检查第二个列表与第一个列表相比是否有任何独特的元素

Question

Trying my own brute force solution. Now, there is a problem with my code I don't know what the problem is. I calculated the problem on my paper perfectly and it works great. It should work as expected. But gives unexpected results.

Problem: Missing Number

Code:

public static List<Integer> getMissingNumber (List<Integer> arr, List<Integer> brr){

Integer[] value = new Integer[brr.size()]; 
value = brr.toArray(value); 

   for(int i=0; i<arr.size();i++){
       for(int j=0; j<brr.size(); j++){
           if(arr.get(i)==brr.get(j)){
               value[j]=0;
               break;
           }
       }
   }
  Arrays.sort(value);
  List<Integer> exect_value = new ArrayList<Integer>();
   for(int i=0;i<value.length;i++) {
       if(value[i]!=-1) {
           exect_value.add(value[i]);
       }
   }

return Arrays.asList(value);
}

Problem I'm Facing here:

for(int i=0; i<arr.size();i++){
               for(int j=0; j<brr.size(); j++){
                   if(arr.get(i)==brr.get(j)){
                       value[j]=0;
                       break;
                   }
               }
           }

When I tested for: Taking 2 different input(Function will pass 2 list):

arr[6] : [7,2,5,3,5,3] 
brr[8] : [7,2,5,4,6,3,5,3]

output: [4,6] you have to print elements that is not founded in list_1

It works perfectly for this test case

But

When I try for: Input:

arr[10] =[11 4 11 7 13 4 12 11 10 14]
brr[15] = [11 4 11 7 3 7 10 13 4 8 12 11 10 14 12]

Output gives: [0, 0, 11, 0, 3, 7, 0, 0, 4, 8, 0, 11, 10, 0, 12] This but I'm expecting -> [0, 0, 0, 0, 3, 7, 0, 0, 0, 8, 0, 0, 10, 0, 12] extra [11,4,11] comes on this array.

Why I'm so confused, please help me.

Answer 1

When I try for: Input:

 arr[10] =[11 4 11 7 13 4 12 11 10 14] brr[15] = [11 4 11 7 3 7 10 13 4 8 12 11 10 14 12]

Output gives: [0, 0, 11, 0, 3, 7, 0, 0, 4, 8, 0, 11, 10, 0, 12] This but I'm expecting -> [0, 0, 0, 0, 3, 7, 0, 0, 0, 8, 0, 0, 10, 0, 12] extra [11,4,11] comes on this array.

That is because here:

 for(int i=0; i<arr.size();i++){ for(int j=0; j<brr.size(); j++){ if(arr.get(i)==brr.get(j)){ value[j]=0; break; } } }

you are comparing arr.get(i) with brr.get(j) instead of value[j] .

---

PS. I seems that the proper solution is to sort and "anti-merge" two lists:

    public static List<Integer> missingNumbers(List<Integer> arr, List<Integer> brr) {
        // Write your code here
        Collections.sort(arr);
        Collections.sort(brr);
        Set<Integer> result = new LinkedHashSet<>();
        int l = 0, r = 0;
        while (l < arr.size() && r < brr.size()) {
            if (arr.get(l).equals(brr.get(r))) {
                l++;
                r++;
            } else {
                result.add(brr.get(r++));
            }
        }
        for (; r < brr.size(); r++) {
            result.add(brr.get(r));
        }
        return new ArrayList<>(result);
    }

Answer 2

Maybe you can do some like that:

1. Change "==" for "equals()"

2. Set in brr array zero when some value has already been stored (because your problem was that you were comparing values ​​of arr that had already been validated). For example, you compared arr in position 2 against brr position 0 (Both values 11), then put 0 back where one already existed.

    Integer[] value = new Integer[brr.size()]; value = brr.toArray(value); for(int i=0; i<arr.size();i++){ for(int j=0; j<brr.size(); j++){ if(arr.get(i).equals(brr.get(j))){ brr.set(j, 0); value[j]=0; break; } } } Arrays.sort(value); List<Integer> exect_value = new ArrayList<Integer>(); for(int i=0;i<value.length;i++) { if(value[i]!=0) { exect_value.add(value[i]); } } return exect_value;