304. Range Sum Query 2D - Immutable

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).

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
sumRegion(1, 1, 2, 2) -> 11
sumRegion(1, 2, 2, 4) -> 12

Note:

1. You may assume that the matrix does not change.
2. There are many calls to sumRegion function.
3. You may assume that row1 ≤ row2 and col1 ≤ col2.

Solution:

The idea is to pre-computer a prefix-sum matrix int[][] sum  = new int[m + 1][n + 1] such that :

sum[i][j] means the sum of a matrix from the origin (0, 0) to coordinate (i - 1, j - 1).

The reason to create m + 1 by n + 1 array is because we do not need to handle the corner case that i == 0 or j == 0.

Thus, to construct this array, we know:

sum[i][j] = sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1] + matrix[i - 1][j - 1].

This takes O(mn) time with O(mn) extra space.

Now, to look up a range sum from (x1, y1) to (x2, y2) is easy.

It only takes constant time.

Therefore, the amortized time complexity is O(1).



Code:

public class NumMatrix {

    public int[][] sum;
    public NumMatrix(int[][] matrix) {
        if (matrix == null || matrix.length == 0) {
            return;
        }
        if (matrix[0] == null || matrix[0].length == 0) {
            return;
        }
        int m = matrix.length;
        int n = matrix[0].length;
        sum = new int[m + 1][n + 1];
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                sum[i][j] = sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1] + matrix[i - 1][j - 1];
            }
        }
    }
    
    public int sumRegion(int row1, int col1, int row2, int col2) {
        int iMax = Math.max(row1, row2);
        int iMin = Math.min(row1, row2);
        int jMax = Math.max(col1, col2);
        int jMin = Math.min(col1, col2);
        return sum[iMax + 1][jMax + 1] - sum[iMin][jMax + 1] - sum[iMax + 1][jMin] + sum[iMin][jMin];
    }
}

/**
 * Your NumMatrix object will be instantiated and called as such:
 * NumMatrix obj = new NumMatrix(matrix);
 * int param_1 = obj.sumRegion(row1,col1,row2,col2);
 */