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Kotlin-based LeetCode algorithm problem solutions, regularly updated
2718\. Sum of Matrix After Queries
Medium
You are given an integer `n` and a **0-indexed** **2D array** `queries` where queries[i] = [typei, indexi, vali]
.
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 typei == 0
, set the values in the row with indexi
to vali
, overwriting any previous values.
* if typei == 1
, set the values in the column with indexi
to vali
, 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 <= typei <= 1
* 0 <= indexi < n
* 0 <= vali <= 105