Scale Engineer

Range Sum Query 2D - Immutable

Med
#0291Time: O(1) for constructor, O(m*n) for sumRegion in worst case where m and n are the dimensions of the query rectangleSpace: O(1) additional space (only storing reference to original matrix)6 companies
Patterns
DesignPrefix Sum
Data structures
ArrayMatrix
Companies
DoorDashNvidiaSnowflakeLyft

Prompt

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
  • -104 <= matrix[i][j] <= 104
  • 0 <= row1 <= row2 < m
  • 0 <= col1 <= col2 < n
  • At most 104 calls will be made to sumRegion.

Approaches

3 approaches with complexity analysis and trade-offs.

The most straightforward approach is to calculate the sum for each query by iterating through all elements in the specified rectangle. For each sumRegion call, we traverse the matrix from (row1, col1) to (row2, col2) and accumulate the sum.

Algorithm

  1. Store the original matrix in the constructor
  2. For each sumRegion query:
    • Initialize sum to 0
    • Iterate through rows from row1 to row2
    • For each row, iterate through columns from col1 to col2
    • Add each element to the sum
    • Return the accumulated sum

Walkthrough

This approach directly calculates the sum for each query without any preprocessing. When sumRegion(row1, col1, row2, col2) is called, we iterate through all rows from row1 to row2 and all columns from col1 to col2, adding each element to our running sum.

class NumMatrix {    private int[][] matrix;        public NumMatrix(int[][] matrix) {        this.matrix = matrix;    }        public int sumRegion(int row1, int col1, int row2, int col2) {        int sum = 0;        for (int i = row1; i <= row2; i++) {            for (int j = col1; j <= col2; j++) {                sum += matrix[i][j];            }        }        return sum;    }}

This approach is simple to implement and understand, but it doesn't meet the O(1) requirement for sumRegion operations.

Complexity

Time

O(1) for constructor, O(m*n) for sumRegion in worst case where m and n are the dimensions of the query rectangle

Space

O(1) additional space (only storing reference to original matrix)

Trade-offs

Pros

  • Simple and straightforward implementation

  • No additional space required beyond storing the original matrix

  • Easy to understand and debug

Cons

  • Does not meet the O(1) requirement for sumRegion

  • Inefficient for multiple queries as it recalculates the same sums

  • Performance degrades significantly with larger query rectangles

Solutions

class NumMatrix { private int [][] s ; public NumMatrix ( int [][] matrix ) { int m = matrix . length , n = matrix [ 0 ]. length ; s = new int [ m + 1 ][ n + 1 ]; for ( int i = 0 ; i < m ; ++ i ) { for ( int j = 0 ; j < n ; ++ j ) { s [ i + 1 ][ j + 1 ] = s [ i + 1 ][ j ] + s [ i ][ j + 1 ] - s [ i ][ j ] + matrix [ i ][ j ]; } } } public int sumRegion ( int row1 , int col1 , int row2 , int col2 ) { return s [ row2 + 1 ][ col2 + 1 ] - s [ row2 + 1 ][ col1 ] - s [ row1 ][ col2 + 1 ] + s [ row1 ][ col1 ]; } } /** * Your NumMatrix object will be instantiated and called as such: * NumMatrix obj = new NumMatrix(matrix); * int param_1 = obj.sumRegion(row1,col1,row2,col2); */

Video walkthrough

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