Array contains element in Java

Question

I've got a task of writing a function that returns integers which are in both two arrays. For example: nums1 [1,2,3] and nums2 [2,3,5] and answer should be [2,3]. I came with this solution:

public static void main(String[] args) {
    System.out.println(arraysIntersection(new int[] {1,2,3}, new int[] {2,3,5}));
}

public static List<Integer> arraysIntersection(int[] nums1, int[] nums2) {
    List<Integer> answer = new ArrayList<>();
    for (int i = 0; i < nums1.length; i++) {
        if (Arrays.asList(nums2).contains(nums1[i])) {
            answer.add(nums1[i]);
        }
    }
    return answer;
}

however it seems this condition doesn't work as intended:

        if (Arrays.asList(nums2).contains(nums1[i]))

It says it doesn't contain the value altough it clearly contains 2 and 3. Any ideas? I know I could just iterate each i over the second array but I thought this solution would be faster. Does anyone knows why it's not working?

Answer 1

1. You can do it in O(NlogN) time complexity and O(n) memory. Just sort arrays and use two pointers technique.

    List<Integer> answer = new ArrayList<>();
    int j = 0;
    Arrays.sort(nums1);
    Arrays.sort(nums2);
    for(int i = 0; i < nums1.length; i++) {
        if(i > 0 && nums1[i] == nums1[i - 1]) //we have already processed this number
            continue;
        while(j < nums2.length && nums2[j] < nums1[i])
            j++;
        if(j < nums2.length && nums1[i] == nums2[j])
            answer.add(nums1[i]);
    }
    return answer;

2. You can do it in O(N) time complexity and O(n) memory (but constant is higher). Just add all elements of nums1 in first HashSet and all elements of nums2 if another HashSet. Then you can for each element in first set check if another set contains this element using O(1)* time.

    List<Integer> answer = new ArrayList<>();
    Set<Integer> set1 = new HashSet<>(), set2 = new HashSet<>();
    set1.addAll(nums1);
    set2.addAll(nums2);
    
    for(var el : set1) {
        if(set2.contains(el)) {
            answer.add(el);
        }
    }
    
    return answer;

*O(1) is middle time of operations with hashset

Answer 2

Maybe instead Arrays, you wished to write Array, not Arrays. Array() could be called as a constructor for an array. And after an object is initialized, can be called asList(). If Arrays is a static object already initialized, or declared at global scope, it may be OK, I don't know for sure.

You can try another way:

public static List<Integer> arraysIntersection(int[] nums1, int[] nums2){ 
   List<Integer> answer = new ArrayList<>();
   for (int i = 0; i < nums1.length; i++) {
      int u = nums1[i]; 
      for (int j = 0; j < nums2.length; j++) { 
         if (u == nums2[j]) {  
            answer.add(u); 
            break; 
         }
      } 
   }
   return answer;
} 

A break could be necessary, if the values must be added only once. ( a number can be found more times into an array ) The break was meant just for ending the inner loop. The outer loop should continue up to the end of the search.

But the same number can be find more times in the first array. Before returning the answer, the result should be checked for duplicate values. And this is another problem.

But I think remove() it's a method for the answer List. A duplicate item can be removed at a certain index. It would be more like a sorting algorithm. (I mean you have to skip the current index).

Best regards, Adrian Brinas. Have a nice evening.