Checking if the 2nd list have any unique elements compared to 1st list-CodePudding

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.

CodePudding user response：

Maybe you can do some like that:

Change "==" for "equals()"

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;

CodePudding user response：

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);
    }