This approach uses a hash map (or dictionary) to group elements that belong to the same diagonal. The sum of indices (i + j) of each element forms a unique key for each diagonal. We iterate over the 2D list, store elements based on their diagonal key, and then read each diagonal to form the result in the required traversal order.
Time Complexity: O(N), where N is the total number of elements, as we iterate over each element, and store them in the diagonals dictionary.
Space Complexity: O(N), due to the storage requirement for the diagonals dictionary.
1def findDiagonalOrder(nums):
2 diagonals = {}
3 for i in range(len(nums)):
4 for j in range(len(nums[i])):
5 if (i + j) not in diagonals:
6 diagonals[i + j] = []
7 diagonals[i + j].append(nums[i][j])
8 result = []
9 for k in sorted(diagonals.keys()):
10 result.extend(reversed(diagonals[k]))
11 return resultThis solution uses a dictionary to store elements by their diagonal key. We compute the diagonal key as the sum of row and column indices. For each diagonal, we store elements in a list. Finally, we sort the keys and concatenate the reversed lists (since lower diagonals need to come first in the result).
Another approach is to flatten the 2D matrix into a list of triples containing the row index, column index, and value of each element. We then sort this list, prioritizing the sum of indices (i + j), followed by reversing elements when processed within the same diagonal, ensuring the correct order for traversing.
Time Complexity: O(N log N) due to sorting operation.
Space Complexity: O(N), storing all elements in a list.
1import java.util.*;
2
3class Solution {
4 public int[] findDiagonalOrder(List<List<Integer>> nums) {
5 List<int[]> elements = new ArrayList<>();
6 for (int i = 0; i < nums.size(); i++) {
7 for (int j = 0; j < nums.get(i).size(); j++) {
8 elements.add(new int[]{i, j, nums.get(i).get(j)});
9 }
10 }
11 Collections.sort(elements, (a, b) -> {
12 int cmp = (a[0] + a[1]) - (b[0] + b[1]);
13 return cmp != 0 ? cmp : b[0] - a[0];
14 });
15 int[] result = new int[elements.size()];
16 for (int k = 0; k < elements.size(); k++) {
17 result[k] = elements.get(k)[2];
18 }
19 return result;
20 }
21}The Java solution collects elements with their indices into a list, which it sorts first by diagonal (i.e., the sum of indices), and secondly by row index, reversed. This permits easy creation of the result as values are extracted in final order.