Java Solution with Comments || HashMap
\npublic List<Integer> intersection(int[][] nums) {\n        HashMap<Integer,Integer> hm=new HashMap(); //To store the count of number\n        List<Integer> li
```\npublic List<Integer> intersection(int[][] nums) {\n        HashMap<Integer,Integer> hm=new HashMap(); //To store the count of number\n        List<Integer> li=new ArrayList(); // To store the result\n        for(int i=0;i<nums.length;i++){\n            for(int j=0;j<nums[i].length;j++){\n                int val=hm.getOrDefault(nums[i][j],0);\n                hm.put(nums[i][j],val+1);\n                if(val+1==nums.length) li.add(nums[i][j]); //whenever we get the count of number equal to row length that mean it is present in every rows.\n            }\n        }\n        Collections.sort(li); //Sort the number in result list.\n        return li;\n    }\n```
['Hash Table', 'Java']