How do you solve Sort Items by Groups Respecting Dependencies on LeetCode?

Sort Items by Groups Respecting Dependencies (LeetCode #1203) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + m) time complexity and O(n + m) space complexity. Key insight: Understanding the problem as a graph with dependencies helps in visualizing the solution.

What is the time complexity of Sort Items by Groups Respecting Dependencies?

The optimal solution for Sort Items by Groups Respecting Dependencies runs in O(n + m) time and uses O(n + m) space. The time complexity is O(n + m) because we perform a topological sort on both items and groups, which involves visiting each node and edge once.

What is the brute force solution for Sort Items by Groups Respecting Dependencies?

The brute force approach for Sort Items by Groups Respecting Dependencies is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible permutations of the items and checking which ones satisfy the group and dependency constraints. This is simple but inefficient. The optimal Optimal Solution approach improves this to O(n + m).

What are the interview tips for Sort Items by Groups Respecting Dependencies?

Clearly explain your thought process and how you visualize the problem as a graph. Discuss edge cases, such as items with no dependencies or groups with no items. Practice implementing topological sorting, as it's a common technique in graph-related problems.

What are common mistakes when solving Sort Items by Groups Respecting Dependencies?

Failing to account for items that belong to no group and treating them separately. Not checking for cycles in the dependency graph, which can lead to incorrect assumptions about the possibility of sorting.

#1203

Sort Items by Groups Respecting Dependencies

Hard

Depth-First SearchBreadth-First SearchGraph TheoryTopological SortGraph TheoryTopological SortDependency Resolution

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n + m)
Space	O(1)	O(n + m)

💡

Intuition

Time O(n + m)Space O(n + m)

The optimal solution uses graph theory concepts, specifically topological sorting, to efficiently order the items while respecting both group and dependency constraints.

⚙️

Algorithm

4 steps

1Step 1: Create a graph for items and their dependencies.
2Step 2: Perform a topological sort on the items to respect the beforeItems constraints.
3Step 3: Group the sorted items based on their group assignments.
4Step 4: Sort each group individually and concatenate the results.

solution.py53 lines

1from collections import defaultdict, deque
2
3def sortItems(n, m, group, beforeItems):
4    # Step 1: Create a graph
5    item_graph = defaultdict(list)
6    in_degree = [0] * n
7    for i in range(n):
8        for before in beforeItems[i]:
9            item_graph[before].append(i)
10            in_degree[i] += 1
11    
12    # Step 2: Topological sort for items
13    item_order = topological_sort(n, in_degree, item_graph)
14    if not item_order:
15        return []
16    
17    # Step 3: Group items
18    group_graph = defaultdict(list)
19    group_in_degree = [0] * m
20    group_items = defaultdict(list)
21    for i in range(n):
22        g = group[i] if group[i] != -1 else m
23        group_items[g].append(i)
24        for before in beforeItems[i]:
25            before_group = group[before] if group[before] != -1 else m
26            if before_group != g:
27                group_graph[before_group].append(g)
28                group_in_degree[g] += 1
29    
30    # Step 4: Topological sort for groups
31    group_order = topological_sort(m + 1, group_in_degree, group_graph)
32    if not group_order:
33        return []
34    
35    # Step 5: Sort items in each group
36    result = []
37    for g in group_order:
38        if g in group_items:
39            result.extend(sorted(group_items[g]))
40    return result
41
42
43def topological_sort(n, in_degree, graph):
44    queue = deque([i for i in range(n) if in_degree[i] == 0])
45    order = []
46    while queue:
47        node = queue.popleft()
48        order.append(node)
49        for neighbor in graph[node]:
50            in_degree[neighbor] -= 1
51            if in_degree[neighbor] == 0:
52                queue.append(neighbor)
53    return order if len(order) == n else []

ℹ

Complexity note: The time complexity is O(n + m) because we perform a topological sort on both items and groups, which involves visiting each node and edge once.

1Understanding the problem as a graph with dependencies helps in visualizing the solution.
2Topological sorting is crucial for resolving dependencies in both items and groups.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.