Counting number of times an int occurs in an array (Java)

Question

I'm trying to count the number of occurences of ints, one to six inclusive, in an array of size 6. I want to return an array with the number of times an int appears in each index, but with one at index zero.
Example:

Input: [3,2,1,4,5,1,3]
Expected output: [2,1,2,1,1,0].

Problem:
It outputs [1,1,3,0,1,0] with the code excerpt below. How can I fix this? I can't find where I'm going wrong.

public static int arrayCount(int[] array, int item) {
    int amt = 0;
    for (int i = 0; i < array.length; i++) {
        if (array[i] == item) {
            amt++;
        }
    }
    return amt;
}

public int[] countNumOfEachScore(){
    int[] scores = new int[6];
    int[] counts = new int[6];
    for (int i = 0; i < 6; i++){
        scores[i] = dice[i].getValue();
    }
    for (int j = 0; j < 6; j++){
        counts[j] = arrayCount(scores, j+1);
    }
    return counts;
}

dice[] is just an array of Die objects, which have a method getValue() which returns an int between 1 and 6, inclusive. counts[] is the int array with the wrong contents.

Why don't you use Collections.frequency() to get the count for each array elements — mhasan, Oct 04 '16 at 18:23

xenteros · Accepted Answer · 2016-10-04T18:23:42.117

4

It'll be faster to write another code instead of debugging yours.

public static int[] count(int[] array) {
    int[] result = new int[6];
    for (int i = 0; i < array.length; i++) {
        try{
            result[array[i]-1]++;
        } catch (IndexOutOfBoundsException e) {
            throw new IllegalArgumentException("The numbers must be between 1 and 6. Was " + String.valueOf(array[i]));
        }
    }
    return result;
}

The above will result in an array of 6 ints. ith number in the result array will store the number of occurences of i+1.

PoC for the OP

edited Oct 04 '16 at 18:23

answered Oct 04 '16 at 18:12

xenteros

15,586
12
56
91

It still seems to be off. With input values of [1,2,4,4,3,5] it gave me [0,1,3,3,4,4], which isn't right. – arnbobo Oct 04 '16 at 18:19
It returns [1,1,1,2,1,0] on my machine. On my `Java Virtual Machine`. – xenteros Oct 04 '16 at 18:21
@arnbobo can you show me more of your code? I'll try to help. If it started working, feel free to mark it as a solution. – xenteros Oct 04 '16 at 19:05
@arnbobo you're welcome. I'll be happy to solve your problems in the future. Feel free to call me in comment. – xenteros Oct 04 '16 at 19:12

score -2 · Answer 2 · answered Oct 04 '16 at 20:23

public static void main(String [] args){
        int []ar=new int[]{3,2,1,4,5,1,3};
        System.out.println(Arrays.toString(counter(ar)));

    }


public static int[] counter(int []ar){
        int []result=new int [6];
        for(int i=0;i<ar.length;i++){
            int c=0;
            for(int j=0;j<ar.length;j++){
                if(j<i && ar[i]==ar[j])
                    break;
                if(ar[i]==ar[j])
                    c++;
                if(j==ar.length-1){
                    result[i]=c;
                }
            }
        }
        return result;
    }

Check your required input outputs, please. – Hasib Oct 04 '16 at 20:25 — Hasib, Oct 04 '16 at 20:25

Counting number of times an int occurs in an array (Java)

2 Answers2