Range Sum Query 2D - Mutable

Description

Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2).

Range Sum Query 2D The above rectangle (with the red border) is defined by (row1, col1) = (2, 1) and (row2, col2) = (4, 3), which contains sum = 8.

Example: Given matrix = [ [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] ]

sumRegion(2, 1, 4, 3) -> 8 update(3, 2, 2) sumRegion(2, 1, 4, 3) -> 10

Hint

Train of Thought

So this further helps us understand 2D-BIT. (6,6) represents one rectangle's sum. But (6,6) only tells us one corner point coordinates. How large is the rectangle? The answer is: we can know the rectangle's size from two parents. (6,4) and (4,6) are actually the other two corner points in rectangle. Now we know the rectangle's size.

Code

public class NumMatrix {

int[][] tree;
int[][] nums;
int m;
int n;

public NumMatrix(int[][] matrix) {
    if (matrix.length == 0 || matrix[0].length == 0) return;
    m = matrix.length;
    n = matrix[0].length;
    tree = new int[m+1][n+1];
    nums = new int[m][n];
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            update(i, j, matrix[i][j]);
        }
    }
}

public void update(int row, int col, int val) {
    if (m == 0 || n == 0) return;
    int delta = val - nums[row][col];
    nums[row][col] = val;
    for (int i = row + 1; i <= m; i += i & (-i)) {
        for (int j = col + 1; j <= n; j += j & (-j)) {
            tree[i][j] += delta;
        }
    }
}

public int sumRegion(int row1, int col1, int row2, int col2) {
    if (m == 0 || n == 0) return 0;
    return sum(row2+1, col2+1) + sum(row1, col1) - sum(row1, col2+1) - sum(row2+1, col1);
}

public int sum(int row, int col) {
    int sum = 0;
    for (int i = row; i > 0; i -= i & (-i)) {
        for (int j = col; j > 0; j -= j & (-j)) {
            sum += tree[i][j];
        }
    }
    return sum;
}

}

Complexity