inListObject.getCol()){ ... }
I'm not sure about the last term. The data you've provided would also produce the expected result if row numbers are used to break the tie of equal distance from the main diagonal. Probably this doesn't matter since you want order from top-left to bottom-right within one diagonal.