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Home/ Questions/Q 7403031
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Editorial Team
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Editorial Team
Asked: May 29, 20262026-05-29T04:56:39+00:00 2026-05-29T04:56:39+00:00

What is the best (or anyway, really) of going through a bi-dimensional ArrayList<ArrayList<Integer>> and

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What is the best (or anyway, really) of going through a bi-dimensional ArrayList<ArrayList<Integer>> and for every Int that is equal to 1 you leave it, otherwise you subtract 1 from it.
i.e. if arrayList.get(i).get(j) == 3 it will now be 2 and so forth, but if it is 1 it stays 1) Only in specific columns of the ARRAYLIST<ARRAYLIST<INTEGER>>.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-29T04:56:41+00:00Added an answer on May 29, 2026 at 4:56 am

    Get a shovel – there’s just one way to do it. Iterate over all the columns in all the rows:

    // example declaration only - initially all zeros until you set them.
    // assumes nrows and ncols are initialized and declared elsewhere
    int [][] matrix = new int[nrows][ncols];
    for (int i = 0; i < matrix.length; ++i) {
        for (int j = 0; j < matrix[i].length; ++j) {
            // operate on the values here.
            if (matrix[i][j] != 1) {
                matrix[i][j] -= 1;
            }
        }
    }
    

    If you’ve got an List of List it looks like this:

    List<List<Integer>> matrix = new ArrayList<List<Integer>>();
    List<Integer> columnIdsToTransform = Arrays.asList({0, 4, 6 });
    // You have to initialize the references; all are null right now.
    for (List<Integer> row : matrix) {
        for (int j = 0; j < row.size(); ++j) {            
            // operate on the values here.
            value = row.get(j);
            if (columnsIdsToTransform.contains(j) && (value != 1)) {
                row.set(value-1, j);
            }
        }
    }
    

    UPDATE:

    Based on your edit, you should add an array or List of columns you want to perform this transformation on. I’ve added an example to the second snippet.

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