Shortest Path in DAG Using Topological Order in DSA C++ - Time & Space Complexity

Time Complexity: Shortest Path in DAG Using Topological Order

O(V + E)

Understanding Time Complexity

We want to understand how the time needed to find the shortest path in a Directed Acyclic Graph (DAG) grows as the graph gets bigger.

Specifically, how the steps increase when using topological order to find shortest paths.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

// Find shortest paths in DAG using topological order
void shortestPath(int src, vector> adj[], int V) {
    vector dist(V, INT_MAX);
    dist[src] = 0;
    vector topo = topologicalSort(V, adj);
    for (int u : topo) {
        if (dist[u] != INT_MAX) {
            for (auto &edge : adj[u]) {
                int v = edge.first, w = edge.second;
                if (dist[u] + w < dist[v])
                    dist[v] = dist[u] + w;
            }
        }
    }
}
    

This code finds shortest distances from a source node to all others in a DAG by first sorting nodes topologically, then relaxing edges in order.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The nested loops: outer loop over all vertices in topological order, inner loop over all edges from each vertex.
• How many times: Outer loop runs V times (number of vertices), inner loop runs once per edge from each vertex, totaling E times (number of edges).

How Execution Grows With Input

As the graph grows, the number of vertices (V) and edges (E) increase, affecting the steps.

Input Size (V, E)	Approx. Operations
10 vertices, 15 edges	About 10 + 15 = 25 steps
100 vertices, 200 edges	About 100 + 200 = 300 steps
1000 vertices, 5000 edges	About 1000 + 5000 = 6000 steps

Pattern observation: The total steps grow roughly in proportion to the sum of vertices and edges.

Final Time Complexity

Time Complexity: O(V + E)

This means the time needed grows linearly with the number of vertices plus the number of edges in the graph.

Common Mistake

[X] Wrong: "The time depends on the number of paths or cycles in the graph."

[OK] Correct: Since the graph is acyclic, there are no cycles, and the algorithm processes each vertex and edge only once, so paths or cycles do not increase time.

Interview Connect

Understanding this time complexity helps you explain why topological sorting is efficient for shortest paths in DAGs, a common interview topic.

Self-Check

"What if the graph had cycles? How would the time complexity and approach change?"

Practice

1. What is the main advantage of using topological order to find the shortest path in a Directed Acyclic Graph (DAG)?

easy

A. It allows processing vertices in a linear order without revisiting nodes.

B. It guarantees the graph has no cycles.

C. It finds the longest path instead of the shortest path.

D. It works only for undirected graphs.

Solution

1. Step 1: Understand topological order in DAG

   Topological order arranges vertices so that all edges go from earlier to later vertices, allowing linear processing.

2. Step 2: Use topological order for shortest path

   Processing vertices in this order ensures each vertex is relaxed once, avoiding revisits and cycles.

3. Final Answer:

   It allows processing vertices in a linear order without revisiting nodes. -> Option A

4. Quick Check:

   Topological order = linear processing without revisits [OK]

Hint: Topological order processes nodes once in DAG [OK]

Common Mistakes:

• Thinking it works for graphs with cycles
• Confusing shortest path with longest path
• Assuming it works for undirected graphs

2. Which of the following C++ code snippets correctly initializes the distance array for shortest path calculation in a DAG with n vertices, assuming source vertex is 0?

easy

A. std::vector<int> dist(n, 0); dist[0] = INT_MAX;

B. std::vector<int> dist(n, INT_MAX); dist[0] = 0;

C. int dist[n]; for(int i=0; i

D. std::vector<int> dist(n); dist[0] = 0;

Solution

1. Step 1: Initialize distances with maximum value

   Distances should start as infinity (INT_MAX) except source which is 0.

2. Step 2: Check code correctness

   std::vector<int> dist(n, INT_MAX); dist[0] = 0; correctly sets all distances to INT_MAX and source distance to 0.

3. Final Answer:

   std::vector<int> dist(n, INT_MAX); dist[0] = 0; -> Option B

4. Quick Check:

   Distance init = INT_MAX except source 0 [OK]

Hint: Initialize all distances to INT_MAX except source 0 [OK]

Common Mistakes:

• Setting source distance to INT_MAX
• Initializing distances to zero
• Not using vector or uninitialized array

3. Consider the following DAG with edges and weights:
0 -> 1 (2), 0 -> 2 (4), 1 -> 2 (1), 1 -> 3 (7), 2 -> 3 (3)
Using topological order starting from vertex 0, what is the shortest distance to vertex 3?

medium

A. 6

B. 7

C. 9

D. 10

Solution

1. Step 1: Find topological order and initialize distances

   Topological order: 0, 1, 2, 3. Initialize dist[0]=0, others = infinity.

2. Step 2: Relax edges in topological order

   From 0: dist[1]=2, dist[2]=4
   From 1: dist[2]=min(4, 2+1=3)=3, dist[3]=2+7=9
   From 2: dist[3]=min(9, 3+3=6)=6

3. Final Answer:

   6 -> Option A

4. Quick Check:

   Shortest path to 3 = 6 [OK]

Hint: Relax edges in topological order to update shortest distances [OK]

Common Mistakes:

• Not updating dist[2] to 3 via vertex 1
• Ignoring relaxation from vertex 2 to 3
• Using wrong topological order

4. The following code snippet attempts to find shortest paths in a DAG but produces incorrect results. What is the bug?

std::vector<int> dist(n, INT_MAX);
dist[source] = 0;
for (int u : topoOrder) {
  for (auto &edge : graph[u]) {
    int v = edge.first, w = edge.second;
    if (dist[v] > dist[u] + w) {
      dist[v] = dist[u] + w;
    }
  }
}

medium

A. The distance array is not initialized properly.

B. The graph edges are not relaxed in topological order.

C. The source vertex distance is not set to zero.

D. The code does not check if dist[u] is INT_MAX before adding w.

Solution

1. Step 1: Analyze relaxation condition

   The code adds w to dist[u] without checking if dist[u] is INT_MAX, causing overflow.

2. Step 2: Identify missing check

   Before relaxing edges, must check dist[u] != INT_MAX to avoid invalid addition.

3. Final Answer:

   The code does not check if dist[u] is INT_MAX before adding w. -> Option D

4. Quick Check:

   Check dist[u] != INT_MAX before relaxation [OK]

Hint: Always check dist[u] != INT_MAX before adding edge weight [OK]

Common Mistakes:

• Assuming initialization is wrong
• Thinking topological order is incorrect
• Forgetting to set source distance

5. Given a DAG with vertices 0 to 4 and edges:
0->1(3), 0->2(6), 1->2(4), 1->3(4), 2->3(8), 3->4(2)
Using topological order, what is the shortest path distance from vertex 0 to vertex 4?

hard

A. 13

B. 11

C. 9

D. 14

Solution

1. Step 1: Determine topological order and initialize distances

   Topological order: 0,1,2,3,4. Initialize dist[0]=0, others = infinity.

2. Step 2: Relax edges in order

   From 0: dist[1]=3, dist[2]=6
   From 1: dist[2]=min(6,3+4=7)=6, dist[3]=3+4=7
   From 2: dist[3]=min(7,6+8=14)=7
   From 3: dist[4]=7+2=9

3. Final Answer:

   9 -> Option C

4. Quick Check:

   Shortest path to 4 = 9 [OK]

Hint: Relax edges stepwise in topological order to find shortest path [OK]

Common Mistakes:

• Not updating dist[3] correctly
• Choosing longer path via vertex 2
• Misordering topological sort