Weighted graphs in Data Structures Theory - Time & Space Complexity

Time Complexity: Weighted graphs

O(E)

Understanding Time Complexity

When working with weighted graphs, it is important to understand how the time to process the graph grows as the graph gets bigger.

We want to know how the number of steps changes when the graph has more nodes and edges.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function sumWeights(graph) {
  let total = 0;
  for (let node in graph) {
    for (let edge of graph[node]) {
      total += edge.weight;
    }
  }
  return total;
}
    

This code adds up all the weights of edges in a weighted graph represented as an adjacency list.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Adding edge weights inside the inner loop.
• How many times: Once for every edge in the graph.

How Execution Grows With Input

As the number of edges increases, the total steps increase roughly the same amount.

Input Size (Edges)	Approx. Operations
10	About 10 additions
100	About 100 additions
1000	About 1000 additions

Pattern observation: The work grows directly with the number of edges.

Final Time Complexity

Time Complexity: O(E)

This means the time to sum weights grows in direct proportion to the number of edges in the graph.

Common Mistake

[X] Wrong: "The time depends mostly on the number of nodes, not edges."

[OK] Correct: Because edges hold the weights, the code must look at each edge, so the number of edges controls the time.

Interview Connect

Understanding how time grows with edges and nodes helps you explain graph algorithms clearly and confidently in interviews.

Self-Check

"What if the graph was represented as an adjacency matrix instead of a list? How would the time complexity change?"

Practice

1. What does the weight on an edge in a weighted graph usually represent?

easy

A. The cost or distance between two connected points

B. The color of the edge

C. The number of vertices in the graph

D. The direction of the edge

Solution

1. Step 1: Understand the role of weights in graphs

   Weights on edges represent values like cost, distance, or time between two connected points (vertices).

2. Step 2: Differentiate weights from other graph properties

   Weights are not about color, number of vertices, or direction but about measurable values on edges.

3. Final Answer:

   The cost or distance between two connected points -> Option A

4. Quick Check:

   Weight = cost/distance [OK]

Hint: Weights show cost or distance between points [OK]

Common Mistakes:

• Confusing weight with edge color
• Thinking weight counts vertices
• Mixing weight with edge direction

2. Which of the following is the correct way to represent a weighted edge between vertices A and B with weight 5?

easy

A. (A, B, 5)

B. {A: B = 5}

C. [A, B, weight=5]

D. A - B : 5

Solution

1. Step 1: Recognize common weighted edge notation

   Weighted edges are often represented as tuples like (vertex1, vertex2, weight).

2. Step 2: Check each option's format

   (A, B, 5) uses tuple format (A, B, 5), which is standard. Others are incorrect syntax or informal.

3. Final Answer:

   (A, B, 5) -> Option A

4. Quick Check:

   Weighted edge = (vertex1, vertex2, weight) [OK]

Hint: Use tuple (A, B, weight) for weighted edges [OK]

Common Mistakes:

• Using incorrect symbols like braces or colons
• Confusing syntax with dictionaries
• Writing weight as a keyword inside list

3. Consider the weighted graph edges: (A, B, 3), (B, C, 4), (A, C, 10). What is the shortest path weight from A to C?

medium

A. 4

B. 10

C. 3

D. 7

Solution

1. Step 1: Identify possible paths from A to C

   Paths: Direct (A to C) with weight 10, or via B: A to B (3) + B to C (4).

2. Step 2: Calculate total weights for each path

   Direct path weight = 10; via B = 3 + 4 = 7.

3. Final Answer:

   7 -> Option D

4. Quick Check:

   Shortest path weight = 7 [OK]

Hint: Sum weights on all paths, pick smallest [OK]

Common Mistakes:

• Choosing direct edge without checking alternatives
• Adding weights incorrectly
• Ignoring intermediate vertices

4. Given the weighted graph edges: (X, Y, 2), (Y, Z, 5), (X, Z, 4), a student claims the shortest path from X to Z is 7 by going through Y. What is wrong with this claim?

medium

A. They confused vertices Y and Z

B. They ignored the direct edge from X to Z with weight 4

C. They added weights incorrectly; 2 + 5 is not 7

D. They assumed edges are unweighted

Solution

1. Step 1: Analyze the paths from X to Z

   Paths: Direct edge (X, Z) with weight 4, and path via Y with weights 2 + 5 = 7.

2. Step 2: Identify the shortest path

   The direct edge weight 4 is less than 7, so shortest path is direct, not via Y.

3. Final Answer:

   They ignored the direct edge from X to Z with weight 4 -> Option B

4. Quick Check:

   Shortest path uses smallest weight edge [OK]

Hint: Check all edges before choosing path [OK]

Common Mistakes:

• Ignoring direct edges
• Incorrectly adding weights
• Mixing up vertex names

5. You have a weighted graph representing cities connected by roads with distances. To find the cheapest route from city A to city D considering toll costs on roads, which approach is best?

hard

A. Select the path with the most edges to maximize tolls

B. Count the number of roads between cities ignoring weights

C. Use a shortest path algorithm like Dijkstra's considering weights as toll costs

D. Use a depth-first search without considering weights

Solution

1. Step 1: Understand the problem context

   We want the cheapest route considering toll costs, which are weights on edges.

2. Step 2: Choose an appropriate algorithm

   Dijkstra's algorithm finds shortest paths in weighted graphs by minimizing total weight (cost).

3. Final Answer:

   Use a shortest path algorithm like Dijkstra's considering weights as toll costs -> Option C

4. Quick Check:

   Weighted shortest path = Dijkstra's algorithm [OK]

Hint: Use Dijkstra's for weighted shortest path problems [OK]

Common Mistakes:

• Ignoring weights and counting edges only
• Using DFS which ignores weights
• Choosing longest path mistakenly