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Provided Code

public class ArrayTester
{
    /** Returns 1D array of column c from arr2D, in same row order. */
        public static int[] getColumn(int[][] arr2D, int c) {
        /* part (a) */
    }
    /** Returns true if every value in arr1 appears in arr2. */
    public static boolean hasAllValues(int[] arr1, int[] arr2) { /* not shown */ }
    /** Returns true if arr contains any duplicate values. */
    public static boolean containsDuplicates(int[] arr) { /* not shown */ }
    /** Returns true if square is a Latin square:
     *  - first row has no duplicates
     *  - each row contains all values from first row
     *  - each column contains all values from first row */
        public static boolean isLatin(int[][] square) {
        /* part (b) */
    }
}

Part A — 4 Points

Write getColumn: return a new 1D array containing the elements of column c in the same row order as they appear in arr2D.

Write Your Solution

Check Answer Clear

public static int[] getColumn(int[][] arr2D, int c)
{
    int[] result = new int[arr2D.length];
    for (int r = 0; r < arr2D.length; r++)
    {
        result[r] = arr2D[r][c];
    }
    return result;
}

Part B — 5 Points

Write isLatin: return true if and only if the 2D square array is a Latin square (no first-row duplicates, all rows match first-row values, all columns match first-row values). Must use getColumn, hasAllValues, and containsDuplicates.

Write Your Solution

Check Answer Clear

public static boolean isLatin(int[][] square)
{
    if (containsDuplicates(square[0]))
    {
        return false;
    }
    for (int r = 1; r < square.length; r++)
    {
        if (!hasAllValues(square[0], square[r]))
        {
            return false;
        }
    }
    for (int c = 0; c < square[0].length; c++)
    {
        if (!hasAllValues(square[0], getColumn(square, c)))
        {
            return false;
        }
    }
    return true;
}

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