Finding the intersection of two arrays [closed]

Answer 1

public static void main(String[] args) {
        int a[] = {3, 10, 4, 2, 8};
        int[] b = {10, 4, 12, 3, 23, 1, 8};
        int[] c = new int[(int)Math.min(a.length, b.length)];
        int i=0;
         for(int f=0;f<a.length;f++){
              for(int k=0;k<b.length;k++){
                    if(a[f]==b[k]){
                    c[i] = a[f];
                    i++;
            }
          }
        }
        for (int x=0; x<i; x++){
           System.out.println(c[x]);
        }
       }
  }

Hope it helps. Or if you have time complexity issue then try Java Set.

Answer 2

This should be an easy way to do.

int a[] = {3, 10, 4, 2, 8};
int[] b = {10, 4, 12, 3, 23, 1, 8};
List<Integer> aList =  Arrays.asList(a);
List<Integer> bList =  Arrays.asList(b);
aList.retainAll(bList);
System.out.println(" a intersection b "+aList);
int[] c = aList.toArray(new int[0]);

Answer 3

if permitted use ArrayList for c, its growable array

ArrayList c = new ArrayList();
.
.
.
.
.
c.add(a[f]);

also if permitted to sort arrays, i recommend you to sort smaller array and then iterate over larger array and binary search in smaller array.

Answer 4

You can take a help of temporary variable (but this is basically reinventing the wheel, if you are not required to do so) -

int[] c = new int[0];
//...
    if(a[f] == b[k]) { 
        int[] temp = c;
        c = new int[c.length + 1];
        for(int i=0; i<temp.length; i++) {
            c[i] = temp[i];
        }
        c[c.length - 1] = a[f];
    }
//...

Answer 5

First you need to allocate space for your array:

int[] c = new int[SOME_SIZE];

The hard part is figuring out how much SOME_SIZE should be. Since you are calculating an intersection, the most it can be is the size of the smallest of a and b.

Finally, to assign an element in the array, you just do

c[idx] = a[f]

Now you need to keep track of where idx goes. I suggest starting with idx = 0 and incrementing it each time you find a new element to add to c.