Sorted Extended Matrix Diagonals

Given a square matrix of characters matrix with a size of n x n, your task is to create a sorted list of matrix's extended diagonals, where each diagonal has a length of n.
A matrix of size n x n has 2n - 1 diagonals parallel to the main diagonal, with each diagonal starting at its upper point and ending at its lower point. Since these diagonals have different lengths, traverse each one cyclically (go back to the starting point after reaching the end point) until it consists of n characters.
Sort the resulting strings of n characters in alphabetical order, and return an array of 2n - 1 integers, representing the diagonals' 1-based indices in their sorted order. In the case of alphabetically equal strings, their indices should be kept in the original order.
Below you can find an example of diagonals' numbering for a 5 x 5 matrix, where the number in the matrix corresponds to the diagonal index:
5 6 7 8 9
4 5 6 7 8
3 4 5 6 7
2 3 4 5 6
1 2 3 4 5
Note: You are not expected to provide the most optimal solution, but a solution with time complexity not worse than O(n^4) will fit within the execution time limit.

Function
solution(matrix: String[][]) → int[]

Examples
Example 1
matrix = [["b","b"],["c","a"]]
return = [2,3,1]
For
matrix = [["b", "b"],
["c", "a"]]
the output should be solution(matrix) = [2, 3, 1].
The diagonal with index 1 is ["c"] and its corresponding cyclic string is "cc".
The diagonal with index 2 is ["b", "a"] and its corresponding cyclic string is "ba".
The diagonal with index 3 is ["b"] and its corresponding cyclic string is "bb".
The alphabetical ordering of the matrix diagonals looks like ["ba", "bb", "cc"], so the answer is [2, 3, 1].
Example 2
matrix = [["a","c","a","b","b"],["c","b","a","c","b"],["a","a","e","c","b"],["b","b","d","a","g"],["a","b","e","b","a"]]
return = [1,5,3,7,2,8,9,6,4]
For
matrix = [["a", "c", "a", "b", "b"],
["c", "b", "a", "c", "b"],
["a", "a", "e", "c", "b"],
["b", "b", "d", "a", "g"],
["a", "b", "e", "b", "a"]]
the output should be solution(matrix) = [1, 5, 3, 7, 2, 8, 9, 6, 4].
The explanation for this example is not visible in the source image.

Constraints
FastPrep execution-adapter constraints (not shown in the source image):
n = matrix.length and n >= 1.
Every row of matrix has exactly n entries.
Every entry of matrix is a one-character string.

The trick is indexing. Diagonal k, from 1 to 2n - 1, starts at the bottom-left and ends at the top-right, with cells where j - i = k - n. Collect characters going down each diagonal in order, then repeat them cyclically until you have n characters. Build the string with a modulo over the diagonal's length, so a length-1 diagonal like "c" becomes "cc". Then sort the indices by string with a stable sort. The stated O(n^4) bound is generous. Building strings costs O(n^2) total and sorting is about O(n^2 log n) comparisons. The common pitfalls are off-by-one errors in the 1-based index, getting the diagonal order backwards, and using an unstable sort so equal strings swap places. Test Example 1 by hand first. StealthCoder is your hedge in the live OA if the index math falls apart under pressure.

The honest play: practice the pattern, and have StealthCoder ready for the one you didn't see coming.