LeetCode #2718 — MEDIUM

Sum of Matrix After Queries

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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:

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:

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

3
[[0,0,1],[1,2,2],[0,2,3],[1,0,4]]

Example 2

3
[[0,0,4],[0,1,2],[1,0,1],[0,2,3],[1,2,1]]

Core Insight

What unlocks the optimal approach

Process queries in reversed order, as the latest queries represent the most recent changes in the matrix.
Once you encounter an operation on some row/column, no further operations will affect the values in this row/column. Keep track of seen rows and columns with a set.
When operating on an unseen row/column, the number of affected cells is the number of columns/rows you haven’t previously seen.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2718: Sum of Matrix After Queries
class Solution {
    public long matrixSumQueries(int n, int[][] queries) {
        Set<Integer> row = new HashSet<>();
        Set<Integer> col = new HashSet<>();
        int m = queries.length;
        long ans = 0;
        for (int k = m - 1; k >= 0; --k) {
            var q = queries[k];
            int t = q[0], i = q[1], v = q[2];
            if (t == 0) {
                if (row.add(i)) {
                    ans += 1L * (n - col.size()) * v;
                }
            } else {
                if (col.add(i)) {
                    ans += 1L * (n - row.size()) * v;
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2718: Sum of Matrix After Queries
func matrixSumQueries(n int, queries [][]int) (ans int64) {
	row, col := map[int]bool{}, map[int]bool{}
	m := len(queries)
	for k := m - 1; k >= 0; k-- {
		t, i, v := queries[k][0], queries[k][1], queries[k][2]
		if t == 0 {
			if !row[i] {
				ans += int64(v * (n - len(col)))
				row[i] = true
			}
		} else {
			if !col[i] {
				ans += int64(v * (n - len(row)))
				col[i] = true
			}
		}
	}
	return
}

# Accepted solution for LeetCode #2718: Sum of Matrix After Queries
class Solution:
    def matrixSumQueries(self, n: int, queries: List[List[int]]) -> int:
        row = set()
        col = set()
        ans = 0
        for t, i, v in queries[::-1]:
            if t == 0:
                if i not in row:
                    ans += v * (n - len(col))
                    row.add(i)
            else:
                if i not in col:
                    ans += v * (n - len(row))
                    col.add(i)
        return ans

// Accepted solution for LeetCode #2718: Sum of Matrix After Queries
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2718: Sum of Matrix After Queries
// class Solution {
//     public long matrixSumQueries(int n, int[][] queries) {
//         Set<Integer> row = new HashSet<>();
//         Set<Integer> col = new HashSet<>();
//         int m = queries.length;
//         long ans = 0;
//         for (int k = m - 1; k >= 0; --k) {
//             var q = queries[k];
//             int t = q[0], i = q[1], v = q[2];
//             if (t == 0) {
//                 if (row.add(i)) {
//                     ans += 1L * (n - col.size()) * v;
//                 }
//             } else {
//                 if (col.add(i)) {
//                     ans += 1L * (n - row.size()) * v;
//                 }
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2718: Sum of Matrix After Queries
function matrixSumQueries(n: number, queries: number[][]): number {
    const row: Set<number> = new Set();
    const col: Set<number> = new Set();
    let ans = 0;
    queries.reverse();
    for (const [t, i, v] of queries) {
        if (t === 0) {
            if (!row.has(i)) {
                ans += v * (n - col.size);
                row.add(i);
            }
        } else {
            if (!col.has(i)) {
                ans += v * (n - row.size);
                col.add(i);
            }
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(m)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.