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308. Range Sum Query 2D - Mutable πŸ”’

Description

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

  1. Update the value of a cell in matrix.
  2. 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.
  • void update(int row, int col, int val) Updates the value of matrix[row][col] to be val.
  • 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", "update", "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], [3, 2, 2], [2, 1, 4, 3]]
Output
[null, 8, null, 10]

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 left red rectangle)
numMatrix.update(3, 2, 2); // matrix changes from left image to right image
numMatrix.sumRegion(2, 1, 4, 3); // return 10 (i.e. sum of the right red rectangle)

 

Constraints:

  • m == matrix.length
  • n == matrix[i].length
  • 1 <= m, n <= 200
  • -1000 <= matrix[i][j] <= 1000
  • 0 <= row < m
  • 0 <= col < n
  • -1000 <= val <= 1000
  • 0 <= row1 <= row2 < m
  • 0 <= col1 <= col2 < n
  • At most 5000 calls will be made to sumRegion and update.

Solutions

Solution 1

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class BinaryIndexedTree:
    def __init__(self, n):
        self.n = n
        self.c = [0] * (n + 1)

    @staticmethod
    def lowbit(x):
        return x & -x

    def update(self, x, delta):
        while x <= self.n:
            self.c[x] += delta
            x += BinaryIndexedTree.lowbit(x)

    def query(self, x):
        s = 0
        while x > 0:
            s += self.c[x]
            x -= BinaryIndexedTree.lowbit(x)
        return s


class NumMatrix:
    def __init__(self, matrix: List[List[int]]):
        self.trees = []
        n = len(matrix[0])
        for row in matrix:
            tree = BinaryIndexedTree(n)
            for j, v in enumerate(row):
                tree.update(j + 1, v)
            self.trees.append(tree)

    def update(self, row: int, col: int, val: int) -> None:
        tree = self.trees[row]
        prev = tree.query(col + 1) - tree.query(col)
        tree.update(col + 1, val - prev)

    def sumRegion(self, row1: int, col1: int, row2: int, col2: int) -> int:
        return sum(
            tree.query(col2 + 1) - tree.query(col1)
            for tree in self.trees[row1 : row2 + 1]
        )


# Your NumMatrix object will be instantiated and called as such:
# obj = NumMatrix(matrix)
# obj.update(row,col,val)
# param_2 = obj.sumRegion(row1,col1,row2,col2)

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class BinaryIndexedTree {
    private int n;
    private int[] c;

    public BinaryIndexedTree(int n) {
        this.n = n;
        c = new int[n + 1];
    }

    public void update(int x, int delta) {
        while (x <= n) {
            c[x] += delta;
            x += lowbit(x);
        }
    }

    public int query(int x) {
        int s = 0;
        while (x > 0) {
            s += c[x];
            x -= lowbit(x);
        }
        return s;
    }

    public static int lowbit(int x) {
        return x & -x;
    }
}

class NumMatrix {
    private BinaryIndexedTree[] trees;

    public NumMatrix(int[][] matrix) {
        int m = matrix.length;
        int n = matrix[0].length;
        trees = new BinaryIndexedTree[m];
        for (int i = 0; i <