Lets say with the array:

array = {1,2,1,2,1,3,2,1};

I want the output to be:

2 pairs of number 1, 1 pair of number 2

For that I created a Hash table. The code :

class Trail{
    static void countFreq(int arr[], int n)
    {
        Map<Integer, Integer> mp = new HashMap<>();
 
        // Insert elements into HashTable while avoiding overwriting:
        for (int i = 0; i < n; i  )
        {
            // This part is to avoid overwriting:
            if (mp.containsKey(arr[i]))
            {
                mp.put(arr[i], mp.get(arr[i])   1);
            }
            else
            {
                mp.put(arr[i], 1);
            }
        }
        // Traverse through map and print frequencies
        for (Map.Entry<Integer, Integer> entry : mp.entrySet())
        { 
            System.out.println(entry.getKey()   " "   entry.getValue());
            
        }
         
    }

    public static void main(String[] args) {

        int arr[] = {1,2,1,2,1,3,2,1};
        int n = arr.length;
        countFreq(arr, n);
    }
}

Not sure what to do next to print out the desired output. Been stuck in this simple part for a long time.

CodePudding user response：

The calculation of frequencies seems to be fine, only printing part needs to be addressed. To get the number of pairs, divide the frequency by 2 (or shift right by 1), and skip if the pair count is 0 (according to the expected output).

Printing should be moved into a separate method:

static void printFreq(Map<Integer, Integer> mp) {
    boolean addComma = false;
    for (Map.Entry<Integer, Integer> entry : mp.entrySet()) {
        int pairs = entry.getValue() / 2;
        if (pairs < 1) {
            continue; // skip 0 pairs
        }
        if (addComma) {
            System.out.print(", ");
        }
        String p = pairs > 1 ? " pairs " : " pair ";
        System.out.print(pairs   p   "of number "   entry.getKey());
        addComma = true;
    }        
    System.out.println();    
}

However, Stream API may be used for such tasks:

• use vararg int ... arr to pass the array of integer values in more convenient way (n as the array length is redundant)
• use Collectors.groupingBy and Collectors.summingInt (or Collectors.counting) to calculate raw frequency
• calculate the number of pairs
• map each key-value pair into String
• join the strings using Collectors.joining

static void countFreq(int ... arr) {
    String message = Arrays.stream(arr)
        .boxed()
        .collect(Collectors.groupingBy(
            x -> x,
            Collectors.summingInt(x -> 1)
        )) // Map<Integer, Integer>
        .entrySet()
        .stream()
        .peek(e -> e.setValue(e.getValue() / 2))
        .filter(e -> e.getValue() > 0)
        .map(e -> String.format("%d %s of number %d", 
            e.getValue(), e.getValue() > 1 ? "pairs" : "pair", e.getKey()
        ))
        .collect(Collectors.joining(", "));
    System.out.println(message);
}

Output (in both cases):

2 pairs of number 1, 1 pair of number 2

CodePudding user response：

It would be easiest do do it with a Map<Integer, Long> and do a frequency count.

• stream the array
• box it (convert to an Integer object)
• group based on the value a
• and count the occurrence of that value
• in the map, k is the value and v is the count
• logic is also added to correct for number plurality.

int[] vals  = {1,2,1,2,1,3,2,1};
Arrays.stream(vals).boxed().collect(
        Collectors.groupingBy(a -> a, Collectors.counting()))
        .forEach((k, v) -> {
            if (v > 1) {
                System.out.println(v / 2   " pair"
                          ((v > 3) ? "s" : "")   " of "   k);
            }
        });

prints

2 pairs of 1
1 pair of 2

Note that this takes advantage of integer division and drops the fraction. So any odd count n will have the value of (n-1)/2 pairs.