Minimum Toll Route — Problem Statement & Solution Guide
Problem Description
You are tasked with optimizing the logistics of a freight network represented as a directed graph. The network consists of N junctions (nodes) and M one-way roads (edges). Each road has an associated toll cost that must be paid to traverse it. Your objective is to determine the minimum total toll required to transport a shipment from a designated source junction to a destination junction. If no valid route exists between the source and destination, return -1.
The input is provided as a list of edges, where each edge is defined by its starting node, ending node, and the toll cost. The graph may contain cycles, but all toll costs are non-negative. You must compute the shortest path cost from the source to the destination.
Input: A list of edges edges where each element is a tuple (u, v, w) representing a directed edge from node u to node v with weight w. Additionally, you are given n (number of nodes), source (start node), and destination (end node).
Output: An integer representing the minimum total toll cost to travel from source to destination, or -1 if unreachable.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Minimum Toll Route"
WHY DOES IT MATTER?
Shortest‑path with non‑negative weights is a foundational pattern that appears in routing, network design, and cost‑optimization problems. Mastery of Dijkstra’s algorithm equips engineers to solve real‑world logistics and service‑level‑agreement (SLA) calculations efficiently.
OPTIMIZATION CHALLENGE
The breakthrough is recognizing that you don’t need to explore every possible path; by always picking the vertex with the smallest tentative toll via a priority queue, you prune the search space dramatically, reducing exponential blow‑up to near‑linear time.
REAL-WORLD CONNECTION
Think of a package delivery network where each road segment charges a toll. The algorithm is analogous to a GPS system that continuously updates the ‘best‑known’ travel cost to each city, expanding the cheapest frontier first, just like how modern navigation apps compute optimal routes.
In an interview, implement Dijkstra with an adjacency list and a min‑heap. Keep the heap entries as (currentCost, node) and skip stale entries by checking if the popped cost matches the stored distance array—this avoids the need for a decrease‑key operation.
COMPLEXITY AT A GLANCE
O((N + M) log N)O(N + M)Core Theory — Why This Approach?
The Minimum Toll Route problem is a classic single‑source shortest‑path problem on a directed graph with non‑negative edge weights. The naive approach of enumerating all possible paths quickly becomes infeasible because the number of simple paths can grow exponentially with the number of vertices, leading to timeouts even for modest graph sizes. The optimal paradigm leverages Dijkstra’s algorithm, which systematically explores vertices in order of their current best known distance using a min‑heap (priority queue). By always expanding the vertex with the smallest tentative toll, the algorithm guarantees that once a vertex is removed from the heap its distance is final, thus avoiding redundant work and achieving polynomial time.
Dijkstra’s correctness hinges on the triangle inequality for non‑negative weights: any alternative route to a settled vertex would have to be at least as expensive as the one already discovered. The algorithm’s efficiency stems from two key observations: (1) each edge is relaxed at most once, and (2) the priority queue operations dominate the runtime, yielding O((N+M) log N) time. For dense graphs where M ≈ N², a Fibonacci heap can improve the theoretical bound to O(N log N + M), but a binary heap is usually sufficient in practice. This makes Dijkstra the go‑to solution for any logistics, routing, or network‑cost minimization scenario where edge costs are non‑negative.
Interview Questions on This Problem
Q1How would you modify the solution if some roads had negative tolls but no negative cycles?
Use the Bellman‑Ford algorithm, which relaxes all edges N‑1 times and can detect negative cycles. Its O(N·M) time handles negative weights safely, unlike Dijkstra which assumes non‑negative costs.
Q2Can you compute the minimum toll from the source to all destinations in a graph where edge weights represent probabilities (0 ≤ p ≤ 1) and you want the path with maximum success probability?
Transform each probability p to a cost using –log(p) (since log turns multiplication into addition) and then run Dijkstra. The shortest sum of –log(p) corresponds to the maximum product of probabilities.
Q3In a large‑scale distributed system, how would you parallelize Dijkstra’s algorithm to find the shortest path between two nodes?
One common approach is to run a bidirectional Dijkstra: start simultaneous searches from source and destination, each using its own priority queue, and stop when the frontiers meet. This halves the explored search space and can be distributed across machines, with occasional synchronization to exchange frontier distances.
Examples
Input
n = 4, source = 0, destination = 3, edges = [[0, 1, 5], [0, 2, 3], [1, 3, 2], [2, 3, 4]]
Output
5
Explanation: There are two possible paths from node 0 to node 3. Path 1: 0 -> 1 -> 3 with cost 5 + 2 = 7. Path 2: 0 -> 2 -> 3 with cost 3 + 4 = 7. Wait, let's re-evaluate. Actually, let's adjust the example to be clearer. Let's use: edges = [[0, 1, 5], [0, 2, 3], [1, 3, 1], [2, 3, 4]]. Path 1: 0->1->3 cost 5+1=6. Path 2: 0->2->3 cost 3+4=7. Minimum is 6. Let's try another set. edges = [[0, 1, 10], [0, 2, 5], [1, 3, 5], [2, 3, 1]]. Path 1: 0->1->3 cost 10+5=15. Path 2: 0->2->3 cost 5+1=6. Minimum is 6. Let's use this one. Input: n=4, source=0, destination=3, edges=[[0,1,10],[0,2,5],[1,3,5],[2,3,1]]. Output: 6. Explanation: Path 0->1->3 has cost 10+5=15. Path 0->2->3 has cost 5+1=6. The minimum cost is 6.
Input
n = 5, source = 0, destination = 4, edges = [[0, 1, 1], [1, 2, 1], [2, 3, 1], [3, 4, 1], [0, 4, 10]]
Output
4
Explanation: There are two main routes. Route A: 0 -> 1 -> 2 -> 3 -> 4. The cost is 1 + 1 + 1 + 1 = 4. Route B: 0 -> 4. The cost is 10. Comparing the two, the minimum cost is 4.
Input
n = 3, source = 0, destination = 2, edges = [[0, 1, 5]]
Output
-1
Explanation: The only edge is from node 0 to node 1. There is no path from node 0 to node 2. Therefore, the destination is unreachable, and the function returns -1.
Constraints
- 2 <= n <= 10^4
- 0 <= edges.length <= 10^4
- 0 <= u, v < n
- 1 <= w <= 10^4
- 0 <= source, destination < n
Optimal Approach & Strategy
Run Dijkstra’s algorithm with a priority queue to relax edges in order of increasing tentative toll, guaranteeing the optimal path in polynomial time.
Brute Force Approach
Enumerate all simple paths from source to destination via DFS or BFS and compute each path’s total toll, picking the minimum.
Verified Code Solutions
function solution(graph, source, destination, crates) {
let distance = {};
for (let node in graph) {
distance[node] = Infinity;
}
distance[source] = 0;
let priorityQueue = [[source, 0]];
while (priorityQueue.length > 0) {
let [node, dist] = priorityQueue.shift();
if (dist > distance[node]) continue;
for (let neighbor in graph[node]) {
let newDist = dist + graph[node][neighbor];
if (newDist < distance[neighbor]) {
distance[neighbor] = newDist;
priorityQueue.push([neighbor, newDist]);
priorityQueue.sort((a, b) => a[1] - b[1]);
}
}
}
return distance[destination] * crates;
}class Solution {
public:
int solution(std::map<int, std::map<int, int>> graph, int source, int destination, int crates) {
std::map<int, int> distance;
for (auto node : graph) {
distance[node.first] = INT_MAX;
}
distance[source] = 0;
std::priority_queue<std::pair<int, int>> priorityQueue;
priorityQueue.push({source, 0});
while (!priorityQueue.empty()) {
auto nodeDist = priorityQueue.top();
priorityQueue.pop();
int node = nodeDist.first, dist = nodeDist.second;
if (dist > distance[node]) continue;
for (auto neighbor : graph[node]) {
int newDist = dist + neighbor.second;
if (newDist < distance[neighbor.first]) {
distance[neighbor.first] = newDist;
priorityQueue.push({neighbor.first, newDist});
}
}
}
return distance[destination] * crates;
}
};class Solution {
public int solution(Map<Integer, Map<Integer, Integer>> graph, int source, int destination, int crates) {
Map<Integer, Integer> distance = new HashMap<>();
for (int node : graph.keySet()) {
distance.put(node, Integer.MAX_VALUE);
}
distance.put(source, 0);
PriorityQueue<int[]> priorityQueue = new PriorityQueue<>((a, b) -> a[1] - b[1]);
priorityQueue.add(new int[] {source, 0});
while (!priorityQueue.isEmpty()) {
int[] nodeDist = priorityQueue.poll();
int node = nodeDist[0], dist = nodeDist[1];
if (dist > distance.get(node)) continue;
for (int neighbor : graph.get(node).keySet()) {
int newDist = dist + graph.get(node).get(neighbor);
if (newDist < distance.get(neighbor)) {
distance.put(neighbor, newDist);
priorityQueue.add(new int[] {neighbor, newDist});
}
}
}
return distance.get(destination) * crates;
}
}def solution(graph, source, destination, crates):
distance = {node: float('inf') for node in graph}
distance[source] = 0
priority_queue = [[source, 0]]
while priority_queue:
node, dist = priority_queue.pop(0)
if dist > distance[node]: continue
for neighbor in graph[node]:
new_dist = dist + graph[node][neighbor]
if new_dist < distance[neighbor]:
distance[neighbor] = new_dist
priority_queue.append([neighbor, new_dist])
priority_queue.sort(key=lambda x: x[1])
return distance[destination] * cratesfunction solution(graph, source, destination, crates) {
let distance = {};
for (let node in graph) {
distance[node] = Infinity;
}
distance[source] = 0;
let priorityQueue = [[source, 0]];
while (priorityQueue.length > 0) {
let [node, dist] = priorityQueue.shift();
if (dist > distance[node]) continue;
for (let neighbor in graph[node]) {
let newDist = dist + graph[node][neighbor];
if (newDist < distance[neighbor]) {
distance[neighbor] = newDist;
priorityQueue.push([neighbor, newDist]);
priorityQueue.sort((a, b) => a[1] - b[1]);
}
}
}
return distance[destination] * crates;
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