Class Solution
-
- All Implemented Interfaces:
public final class Solution1203 - Sort Items by Groups Respecting Dependencies.
Hard
There are
nitems each belonging to zero or one ofmgroups wheregroup[i]is the group that thei\-th item belongs to and it's equal to-1if thei\-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 thei\-th item in the sorted array (to the left of thei\-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:
<code>1 <= m <= n <= 3 * 10<sup>4</sup></code>
group.length == beforeItems.length == n-1 <= group[i] <= m - 10 <= beforeItems[i].length <= n - 10 <= beforeItems[i][j] <= n - 1i != beforeItems[i][j]beforeItems[i]does not contain duplicates elements.
-
-
Constructor Summary
Constructors Constructor Description Solution()
-