Class Solution
java.lang.Object
g1201_1300.s1203_sort_items_by_groups_respecting_dependencies.Solution
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:**
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**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:**
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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 Summary
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Solution
public Solution()
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Method Details
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sortItems
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