所有号码都被重复少于 N 次

Question

i'm tring to return an array that have all number have been rebeated less than N times.

expected output : 1,4

this is my cod :

-- Main --

 int[] data = {1, 2, 2, 3, 3, 3, 4, 5, 5};
    int n = 1;
        Solution.solution(data, n);

-- class Solution --

public static int[] solution(int[] data, int n) {
        
        int l =  data.length, count = 0;
        int[] Narr = new int[l];
        
        for(int i =0 ; i < l; i++){
            count = 0;
            
            for(int j=1 ; j < l ; j++){
                if(data[i] == data[j]){
                    count++;
                }
                
                if(j == l-1){
                    
                    if(count < n){
                         Narr[i] = data[i];
                         System.out.println(Narr[i]);
                     } 
                }
            }
            
        }
        
        return Narr;
    }

Answer 1

I would break this down to multiple steps.

First lets create a countMap that holds track of all the occurrences:

int[] data = {1, 2, 2, 3, 3, 3, 4, 5, 5};
int n = 1;

Map<Integer, Long> countMap = Arrays.stream(data).boxed()
        .collect(Collectors.groupingBy(s -> s, Collectors.counting()));

System.out.println(countMap); // {1=1, 2=2, 3=3, 4=1, 5=2}

Now all we need to do is to check for the condition count <= n :

List<Integer> resultList = new ArrayList<>();
countMap.forEach((integer, count) -> {
    if (count <= n) resultList.add(integer);
});

System.out.println(resultList); // [1, 4]

And last but not least converting the list back to array :

// {1, 4}
int[] resultArray = resultList.stream()
        .mapToInt(Integer::intValue)
        .toArray();