Solution in C#
\n# Approach\nThe code finds all distinct numbers that appear in one list but not the other. It achieves this by iterating through each list and checking if the
\n# Approach\nThe code finds all distinct numbers that appear in one list but not the other. It achieves this by iterating through each list and checking if the number exists in the other list and hasn\'t already been added to a collection of distinct numbers for the current list. Finally, it combines these distinct numbers into a single result structure.\n\n# Complexity\n- Time complexity:\n***O(n * m)***\n**Beats 65.54% of users with C#**\n\n- Space complexity:\n***O(n + m)***\n**Beats 95.16% of users with C#**\n\n# Code\n```\npublic class Solution {\n    public IList<IList<int>> FindDifference(int[] nums1, int[] nums2) {\n        IList<int> distinctNums1 = new List<int>();\n            IList<int> distinctNums2 = new List<int>();\n            foreach (int i in nums1)\n            {\n                if(Array.IndexOf(nums2, i) == -1 && !distinctNums1.Contains(i))\n                {\n                    distinctNums1.Add(i);\n                }\n            }\n            foreach (int i in nums2)\n            {\n                if (Array.IndexOf(nums1, i) == -1 && !distinctNums2.Contains(i))\n                {\n                    distinctNums2.Add(i);\n                }\n            }\n            IList<IList<int>> result = new List<IList<int>>();\n            result.Add(distinctNums1);\n            result.Add(distinctNums2);\n            return result;\n    }\n}\n```
['C#']