Java counting duplicate numbers In a 2d Array

Question

Im trying to create a java method that will look at a 2d array and see how many times a number has been entered more than once and then output the count.

So given this array

1   2   3   3   
5   6   7   8   
8   45  9   45  
10  17  18  13

the method will return a count of 3.

So far this is what i have

int dupe=0;
    int chk1=0, chk2=0;

    for (int row =0; row < dataTable.length; row ++)
    {//for 1

        for ( int col = 0; col < dataTable[row].length; col++)
        {//for 2





            if (dataTable[row][col] ==  dataTable [chk1][chk2])
            {//if

             dupe++;
            }//end if 

        chk1++;
        chk2++;



    }//end for 2 

}//end for 1
    dupe=dupe-1;
return dupe;

however it will not run unless i declare chk1 and chk2 inside the second 4 which just re declares them every time the check is run.

Are you counting the global total of duplicates, or the individual duplicates? It seems that you're getting the global duplicates. — Makoto
– Makoto, Commented Apr 18, 2012 at 0:45

mellamokb · Accepted Answer · 2012-04-18 00:52:16Z

4

How it might be implemented:

Create a map to store seen numbers.
Loop through each row in 2d array.
Loop through each column in 2d array
If number has not been seen before (not in map), insert into map with value 1
If number has been seen before (in map), increment value at map entry
Loop through values in map, and count number > 1; this value is your answer

answered Apr 18, 2012 at 0:52

mellamokb

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Comments

Makoto · Accepted Answer · 2012-04-18 00:52:23Z

1

Strategy: Iterate over the entire array, and compare each element to all other elements. Gather the results in a variable. Since I think that this is homework, I can't give you the code, but the strategy should be enough to get you started.

answered Apr 18, 2012 at 0:52

Makoto

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Comments

ppepper · Accepted Answer · 2012-04-18 00:56:09Z

0

Well you could loop through the array by looking through every row...

for(int i = 0; i < rows; i++) {
    for(int j = 0; j < columns; j++) {
        // check
    }
}

And then check if a number has been repeated by comparing it to all the other numbers, which you can store somewhere else.

answered Apr 18, 2012 at 0:56

ppepper

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2 Comments

Vitaly Olegovitch Over a year ago

That somewhere else could be a HashSet or a TreeSet of the numbers that you have already found.

arkon Over a year ago

It looks as if OP already understands how to iterate through a 2d array. This 'solution' does nothing to answer OP's question.

Joop Eggen · Accepted Answer · 2012-04-18 01:17:04Z

Two interpretations:

/**
 * The number of entries that are double of others.
 * When 4 threes occure, 3 threes are counted.
 */
public static int moreThanOnce(int[][] matrix) {
    Set<Integer> uniqueNumbers = new HashSet<Integer>();
    int n = 0;
    for (int row = 0; row < matrix.length; ++i) {
        n += matrix[i].length;
        Collections.addAll(uniqueNumbers, matrix[i]);
    }
    return n - uniqueNumbers.size();
}

/**
 * The number of different numbers appearing more than once.
 * When four appears 3 times and eight 4 times, then the result is 2.
 */
public static int moreThanOnce2(int[][] matrix) {
    Set<Integer> usedNumbers = new HashSet<Integer>();
    Set<Integer> doubleNumbers = new HashSet<Integer>();
    for (int[] row : matrix) {
        for (int x : row) {
            if (!usedNumbers.add(x)) { // Not added, already used
                doubleNumbers.add(x);
            }
        }
    }
    return doubleNumbers.size();
}

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