java.lang.Object
g1201_1300.s1203_sort_items_by_groups_respecting_dependencies.Solution

public class Solution extends Object
1203 - Sort Items by Groups Respecting Dependencies\. Hard There are `n` items each belonging to zero or one of `m` groups where `group[i]` is the group that the `i`\-th item belongs to and it's equal to `-1` if the `i`\-th item belongs to no group. The items and the groups are zero indexed. A group can have no item belonging to it. Return a sorted list of the items such that: * The items that belong to the same group are next to each other in the sorted list. * There are some relations between these items where `beforeItems[i]` is a list containing all the items that should come before the `i`\-th item in the sorted array (to the left of the `i`\-th item). Return any solution if there is more than one solution and return an **empty list** if there is no solution. **Example 1:** **![](https://assets.leetcode.com/uploads/2019/09/11/1359_ex1.png)** **Input:** n = 8, m = 2, group = [-1,-1,1,0,0,1,0,-1], beforeItems = \[\[],[6],[5],[6],[3,6],[],[],[]] **Output:** [6,3,4,1,5,2,0,7] **Example 2:** **Input:** n = 8, m = 2, group = [-1,-1,1,0,0,1,0,-1], beforeItems = \[\[],[6],[5],[6],[3],[],[4],[]] **Output:** [] **Explanation:** This is the same as example 1 except that 4 needs to be before 6 in the sorted list. **Constraints:** * 1 <= m <= n <= 3 * 104 * `group.length == beforeItems.length == n` * `-1 <= group[i] <= m - 1` * `0 <= beforeItems[i].length <= n - 1` * `0 <= beforeItems[i][j] <= n - 1` * `i != beforeItems[i][j]` * `beforeItems[i] `does not contain duplicates elements.
  • Constructor Details

    • Solution

      public Solution()
  • Method Details

    • sortItems

      public int[] sortItems(int n, int m, int[] group, List<List<Integer>> beforeItems)