350. Intersection of Two Arrays II

Given two arrays, write a function to compute their intersection.

Example:
Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2, 2].

Note:

• Each element in the result should appear as many times as it shows in both arrays.
• The result can be in any order.

Follow up:

• What if the given array is already sorted? How would you optimize your algorithm?
• What if nums1's size is small compared to nums2's size? Which algorithm is better?
• What if elements of nums2 are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once?

Solution:

Use a hashmap to maintain all the number and its occurrences. Iterate nums2, if map contains the number and the occurrences is greater than 0, add to the result list and decrease the occurrences in the map.

If the array is sorted, we can use two pointers and increment the smaller one until the two numbers are equal, add the number to the result list, increment both pointers.

 public int[] intersect(int[] nums1, int[] nums2) {  
     HashMap<Integer,Integer> map=new HashMap<Integer,Integer>();  
     for(int i=0;i<nums1.length;i++){  
       map.put(nums1[i],map.getOrDefault(nums1[i],0)+1);  
     }  
     List<Integer> list=new ArrayList<Integer>();  
     for(int i=0;i<nums2.length;i++){  
       if(map.containsKey(nums2[i]) && map.get(nums2[i])>0){  
         list.add(nums2[i]);  
         map.put(nums2[i],map.get(nums2[i])-1);  
       }  
     }  
     int[] res=new int[list.size()];  
     for(int i=0;i<list.size();i++){  
       res[i]=list.get(i);  
     }  
     return res;  
   }