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.

----------  Solutions-------------
To solve this problem, you can follow these steps:

1. Initialize a 2D array `matrix` of size `n x n` filled with zeros.
2. Iterate through each query in the `queries` array.
3. For each query:
   - Check if `typei` is 0 or 1.
   - If `typei` is 0, update the row with index `indexi` to have all values as `vali`.
   - If `typei` is 1, update the column with index `indexi` to have all values as `vali`.
4. After applying all queries, calculate the sum of integers in the matrix.
5. Return the sum.

Here's a JavaScript function implementing the above steps:

```javascript
function sumAfterQueries(n, queries) {
    let matrix = Array.from({ length: n }, () => Array(n).fill(0)); // Step 1

    for (let [type, index, val] of queries) { // Step 2
        if (type === 0) { // Step 3
            matrix[index].fill(val);
        } else if (type === 1) {
            for (let i = 0; i < n; i++) {
                matrix[i][index] = val;
            }
        }
    }

    let sum = 0;
    for (let row of matrix) { // Step 4
        sum += row.reduce((acc, curr) => acc + curr, 0);
    }

    return sum; // Step 5
}
```

You can then call this function with the given `n` and `queries` array to get the sum of integers in the matrix after all queries are applied.