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:
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 * 104group.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.This approach involves processing the items and grouping them based on dependency relationships. We'll use topological sorting for individual items within a group. The main steps are:
This Python solution implements the approach of topological sorting using graph algorithms. We first assign temporary groups to items without any group. Then, two graphs (one for items and another for groups) are built based on given dependencies. We perform topological sorting on item and group graphs to determine the correct order. The results are gathered by reorganizing items according to their group order and finally returning the sorted list. If any of the topological sort fails to return a complete order, we return an empty list.
Java
The time complexity of this solution is O(n + m) where n is the number of items and m is the number of groups, due to the topological sort operation. The space complexity is also O(n + m) for storing the adjacency list, indegree count, and result list.
This approach incorporates priority queues (heaps) to perform topological sorting on items based on their dependencies, akin to a 'priority' based execution. It works by:
The C++ solution utilizes a min-priority queue for managing items with zero dependencies. This ensures items are processed in order of their natural indices, which mimics the behavior of prioritizing certain operations. After identifying initial independent items, it iterates through the graph, managing dependencies and constructing the correct order through traditional topological sorting mechanisms. The heap structure adds an extra layer of determinism by processing the lowest index 'ready' item first.
C#
The time complexity remains O(n + m) due to the heap operations and graph traversal. Space complexity is O(n + m) to store necessary elements and dependency structures.
| Approach | Complexity |
|---|---|
| Topological Sort with Grouped Items | The time complexity of this solution is O(n + m) where n is the number of items and m is the number of groups, due to the topological sort operation. The space complexity is also O(n + m) for storing the adjacency list, indegree count, and result list. |
| Using Priority Queues for Topological Sorting | The time complexity remains O(n + m) due to the heap operations and graph traversal. Space complexity is O(n + m) to store necessary elements and dependency structures. |
Sort Items by Groups Respecting Dependencies | Broken into Simple Steps | Intuition | Leetcode-1203 • codestorywithMIK • 6,603 views views
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