304. Range Sum Query 2D Immutable - Explanation

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Description

You are given a 2D matrix matrix, handle multiple queries of the following type:

• Calculate the sum of the elements of matrix inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2).

Implement the NumMatrix class:

• NumMatrix(int[][] matrix) Initializes the object with the integer matrix matrix.
• int sumRegion(int row1, int col1, int row2, int col2) Returns the sum of the elements of matrix inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2).
  You must design an algorithm where sumRegion works on O(1) time complexity.

Example 1:

Input: ["NumMatrix", "sumRegion", "sumRegion", "sumRegion"]
[[[[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]]], [2, 1, 4, 3], [1, 1, 2, 2], [1, 2, 2, 4]]

Output: [null, 8, 11, 12]

Explanation:
NumMatrix numMatrix = new NumMatrix([[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]]);
numMatrix.sumRegion(2, 1, 4, 3); // return 8 (i.e sum of the red rectangle)
numMatrix.sumRegion(1, 1, 2, 2); // return 11 (i.e sum of the green rectangle)
numMatrix.sumRegion(1, 2, 2, 4); // return 12 (i.e sum of the blue rectangle)

Constraints:

• m == matrix.length
• n == matrix[i].length
• 1 <= m, n <= 200
• -10,000 <= matrix[i][j] <= 10,000
• 0 <= row1 <= row2 < m
• 0 <= col1 <= col2 < n
• At most 10,000 calls will be made to sumRegion.

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1. Brute Force

class NumMatrix:

    def __init__(self, matrix: list[list[int]]):
        self.matrix = matrix

    def sumRegion(self, row1: int, col1: int, row2: int, col2: int) -> int:
        res = 0
        for r in range(row1, row2 + 1):
            for c in range(col1, col2 + 1):
                res += self.matrix[r][c]
        return res

Time & Space Complexity

• Time complexity: $O(m * n)$ for each query.
• Space complexity: $O(1)$

Where $m$ is the number of rows and $n$ is the number of columns in the matrix.

---

2. One Dimensional Prefix Sum

class NumMatrix:

    def __init__(self, matrix: list[list[int]]):
        self.prefixSum = [[0] * len(matrix[0]) for _ in range(len(matrix))]

        for row in range(len(matrix)):
            self.prefixSum[row][0] = matrix[row][0]
            for col in range(1, len(matrix[0])):
                self.prefixSum[row][col] = self.prefixSum[row][col - 1] + matrix[row][col]

    def sumRegion(self, row1: int, col1: int, row2: int, col2: int) -> int:
        res = 0
        for row in range(row1, row2 + 1):
            if col1 > 0:
                res += self.prefixSum[row][col2] - self.prefixSum[row][col1 - 1]
            else:
                res += self.prefixSum[row][col2]
        return res

Time & Space Complexity

• Time complexity: $O(m)$
• Space complexity: $O(m * n)$

Where $m$ is the number of rows and $n$ is the number of columns in the matrix.

---

3. Two Dimensional Prefix Sum

class NumMatrix:

    def __init__(self, matrix: list[list[int]]):
        ROWS, COLS = len(matrix), len(matrix[0])
        self.sumMat = [[0] * (COLS + 1) for _ in range(ROWS + 1)]

        for r in range(ROWS):
            prefix = 0
            for c in range(COLS):
                prefix += matrix[r][c]
                above = self.sumMat[r][c + 1]
                self.sumMat[r + 1][c + 1] = prefix + above

    def sumRegion(self, row1: int, col1: int, row2: int, col2: int) -> int:
        row1, col1, row2, col2 = row1 + 1, col1 + 1, row2 + 1, col2 + 1
        bottomRight = self.sumMat[row2][col2]
        above = self.sumMat[row1 - 1][col2]
        left = self.sumMat[row2][col1 - 1]
        topLeft = self.sumMat[row1 - 1][col1 - 1]
        return bottomRight - above - left + topLeft

Time & Space Complexity

• Time complexity: $O(1)$ for each query.
• Space complexity: $O(m * n)$

Where $m$ is the number of rows and $n$ is the number of columns in the matrix.