Java Solution for Find the Difference of Two Arrays Problem
Intuition\n Describe your first thoughts on how to solve this problem. \nThe intuition behind the solution is to use two sets to store the distinct elements of 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind the solution is to use two sets to store the distinct elements of the two input arrays nums1 and nums2, respectively. Then, the difference between the two sets can be computed to find the elements that are present in one set but not the other. Finally, these elements are added to separate lists, which are returned as the result.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThe given solution is for the problem of querying the sum of elements in a rectangular region of a matrix. The intuition behind the solution is to precompute the prefix sum matrix, where the value at each cell represents the sum of all elements in the submatrix from (0,0) to that cell. Then for each query, the sum of the required region can be calculated using the precomputed values in constant time.\n\nThe approach used is to compute the prefix sum matrix in the constructor of the `NumMatrix` class. For this, two nested loops are used to iterate over all cells of the matrix. At each cell, the sum of the current row and all previous rows up to the current row, and the sum of the current column and all previous columns up to the current column, is calculated using the precomputed value from the previous row or column. Finally, the sum of the current cell is calculated by subtracting the value of the top-left submatrix, which was added twice.\n\nFor each query of a rectangular region, the sum of the submatrix is calculated using the precomputed values of the prefix sum matrix. Depending on the location of the top-left and bottom-right corners of the submatrix, different cases are handled to calculate the final result using the precomputed values.\n\n\n\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nThe time complexity of the constructor is O(mn), where m and n are the dimensions of the matrix, as it requires iterating over all cells of the matrix. The time complexity of each query is O(1), as it only requires calculating the sum of four cells using the precomputed values.\n\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nThe space complexity of the solution is O(mn), as the prefix sum matrix requires the same amount of space as the original matrix.\n\n# Code\n```\nclass Solution {\n    public List<List<Integer>> findDifference(int[] nums1, int[] nums2) {\n        Set<Integer> set1 = new HashSet<>();\n        Set<Integer> set2 = new HashSet<>();\n        for (int num : nums1) \n        {\n            set1.add(num);\n        }\n        for (int num : nums2) \n        {\n            set2.add(num);\n        }\n        List<Integer> diff1 = new ArrayList<>();\n        List<Integer> diff2 = new ArrayList<>();\n        for (int num : set1) \n        {\n            if (!set2.contains(num)) \n            {\n                diff1.add(num);\n            }\n        }\n        for (int num : set2) \n        {\n            if (!set1.contains(num)) \n            {\n                diff2.add(num);\n            }\n        }\n        List<List<Integer>> answer = new ArrayList<>();\n        answer.add(diff1);\n        answer.add(diff2);\n        return answer;\n    }\n}\n```
['Java']