Sort Items by Groups Respecting Dependencies
Problem Description
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 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:
- 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.
Solve it on LeetCode
Approaches
1. Basic Topological Sort with Two Separate Graphs
Intuition:
The problem can be broken down into two separate topological sort problems: one for the items and another for the groups. Each item belongs to a group, and items and groups must be reordered to respect their respective dependencies.
The plan is to:
- Separate items into groups and form two graphs: one for item dependencies and one for group dependencies.
- Use topological sorting to sort items within a group and then sort groups themselves.
Steps:
- We will first set up the item graph and group graph using adjacency lists.
- Perform topological sort on the items within each group.
- Perform topological sort on the groups themselves.
Code:
- Time Complexity: Building the graphs takes O(n + m), where n is the number of items and m is the number of relations. Topological sort for each graph takes O(n + m) as well. Overall complexity is O(n + m).
- Space Complexity: Storing the graphs requires O(n + m) space.
2. Advanced Topological Sort with Single Graph
Intuition:
To optimize further, we combine the management of items and groups into a single graph. We treat groups as virtual nodes and manage dependencies among them inline with item dependencies. This removes the dependency on two separate graph traversals.
Steps:
- Items are assigned to their respective groups or created as standalone groups for unassigned items.
- A single graph tracks all group and item dependencies.
- Perform a single topological sort to create a valid order of groups and items interspersed.
Code:
- Time Complexity: O(n + m) due to the graph setup and traversal.
- Space Complexity: O(n + m) for storing the graph.