Leetcode #2642: Design Graph With Shortest Path Calculator
In this guide, we solve Leetcode #2642 Design Graph With Shortest Path Calculator 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 is a directed weighted graph that consists of n nodes numbered from 0 to n - 1. The edges of the graph are initially represented by the given array edges where edges[i] = [fromi, toi, edgeCosti] meaning that there is an edge from fromi to toi with the cost edgeCosti.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Graph, Design, Shortest Path, Heap (Priority Queue)
Intuition
We need to repeatedly access the smallest or largest element as the input changes.
A heap provides fast insertions and removals while keeping order.
Approach
Push candidates into the heap as you scan, and pop when you need the best element.
Keep the heap size bounded if the problem requires a top-k structure.
Steps:
- Push candidates into a heap.
- Pop the best candidate when needed.
- Maintain heap size or invariants.
Example
Input
["Graph", "shortestPath", "shortestPath", "addEdge", "shortestPath"]
[[4, [[0, 2, 5], [0, 1, 2], [1, 2, 1], [3, 0, 3]]], [3, 2], [0, 3], [[1, 3, 4]], [0, 3]]
Output
[null, 6, -1, null, 6]
Explanation
Graph g = new Graph(4, [[0, 2, 5], [0, 1, 2], [1, 2, 1], [3, 0, 3]]);
g.shortestPath(3, 2); // return 6. The shortest path from 3 to 2 in the first diagram above is 3 -> 0 -> 1 -> 2 with a total cost of 3 + 2 + 1 = 6.
g.shortestPath(0, 3); // return -1. There is no path from 0 to 3.
g.addEdge([1, 3, 4]); // We add an edge from node 1 to node 3, and we get the second diagram above.
g.shortestPath(0, 3); // return 6. The shortest path from 0 to 3 now is 0 -> 1 -> 3 with a total cost of 2 + 4 = 6.
Python Solution
class Graph:
def __init__(self, n: int, edges: List[List[int]]):
self.n = n
self.g = [[inf] * n for _ in range(n)]
for f, t, c in edges:
self.g[f][t] = c
def addEdge(self, edge: List[int]) -> None:
f, t, c = edge
self.g[f][t] = c
def shortestPath(self, node1: int, node2: int) -> int:
dist = [inf] * self.n
dist[node1] = 0
vis = [False] * self.n
for _ in range(self.n):
t = -1
for j in range(self.n):
if not vis[j] and (t == -1 or dist[t] > dist[j]):
t = j
vis[t] = True
for j in range(self.n):
dist[j] = min(dist[j], dist[t] + self.g[t][j])
return -1 if dist[node2] == inf else dist[node2]
# Your Graph object will be instantiated and called as such:
# obj = Graph(n, edges)
# obj.addEdge(edge)
# param_2 = obj.shortestPath(node1,node2)
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.