Remove N first occurrences of a number in an ArrayList

Question

I have an ArrayList and I have to remove the first 2 instances of number 1

public void remove(int num, int element) {

    ArrayList<Integer> arr = [1, 2, 1, 2, 1];
    // ...
}

element is the number inside myArray, num is how many times I have to remove the element.

So if num is 2 and element is 1, the ArrayList will contain [2, 2, 1]

How can I do it? I've tried a for loop but I don't know how to write the 'how many times'. Thank you

Answer 1

First, it may be needed to make sure that the input list contains at least num occurrences of the needed elements, that is, the appropriate indexes should be tracked and as soon as the num elements is detected, the elements at these indexes are removed:

public static void removeNFirst(ArrayList<Integer> list, int num, int element) {
    int[] indexes = IntStream.range(0, list.size())
        .filter(i -> list.get(i) == element)
        .limit(num)
        .toArray();
    
    // make sure at least num elements are found
    if (indexes.length < num) {
        return;
    }

    for (int i = indexes.length; i-- > 0; ) {
        list.remove(indexes[i]);
    }
}

Test:

ArrayList<Integer> list = new ArrayList<>(Arrays.asList(1, 2, 1, 2, 1, 3));
System.out.println("Initial: " + list);
removeNFirst(list, 3, 1);
System.out.println("Removed three first 1: " + list);
removeNFirst(list, 2, 2);
System.out.println("Removed two first 2: " + list);

Output:

Initial: [1, 2, 1, 2, 1, 3]
Removed three first 1: [2, 2, 3]
Removed two first 2: [3]

Answer 2

SHORT ANSWER TO YOUR QUESTION:

You can keep on reducing the number till it reaches 0. If the number matches the element, then you can remove it.

QUESTION THAT NEEDS TO BE THOUGHT UPON:

What to do in case the number of elements to be removed are more than those present in the list . for example in case of your array, what if it had been element 1 to be removed more than 4 times.

SAMPLE SOLUTION:

The logic i'm providing below works for the case when the number of elements to be removed are always less than or equal to the elements present in the list.

public class Swtich {
    public static void main(String[] args) {

        remove(2,1);
    }

    public static void remove(int num, int element) {

        ArrayList<Integer> arr = new ArrayList<>();
        arr.add(1);
        arr.add(2);
        arr.add(1);
        arr.add(2);
        arr.add(1);

        System.out.println(arr);
        int i = 0;

        while (num > 0) {
//          System.out.println(arr);
            if (arr.get(i) == element) {
                arr.remove(i);
                num--;
            } else {
                i++;
            }
        }

        System.out.println(arr);
    }
}

and the output is as follows :

[1, 2, 1, 2, 1]
[2, 2, 1]

Answer 3

public static void remove(ArrayList arr, int num, int element){
    int removeCounter = 0;
    while (removeCounter != num) {
        arr.remove(arr.indexOf(element));
        removeCounter++;
}

You can use a counter like in my code... Also you should create ArrayList outside of the method block so you can print it in main method.

Answer 4

Combine an iterator with an counter.

public void remove(int num, int element) {
    ArrayList<Integer> arr = [1, 2, 1, 2, 1];
    Iterator<Integer> arrIterator = arr.iterator();
    int count = 0;
    while(arrIterator.hasNext() && num > count) {
        if (arrIterator.next().equals(element)) {
            arrIterator.remove();
            count++;
        }
    }
}

Answer 5

Call remove(Object), which removes the first instance found, n times:

Integer numberToRemove = 1; // note: must be Integer
for (int i = 0; i < n; i++) { 
    arr.remove(numberToRemove);
}

Note that for List<Integer>, you must remove an Integer, otherwise the remove(int)` method will be called which removes the element at position, rather than by value.