Sorting through entire 2D array in ascending order

Using streams, it is possible to convert the 2D array into a stream of integers and then recollect it to 2D array:

public static int[][] sort2D(int[][] arr) {
    int rows = arr.length;
    int cols = arr[0].length;
    int[] row = {0};
    
    return Arrays.stream(arr)
                 .flatMapToInt(Arrays::stream)
                 .sorted()
                 .boxed()
                 .collect(Collectors.groupingBy(x -> row[0]++ / cols))
                 .values()
                 .stream()
                 .map(r -> r.stream().mapToInt(Integer::intValue).toArray())
                 .toArray(int[][]::new);
}

Test

int[][] arr = {
  {9, 7, 5, 4},
  {3, 12, 11, 8},
  {6, 1, 2, 10}
};
int[][] sorted = sort2D(arr);
Arrays.stream(sorted)
      .map(Arrays::toString)
      .forEach(System.out::println);

Output

[1, 2, 3, 4]
[5, 6, 7, 8]
[9, 10, 11, 12]

Update
A simpler solution is possible using method Stream::collect(Supplier supplier, BiConsumer accumulator, BiConsumer combiner):

public static int[][] sort2D(int[][] arr) {
    int rows = arr.length;
    int cols = arr[0].length;
    AtomicInteger ix = new AtomicInteger(0); // int[] ix = {0}; can be used too
    
    return Arrays
        .stream(arr)
        .flatMapToInt(Arrays::stream)
        .sorted()
        .collect(
            () -> new int[rows][cols], // supplier creating the result array
            (a, x) -> a[ix.get() / cols][ix.getAndIncrement() % cols] = x, // populate with sorted data
            (arr1, arr2) -> {} // nothing to combine
        );
}

The problem can be broken down into three parts.

1. Put all the elements of each row of size n in a single vector
2. Order that vector
3. Break it up into vectors of size n

With your example input

int[][] arr = { {3,8,5},{1,6,7},{2,4,9} };

and method:

int[][] sorted(int[][] in) {
    int[] all = new int[in.length * in.length]; // assumption: square matrix
    int i = 0;
    for (int[] row : in) {
        for (int el : row) {
            all[i] = el;
            i++;
        }
    }
    Arrays.sort(all); // sort all elements in single vector
    // put sorted elements in your "output" array
    int[][] out = new int[in.length][in.length];
    int row = 0;
    int col = 0;
    for (int el : all) {
        // note col,row and not row,col!
        out[col][row] = el;
        row++;
        if (row == in.length) {
            row = 0;
            col++;
        }
    }
    return out;
}

You have

int[][] sorted = sorted(arr);

You can try this. got my reference here java Arrays.sort 2d array

Arrays.sort(matrix, (a, b) -> a[0] - b[0]);

You can first flatten this 2d array to a one dimension, then sort a flat array with Arrays.sort method, and then assemble back a 2d array of the same dimensions with elements from the sorted flat array:

// original array
int[][] arr = {
        {3, 8, 5},
        {1, 6, 7},
        {2, 4, 9}};

// flatten a 2d array
int[] flatArr = Arrays.stream(arr).flatMapToInt(Arrays::stream).toArray();

// sorting
Arrays.sort(flatArr);

// reassemble a 2d array of the same dimensions from a flat array
AtomicInteger flatIndex = new AtomicInteger(0);
// iterate over the rows and their elements of
// the original array and sequentially fill the
// new array with elements from the flat array
int[][] sorted = Arrays.stream(arr)
        .map(row -> Arrays.stream(row)
                .map(j -> flatArr[flatIndex.getAndAdd(1)])
                .toArray())
        .toArray(int[][]::new);

// output
Arrays.stream(sorted).map(Arrays::toString).forEach(System.out::println);
//[1,2,3]
//[4,5,6]
//[7,8,9]

^{See also:
• How do you rotate an array 90 degrees without using a storage array?
• Filling a jagged 2d array first by columns}