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Java-based LeetCode algorithm problem solutions, regularly updated

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package g2501_2600.s2536_increment_submatrices_by_one;

// #Medium #Array #Matrix #Prefix_Sum #2023_04_22_Time_12_ms_(88.15%)_Space_50.9_MB_(65.40%)

/**
 * 2536 - Increment Submatrices by One\.
 *
 * Medium
 *
 * 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] = [row1_i, col1_i, row2_i, col2_i], you should do the following operation:
 *
 * *   Add `1` to **every element** in the submatrix with the **top left** corner (row1_i, col1_i) and the **bottom right** corner (row2_i, col2_i). That is, add `1` to `mat[x][y]` for all row1_i <= x <= row2_i and col1_i <= y <= col2_i.
 *
 * Return _the matrix_ `mat` _after performing every query._
 *
 * **Example 1:**
 *
 * ![](https://assets.leetcode.com/uploads/2022/11/24/p2example11.png)
 *
 * **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:**
 *
 * ![](https://assets.leetcode.com/uploads/2022/11/24/p2example22.png)
 *
 * **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 <= row1_i <= row2_i < n
 * *   0 <= col1_i <= col2_i < n
**/
public class Solution {
    public int[][] rangeAddQueries(int n, int[][] queries) {
        int[][] g = new int[n][n];
        for (int[] q : queries) {
            int r1 = q[0];
            int c1 = q[1];
            int r2 = q[2];
            int c2 = q[3];
            g[r1][c1]++;
            if (c2 < n - 1) {
                g[r1][c2 + 1]--;
            }
            if (r2 < n - 1) {
                g[r2 + 1][c1]--;
            }
            if (c2 < n - 1 && r2 < n - 1) {
                g[r2 + 1][c2 + 1]++;
            }
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (i > 0) {
                    g[i][j] += g[i - 1][j];
                }
                if (j > 0) {
                    g[i][j] += g[i][j - 1];
                }
                if (i > 0 && j > 0) {
                    g[i][j] -= g[i - 1][j - 1];
                }
            }
        }
        return g;
    }
}