Class 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:

    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 Detail

      • Solution

        public Solution()
    • Method Detail

      • sortItems

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