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Java-based LeetCode algorithm problem solutions, regularly updated

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package g2701_2800.s2718_sum_of_matrix_after_queries;

// #Medium #Array #Hash_Table #2023_09_18_Time_3_ms_(100.00%)_Space_61.2_MB_(84.49%)

/**
 * 2718 - Sum of Matrix After Queries\.
 *
 * Medium
 *
 * You are given an integer `n` and a **0-indexed** **2D array** `queries` where queries[i] = [type_i, index_i, val_i].
 *
 * Initially, there is a **0-indexed** `n x n` matrix filled with `0`'s. For each query, you must apply one of the following changes:
 *
 * *   if type_i == 0, set the values in the row with index_i to val_i, overwriting any previous values.
 * *   if type_i == 1, set the values in the column with index_i to val_i, overwriting any previous values.
 *
 * Return _the sum of integers in the matrix after all queries are applied_.
 *
 * **Example 1:**
 *
 * ![](https://assets.leetcode.com/uploads/2023/05/11/exm1.png)
 *
 * **Input:** n = 3, queries = \[\[0,0,1],[1,2,2],[0,2,3],[1,0,4]]
 *
 * **Output:** 23
 *
 * **Explanation:** The image above describes the matrix after each query. The sum of the matrix after all queries are applied is 23.
 *
 * **Example 2:**
 *
 * ![](https://assets.leetcode.com/uploads/2023/05/11/exm2.png)
 *
 * **Input:** n = 3, queries = \[\[0,0,4],[0,1,2],[1,0,1],[0,2,3],[1,2,1]]
 *
 * **Output:** 17
 *
 * **Explanation:** The image above describes the matrix after each query. The sum of the matrix after all queries are applied is 17.
 *
 * **Constraints:**
 *
 * *   1 <= n <= 10⁴
 * *   1 <= queries.length <= 5 * 10⁴
 * *   `queries[i].length == 3`
 * *   0 <= type_i <= 1
 * *   0 <= index_i < n
 * *   0 <= val_i <= 10⁵
**/
public class Solution {
    public long matrixSumQueries(int n, int[][] queries) {
        boolean[] queriedRow = new boolean[n];
        boolean[] queriedCol = new boolean[n];
        long sum = 0;
        int remainingRows = n;
        int remainingCols = n;
        for (int i = queries.length - 1; i >= 0; i--) {
            int type = queries[i][0];
            int index = queries[i][1];
            int value = queries[i][2];
            if ((type == 0 && !queriedRow[index]) || (type == 1 && !queriedCol[index])) {
                sum += (long) value * (type == 0 ? remainingCols : remainingRows);
                if (type == 0) {
                    remainingRows--;
                    queriedRow[index] = true;
                } else {
                    remainingCols--;
                    queriedCol[index] = true;
                }
            }
        }
        return sum;
    }
}