Java - count duplicates in array without using outside classes or methods?

Question

I need to count the number of dupes in an array, but I'm constrained to only using arrays, no HashSets or ArrayLists.

Here are example inputs with their expected outputs:

numDuplicates(new double[] { }) --> 0
numDuplicates(new double[] { 11.0 }) --> 0
numDuplicates(new double[] { 11.0, 11.0, 11.0, 11.0 }) --> 3
numDuplicates(new double[] { 11.0, 11.0, 11.0, 11.0, 22.0, 33.0, 44.0, 44.0, 44.0, 44.0, 44.0, 55.0, 55.0, 66.0, 77.0, 88.0, 88.0 }) --> 9
numDuplicates(new double[] { 11.0, 22.0, 33.0, 44.0, 44.0, 44.0, 44.0, 44.0, 55.0, 55.0, 66.0, 77.0, 88.0 }) --> 5

This is the code I have, but it counts duplicates after each number, returning an inflated count, i.e. {11.0, 11.0, 11.0, 11.0} returns 6 instead of 3:

public static int numDuplicates (double[] list) {

    int dupCount = 0;
    for (int i = 0; i < list.length; i++) {
        for (int j = i + 1; j < list.length; j++) {
            if (list[i] == list[j]) {
                dupCount++;
            }
        }
    }

    return dupCount; //TODO1: fix this
}

Note: I'm new on Stack, I tried searching thoroughly but couldn't find an array duplicate question that had similar input/output to mine, but I'm sorry if someone has already asked this.

Based on the example you gave, it looks like you are supposed to find the longest duplicate not just count all the duplicates — smac89
– smac89, Commented Apr 21, 2018 at 23:05
Perhaps this may answer your question: stackoverflow.com/a/31738825/9488265 — J Duh
– J Duh, Commented Apr 21, 2018 at 23:07
No actually it looks like you're right ignore previous comment — smac89
– smac89, Commented Apr 21, 2018 at 23:07
All you need to do to fix your code is insert a break; after you increment dupCount. That way you only count each duplicate once. — castletheperson
– castletheperson, Commented Apr 21, 2018 at 23:08

smac89 · Accepted Answer · 2018-04-22 23:39:53Z

1

Simple fix:

public static int numDuplicates (double[] list) {

    int dupCount = 0;
    for (int i = 0; i < list.length; i++) {
        for (int j = i + 1; j < list.length; j++, i++) { // HERE it is
            if (list[i] == list[j]) {
                dupCount++;
            }
        }
    }

    return dupCount;
}

What I did was to increment i in addition to j. What this did was to start i at the place where j stopped. The reason I thought to do that was because you said it was returning an inflated count, so I figured it must be because you are doing too many counts and that was exactly what was happening.

i was always following in increments of 1 rather than increments of the size of the last duplicate counts found, so it was always repeating counts.

As to how the j++, i++ works - it is just a series of expressions. Java allows you to separate expressions that evaluate to the same type, using a comma.

As per the comments, you can remove the outer loop:

public static int numDuplicates (double[] list) {

    int dupCount = 0;
    for (int i = 1; i < list.length; i++) {
        if (list[i] == list[i - 1]) {
            dupCount++;
        }
    }
    return dupCount;
}

edited Apr 22, 2018 at 23:39

answered Apr 21, 2018 at 23:11

smac89

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5 Comments

Raymo111 Over a year ago

Wow, I never knew you could do this! But how does j++, i++ work?

DejaEntendouche Over a year ago

That worked! I'm new to Java, would you mind explaining what that changed functionally?

smac89 Over a year ago

Explanation coming guys, hang tight

castletheperson Over a year ago

If you increment i and j at the same time, they will always be adjacent indexes. You might as well remove the outer loop and calculate i as j - 1.

smac89 Over a year ago

@4castle, noted

Raymo111 · Accepted Answer · 2018-04-21 23:14:08Z

0

You could use another array like so:

public static int numDuplicates(double[] list) {

    double[] duparray = new double[list.length];
    int dupCount = 0;
    for (int i = 0; i < list.length; i++) {
        for (int j = i + 1; j < list.length; j++) {
            if (list[i] == list[j] && !(list[i] == duparray[i])) {
                duparray[i] = list[i];
                dupCount++;
            }
        }
    }

    return dupCount;
}

answered Apr 21, 2018 at 23:14

Raymo111

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2 Comments

castletheperson Over a year ago

This won't work if the input array contains any zeros. A better solution would be to use a boolean[] array instead that indicates if the value at that index has already been counted as a duplicate.

castletheperson Over a year ago

If the input array was new double[]{ 0, 0 } for example, it would already count the values as equal to the default values in dupArray, so it would return 0 duplicates instead of 1.

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