Leetcode #1203: Sort Items by Groups Respecting Dependencies
In this guide, we solve Leetcode #1203 Sort Items by Groups Respecting Dependencies in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Depth-First Search, Breadth-First Search, Graph, Topological Sort
Intuition
The data forms a graph, so we should explore nodes and edges systematically.
A traversal ensures we visit each node once while maintaining the needed state.
Approach
Build an adjacency list and traverse with BFS or DFS.
Aggregate results as you visit nodes.
Steps:
- Build the graph.
- Traverse with BFS/DFS.
- Accumulate the required output.
Example
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]
Python Solution
class Solution:
def sortItems(
self, n: int, m: int, group: List[int], beforeItems: List[List[int]]
) -> List[int]:
def topo_sort(degree, graph, items):
q = deque(i for _, i in enumerate(items) if degree[i] == 0)
res = []
while q:
i = q.popleft()
res.append(i)
for j in graph[i]:
degree[j] -= 1
if degree[j] == 0:
q.append(j)
return res if len(res) == len(items) else []
idx = m
group_items = [[] for _ in range(n + m)]
for i, g in enumerate(group):
if g == -1:
group[i] = idx
idx += 1
group_items[group[i]].append(i)
item_degree = [0] * n
group_degree = [0] * (n + m)
item_graph = [[] for _ in range(n)]
group_graph = [[] for _ in range(n + m)]
for i, gi in enumerate(group):
for j in beforeItems[i]:
gj = group[j]
if gi == gj:
item_degree[i] += 1
item_graph[j].append(i)
else:
group_degree[gi] += 1
group_graph[gj].append(gi)
group_order = topo_sort(group_degree, group_graph, range(n + m))
if not group_order:
return []
ans = []
for gi in group_order:
items = group_items[gi]
item_order = topo_sort(item_degree, item_graph, items)
if len(items) != len(item_order):
return []
ans.extend(item_order)
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.