easy java solution only using loop beats 88% in runtime 🤞🤞
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n    public   int counting(int [] nums1,int[] nums2){\n        int length1=nums1.length;\n        int length2=nums2.length;\n        int count=0;\n        for(int i=0;i<length1;i++){\n            for(int j=0;j<length2;j++){\n                if(nums1[i]==nums2[j]){\n                    count++;\n                    break;\n                }\n            }\n        }\n        return count;\n    }\n    public int[] findIntersectionValues(int[] nums1, int[] nums2) {\n        //creating an array size of two which we will return\n        int arr[]=new int[2];\n        //find how many numbers of nums1 array present in nums2\n        arr[0]=counting(nums1,nums2);\n        //find how many numbers of nums2 array present in nums1\n        arr[1]=counting(nums2,nums1);\n        return arr;\n\n    }\n}\n```
['Java']