|
| 1 | +# Sort Items by Groups Respecting Dependencies |
| 2 | + |
| 3 | +You are given n items indexed from 0 to n−1. Each item belongs to 0 or one of m groups, described by the array group, |
| 4 | +where: |
| 5 | +- `group[i]` represents the group of the ith item. |
| 6 | +- If `group[i]==−1`, the item isn’t assigned to any existing group and should be treated as belonging to its own unique |
| 7 | + group. |
| 8 | + |
| 9 | +You’re also given a list, `beforeItems`, where `beforeItems[i]` contains all items that must precede item `i` in the final |
| 10 | +ordering. |
| 11 | + |
| 12 | +Your goal is to arrange all n items in a list that satisfies both of the following rules: |
| 13 | + |
| 14 | +- **Dependency order**: Every item must appear after all the items listed in `beforeItems[i]`. |
| 15 | +- **Group continuity**: All items that belong to the same group must appear next to each other in the final order. |
| 16 | + |
| 17 | +If there are multiple valid orderings, return any of them. If there’s no possible ordering, return an empty list. |
| 18 | + |
| 19 | +## Constraints |
| 20 | + |
| 21 | +- 1 <= `m` <= `n` <= 3 * 10^4 |
| 22 | +- `group.length` == `beforeItems.length` == n |
| 23 | +- -1 <= `group[i]` <= m - 1 |
| 24 | +- 0 <= `beforeItems[i].length` <= n - 1 |
| 25 | +- 0 <= `beforeItems[i][j]` <= n - 1 |
| 26 | +- i != `beforeItems[i][j]` |
| 27 | +- `beforeItems[i]` does not contain duplicates elements. |
| 28 | + |
| 29 | +## Examples |
| 30 | + |
| 31 | + |
| 32 | + |
| 33 | + |
| 34 | + |
| 35 | +Example 4 |
| 36 | + |
| 37 | +```text |
| 38 | +Input: n = 8, m = 2, group = [-1,-1,1,0,0,1,0,-1], beforeItems = [[],[6],[5],[6],[3,6],[],[],[]] |
| 39 | +Output: [6,3,4,1,5,2,0,7] |
| 40 | +``` |
| 41 | + |
| 42 | +Example 5 |
| 43 | + |
| 44 | +```text |
| 45 | +Input: n = 8, m = 2, group = [-1,-1,1,0,0,1,0,-1], beforeItems = [[],[6],[5],[6],[3],[],[4],[]] |
| 46 | +Output: [] |
| 47 | +Explanation: This is the same as example 4 except that 4 needs to be before 6 in the sorted list. |
| 48 | +``` |
| 49 | + |
| 50 | +## Topics |
| 51 | + |
| 52 | +- Depth-First Search |
| 53 | +- Breadth-First Search |
| 54 | +- Graph Theory |
| 55 | +- Topological Sort |
| 56 | + |
| 57 | +## Hints |
| 58 | + |
| 59 | +- Think of it as a graph problem. |
| 60 | +- We need to find a topological order on the dependency graph. |
| 61 | +- Build two graphs, one for the groups and another for the items. |
| 62 | + |
| 63 | +## Solution |
| 64 | + |
| 65 | +The essence of this problem lies in managing two-level dependencies: |
| 66 | + |
| 67 | +- Item-level dependencies: Each item must appear only after all items in its beforeItems list. |
| 68 | +- Group-level contiguity: All items belonging to the same group must appear consecutively in the final order. |
| 69 | + |
| 70 | +This dual requirement naturally maps to a hierarchical topological sorting approach on two directed acyclic graphs (DAGs). |
| 71 | +One for groups and one for items. First, we normalize the input by assigning a unique group to every item that originally |
| 72 | +has group[i] = -1, so each item belongs to exactly one group. Then, we build two directed graphs: |
| 73 | + |
| 74 | +- An item-level dependency graph is built from beforeItems, where an edge u → v indicates that item u must come before |
| 75 | + item v. |
| 76 | +- A group-level dependency graph is constructed by examining inter-group dependencies: if an item u depends on an item v |
| 77 | + and they belong to different groups, an edge is added from group[v] to group[u], meaning group[v] must come before |
| 78 | + group[u]. |
| 79 | + |
| 80 | +To solve the problem, we: |
| 81 | + |
| 82 | +- Sort the groups topologically using the group-level graph to determine a valid global sequence of groups. If a cycle |
| 83 | + exists (i.e., no valid ordering), return an empty list. |
| 84 | + |
| 85 | +- For each group in that global order, perform a topological sort on its internal items using only the relevant subgraph |
| 86 | + of item dependencies. If any group contains a cyclic dependency among its items, return an empty list. |
| 87 | + |
| 88 | +- Concatenate the sorted items group by group, following the global order of groups from step 1. |
| 89 | + |
| 90 | +This hierarchical approach cleanly addresses both concerns: the outer sort enforces inter-group dependency and contiguity, |
| 91 | +while the inner sorts respect intra-group ordering. The result is a single linear sequence that satisfies all constraints, |
| 92 | +or correctly reports impossibility if cycles prevent a valid schedule. |
| 93 | + |
| 94 | +Using the intuition above, we implement the algorithm as follows: |
| 95 | + |
| 96 | +1. We iterate through all items in the group: |
| 97 | + - If an item has no assigned group (indicated by -1): |
| 98 | + - We assign it a new unique group number equal to the current value of m. |
| 99 | + - We increment m by one. |
| 100 | +2. We create two adjacency lists (graphs): item_graph to store item-level dependencies and group_graph to store |
| 101 | + group-level dependencies. |
| 102 | +3. We create two in-degree arrays initialized with zero: item_indegree of size n to count the number of prerequisites |
| 103 | + each item has, and group_indegree of size m to count the number of prerequisite groups each group has. |
| 104 | +4. We loop through each item, curr from 0 to n - 1: |
| 105 | + - For each prerequisite prev in beforeItems[curr], we: |
| 106 | + - Add a directed edge from prev to curr in item_graph. |
| 107 | + - Increment item_indegree[curr]. |
| 108 | + - If group[prev] is not equal to group[curr], we: |
| 109 | + - Add an edge from group[prev] to group[curr] in group_graph. |
| 110 | + - Increment group_indegree[group[curr]]. |
| 111 | +5. We perform a topological sort using the helper function, topoSort, on the items and store the ordered items in |
| 112 | + item_order. |
| 113 | +6. We perform another topological sort using the same helper function on groups and store the ordered groups in the |
| 114 | + group_order. |
| 115 | +7. If either the item_order or group_order is empty, we return an empty array because it is impossible to create a |
| 116 | + valid ordering. |
| 117 | +8. We initialize a map, group_to_items, to map each group number to a list of its items. |
| 118 | +9. For each item in item_order: |
| 119 | + - We append each item to its corresponding group’s list in the group_to_items map. This preserves the topological |
| 120 | + order of items within each group. |
| 121 | +10. We define an array, result, to store the final valid ordering. |
| 122 | +11. For each group g in group_order: |
| 123 | + - We add all items in group_to_items[g] to the result array. |
| 124 | +12. We return the result array containing all items in a valid order. |
| 125 | + |
| 126 | +The topo_sort helper function receives a list of nodes, a graph (adjacency list), and an indegree array, and returns the |
| 127 | +topological order using Kahn’s algorithm. It performs the following steps: |
| 128 | + |
| 129 | +1. We initialize a queue q with all nodes in nodes that have an indegree[node] equal to 0. |
| 130 | +2. We also initialize an empty array, order, to store the topological order. |
| 131 | +3. We iterate through q and: |
| 132 | + - Repeatedly remove a node from q. |
| 133 | + - Append this node to the order array. |
| 134 | + - For each neighbor, nei, of node, we: |
| 135 | + - Decrement the indegree of nei. |
| 136 | + - If the indegree of nei is 0: |
| 137 | + - We add nei to q. |
| 138 | +4. We return the order array if all nodes are processed; otherwise, we return an empty array (indicating a cycle exists). |
| 139 | + |
| 140 | +### Time Complexity |
| 141 | + |
| 142 | +The solution’s time complexity is O(n.2^n) because in the worst case, when the array contains |
| 143 | +n distinct elements, the recursion tree generates all 2^n possible subsets, resulting in 2^n recursive calls. In each call, |
| 144 | +we create a copy of the current subset, which can take up to O(n) time when the subset is at its largest. Therefore, the |
| 145 | +overall time complexity of this approach is O(n.2^n). |
| 146 | + |
| 147 | +### Space Complexity |
| 148 | + |
| 149 | +The space complexity of the sorting step depends on the specific sorting algorithm used by the language or library, but |
| 150 | +most standard comparison-based sorts require O(log(n)) auxiliary space. The recursive backtracking process uses up to |
| 151 | +O(n) space on the call stack, as the maximum depth of recursion corresponds to building subsets of length n. The space |
| 152 | +used to store the final list of subsets is not counted toward auxiliary space. |
| 153 | + |
| 154 | +Therefore, the overall auxiliary space complexity of this approach is O(n). |
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