Range Sum Query 2D Immutable Problem LeetCode

Problem:

Given a 2D matrix matrix, handle multiple queries of the following type:

• Calculate the sum of the elements of matrix inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2).

Implement the NumMatrix class:

• NumMatrix(int[][] matrix) Initializes the object with the integer matrix matrix.
• int sumRegion(int row1, int col1, int row2, int col2) Returns the sum of the elements of matrix inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2).

 Example 1:

Input
["NumMatrix", "sumRegion", "sumRegion", "sumRegion"]
[[[[3, 0, 1, 4, 2], [5, 6, 3, 2, 1], [1, 2, 0, 1, 5], [4, 1, 0, 1, 7], [1, 0, 3, 0, 5]]], [2, 1, 4, 3], [1, 1, 2, 2], [1, 2, 2, 4]]
Output
[null, 8, 11, 12]

Explanation
NumMatrix numMatrix = new NumMatrix([[3, 0, 1, 4, 2], [5, 6, 3, 2, 1], [1, 2, 0, 1, 5], [4, 1, 0, 1, 7], [1, 0, 3, 0, 5]]);
numMatrix.sumRegion(2, 1, 4, 3); // return 8 (i.e sum of the red rectangle)
numMatrix.sumRegion(1, 1, 2, 2); // return 11 (i.e sum of the green rectangle)
numMatrix.sumRegion(1, 2, 2, 4); // return 12 (i.e sum of the blue rectangle)

 Constraints:

• m == matrix.length
• n == matrix[i].length
• 1 <= m, n <= 200
• -105 <= matrix[i][j] <= 105
• 0 <= row1 <= row2 < m
• 0 <= col1 <= col2 < n
• At most 104 calls will be made to sumRegion.

HINTS:

• precompute the value on each cell using the previous cells value
• when a sum of a region is required, use the accumulated value on the corners but deduct the sum of the overlapping regions

O--------------------
| | |
| | |
--------A-----------B
| | |
--------C-----------D
If O is the origin, we can note that the region Sum(OA) is covered twice by both Sum(OB) and Sum(OC). We could use the principle of inclusion-exclusion to calculate Sum(ABCD) as following:

Sum(ABCD) = Sum(OD) - Sum(OB) - Sum(OC) + Sum(OA)

Complexity: O(mn) time and O(mn) space

Solution: