Leetcode #2203: Minimum Weighted Subgraph With the Required Paths
In this guide, we solve Leetcode #2203 Minimum Weighted Subgraph With the Required Paths 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
You are given an integer n denoting the number of nodes of a weighted directed graph. The nodes are numbered from 0 to n - 1.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Graph, Shortest Path
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 = 6, edges = [[0,2,2],[0,5,6],[1,0,3],[1,4,5],[2,1,1],[2,3,3],[2,3,4],[3,4,2],[4,5,1]], src1 = 0, src2 = 1, dest = 5
Output: 9
Explanation:
The above figure represents the input graph.
The blue edges represent one of the subgraphs that yield the optimal answer.
Note that the subgraph [[1,0,3],[0,5,6]] also yields the optimal answer. It is not possible to get a subgraph with less weight satisfying all the constraints.
Python Solution
class Solution:
def minimumWeight(
self, n: int, edges: List[List[int]], src1: int, src2: int, dest: int
) -> int:
def dijkstra(g, u):
dist = [inf] * n
dist[u] = 0
q = [(0, u)]
while q:
d, u = heappop(q)
if d > dist[u]:
continue
for v, w in g[u]:
if dist[v] > dist[u] + w:
dist[v] = dist[u] + w
heappush(q, (dist[v], v))
return dist
g = defaultdict(list)
rg = defaultdict(list)
for f, t, w in edges:
g[f].append((t, w))
rg[t].append((f, w))
d1 = dijkstra(g, src1)
d2 = dijkstra(g, src2)
d3 = dijkstra(rg, dest)
ans = min(sum(v) for v in zip(d1, d2, d3))
return -1 if ans >= inf else ans
Complexity
The time complexity is O(V+E). The space complexity is O(V).
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.