1. 程式人生 > >[Coding Made Simple] Sum Query in 2D Immutable Array

[Coding Made Simple] Sum Query in 2D Immutable Array

program swe mut imp ini cor private Coding mat

Given a 2D immutable array, Write an efficient program to support any given sub-rectangle sum query in this 2D matrix.

A simple solution is to add each entry inside the sub-rectangle. The runtime is O(n * m).

The problem of this solution is that for a given 2D matrix, we always start from scratch to compute the sum of a sub-rectangle.

Dynamic programming should be used to optimize the runtime.

State: T[i][j]: the sum of sub rectangle with top left corner (0, 0) and bottom right corner (i - 1, j - 1).

Function: T[i][j] = T[i - 1][j] + T[i][j - 1] + matrix[i - 1][j - 1] - T[i - 1][j - 1].

Initialization: T[0][j] = T[i][0] = 0;

Answer: given top left corner(x1, y1) and bottom right corner(x2, y2),

    Sum of sub rectangle(x1, y1, x2, y2) is T[x2 + 1][y2 + 1] - T[x1][y2 + 1] - T[x2 + 1][y1] + T[x1][y1].

 1 public class SumQuery {
 2     private int[][] T;
 3     public SumQuery(int[][] matrix) {
 4         if(matrix == null || matrix.length == 0 || matrix[0].length == 0) {
 5             T = null
; 6 } 7 T = new int[matrix.length + 1][matrix[0].length + 1]; 8 for(int i = 0; i < T.length; i++) { 9 T[i][0] = 0; 10 } 11 for(int j = 0; j < T[0].length; j++) { 12 T[0][j] = 0; 13 } 14 for(int i = 1; i < T.length; i++) { 15 for(int j = 1; j < T[0].length; j++) { 16 T[i][j] = T[i - 1][j] + T[i][j - 1] + matrix[i - 1][j - 1] - T[i - 1][j - 1]; 17 } 18 } 19 } 20 public int querySubRectangleSum(int x1, int y1, int x2, int y2) { 21 x1++; y1++; x2++; y2++; 22 return T[x2][y2] - T[x1 - 1][y2] - T[x2][y1 - 1] + T[x1 - 1][y1 - 1]; 23 } 24 }

[Coding Made Simple] Sum Query in 2D Immutable Array