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).
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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
Note:
- The matrix is only modifiable by the update function.
- You may assume the number of calls to update and sumRegion function is distributed evenly.
- You may assume that row1 ≤ row2 and col1 ≤ col2.
Solution:
We use int[][] colSum to store the prefix sum of each column.
colSum[i][j] denotes the sum of first i elements in column j.
To update an element, we find the corresponding column and update all the prefix sums that rely on this element.
Code:
public class NumMatrix { public int[][] colSum; public int[][] matrix; public NumMatrix(int[][] matrix) { this.matrix = 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; colSum = new int[m + 1][n]; for (int i = 1; i <= m; i++) { for (int j = 0; j < n; j++) { colSum[i][j] = colSum[i - 1][j] + matrix[i - 1][j]; } } } public void update(int row, int col, int val) { for (int i = row + 1; i < colSum.length; i++) { colSum[i][col] = colSum[i][col] - matrix[row][col] + val; } matrix[row][col] = val; } public int sumRegion(int row1, int col1, int row2, int col2) { int sum = 0; for (int j = col1; j <= col2; j++) { sum += colSum[row2 + 1][j] - colSum[row1][j]; } return sum; } } /** * Your NumMatrix object will be instantiated and called as such: * NumMatrix obj = new NumMatrix(matrix); * obj.update(row,col,val); * int param_2 = obj.sumRegion(row1,col1,row2,col2); */