2536. Increment Submatrices by One

Description

You are given a positive integer n, indicating that we initially have an n x n 0-indexed integer matrix mat filled with zeroes.

You are also given a 2D integer array query. For each query[i] = [row1i, col1i, row2i, col2i], you should do the following operation:

• Add 1 to every element in the submatrix with the top left corner (row1i, col1i) and the bottom right corner (row2i, col2i). That is, add 1 to mat[x][y] for all row1i <= x <= row2i and col1i <= y <= col2i.

Return the matrix mat after performing every query.

 

Example 1:

Input: n = 3, queries = [[1,1,2,2],[0,0,1,1]]
Output: [[1,1,0],[1,2,1],[0,1,1]]
Explanation: The diagram above shows the initial matrix, the matrix after the first query, and the matrix after the second query.
- In the first query, we add 1 to every element in the submatrix with the top left corner (1, 1) and bottom right corner (2, 2).
- In the second query, we add 1 to every element in the submatrix with the top left corner (0, 0) and bottom right corner (1, 1).

Example 2:

Input: n = 2, queries = [[0,0,1,1]]
Output: [[1,1],[1,1]]
Explanation: The diagram above shows the initial matrix and the matrix after the first query.
- In the first query we add 1 to every element in the matrix.

 

Constraints:

• 1 <= n <= 500
• 1 <= queries.length <= 104
• 0 <= row1i <= row2i < n
• 0 <= col1i <= col2i < n

 

Solutions

Solution: Difference Array

• Time complexity: O(n²+queries.length)
• Space complexity: O(n²)

 

JavaScript

/**
 * @param {number} n
 * @param {number[][]} queries
 * @return {number[][]}
 */
const rangeAddQueries = function (n, queries) {
  const diffs = Array.from({ length: n + 1 }, () => {
    return new Array(n + 1).fill(0);
  });
  const result = Array.from({ length: n }, () => new Array(n).fill(0));

  for (const [row1, col1, row2, col2] of queries) {
    diffs[row1][col1] += 1;
    diffs[row1][col2 + 1] -= 1;
    diffs[row2 + 1][col1] -= 1;
    diffs[row2 + 1][col2 + 1] += 1;
  }

  for (let row = 0; row < n; row++) {
    for (let col = 0; col < n; col++) {
      diffs[row][col] += diffs[row - 1]?.[col] ?? 0;
      diffs[row][col] += diffs[row][col - 1] ?? 0;
      diffs[row][col] -= diffs[row - 1]?.[col - 1] ?? 0;
      result[row][col] = diffs[row][col];
    }
  }

  return result;
};