LeetCode: Range Sum Query 2d Immutable Solution

1/**
2 * Your NumMatrix object will be instantiated and called as such:
3 * var obj = new NumMatrix(matrix)
4 * var param_1 = obj.sumRegion(row1,col1,row2,col2)
5 */
6
7// row-based prefix sums
8/**
9 * @param {number[][]} matrix
10 */
11var NumMatrix = function (matrix) {
12  ;[this.M, this.N] = [matrix.length, matrix[0].length]
13
14  this.prefixSum = Array.from({ length: this.M }, _ => Array(this.N).fill())
15
16  for (let row = 0; row < this.M; row++) {
17    for (let col = 0; col < this.N; col++) {
18      this.prefixSum[row][col] =
19        matrix[row][col] + (this.prefixSum[row][col - 1] || 0)
20    }
21  }
22}
23
24/**
25 * @param {number} row1
26 * @param {number} col1
27 * @param {number} row2
28 * @param {number} col2
29 * @return {number}
30 */
31NumMatrix.prototype.sumRegion = function (row1, col1, row2, col2) {
32  let sum = 0
33  for (let row = row1; row <= row2; row++) {
34    sum += this.prefixSum[row][col2] - (this.prefixSum[row][col1 - 1] || 0)
35  }
36  return sum
37}
38
39// sub matrix
40/**
41 * @param {number[][]} matrix
42 */
43var NumMatrix = function (matrix) {
44  ;[this.M, this.N] = [matrix.length, matrix[0].length]
45
46  this.prefixSum = Array.from({ length: this.M + 1 }, _ =>
47    Array(this.N + 1).fill(0)
48  )
49
50  for (let row = 0; row < this.M; row++) {
51    for (let col = 0; col < this.N; col++) {
52      this.prefixSum[row + 1][col + 1] =
53        this.prefixSum[row + 1][col] +
54        this.prefixSum[row][col + 1] -
55        this.prefixSum[row][col] +
56        matrix[row][col]
57    }
58  }
59}
60
61/**
62 * @param {number} row1
63 * @param {number} col1
64 * @param {number} row2
65 * @param {number} col2
66 * @return {number}
67 */
68NumMatrix.prototype.sumRegion = function (row1, col1, row2, col2) {
69  return (
70    this.prefixSum[row2 + 1][col2 + 1] -
71    this.prefixSum[row1][col2 + 1] -
72    this.prefixSum[row2 + 1][col1] +
73    this.prefixSum[row1][col1]
74  )
75}