Minimum Spanning Tree Prim's Algorithm in DSA Go - Time & Space Complexity

Time Complexity: Minimum Spanning Tree Prim's Algorithm

O(n²)

Understanding Time Complexity

We want to understand how the time needed to find a minimum spanning tree using Prim's algorithm changes as the graph grows.

Specifically, how does the number of steps grow when the graph has more nodes and edges?

Scenario Under Consideration

Analyze the time complexity of the following Prim's algorithm code snippet.

func primMST(graph [][]int) int {
    n := len(graph)
    visited := make([]bool, n)
    dist := make([]int, n)
    for i := range dist {
        dist[i] = 1_000_000_000
    }
    dist[0] = 0

    totalWeight := 0
    for i := 0; i < n; i++ {
        u := -1
        for v := 0; v < n; v++ {
            if !visited[v] && (u == -1 || dist[v] < dist[u]) {
                u = v
            }
        }
        visited[u] = true
        totalWeight += dist[u]

        for v := 0; v < n; v++ {
            if graph[u][v] != 0 && !visited[v] && graph[u][v] < dist[v] {
                dist[v] = graph[u][v]
            }
        }
    }
    return totalWeight
}
    

This code finds the minimum spanning tree weight of a graph using Prim's algorithm with an adjacency matrix.

Identify Repeating Operations

Look at the loops that repeat work:

• Primary operation: The outer loop runs once for each node (n times).
• Inside it: There are two inner loops each running n times.
• Dominant work: The nested loops checking all nodes to find the minimum distance and update distances.

How Execution Grows With Input

For each of the n nodes, we do two loops over n nodes, so roughly n * (n + n) = 2n² operations.

Input Size (n)	Approx. Operations
10	About 200
100	About 20,000
1000	About 2,000,000

Pattern observation: The operations grow roughly with the square of the number of nodes.

Final Time Complexity

Time Complexity: O(n²)

This means if the number of nodes doubles, the time to run the algorithm roughly quadruples.

Common Mistake

[X] Wrong: "Prim's algorithm always runs in linear time O(n)."

[OK] Correct: Because Prim's algorithm checks all nodes repeatedly, the time grows with the square of nodes when using a simple adjacency matrix and no advanced data structures.

Interview Connect

Understanding how Prim's algorithm scales helps you explain your choices and shows you know how to handle graphs efficiently in real problems.

Self-Check

"What if we used a priority queue (min-heap) to select the next node instead of scanning all nodes? How would the time complexity change?"

Practice

1. What is the main goal of Prim's Algorithm in graph theory?

easy

A. To find the minimum spanning tree connecting all vertices with minimum total edge weight

B. To find the shortest path between two vertices

C. To detect cycles in a graph

D. To sort the vertices in topological order

Solution

1. Step 1: Understand Prim's Algorithm purpose

   Prim's Algorithm builds a tree connecting all vertices with the smallest sum of edge weights.

2. Step 2: Compare with other graph algorithms

   Shortest path and cycle detection are different problems; Prim's focuses on minimum spanning tree.

3. Final Answer:

   To find the minimum spanning tree connecting all vertices with minimum total edge weight -> Option A

4. Quick Check:

   Prim's Algorithm = Minimum Spanning Tree [OK]

Hint: Prim's always finds minimum spanning tree, not shortest path [OK]

Common Mistakes:

• Confusing Prim's with Dijkstra's algorithm
• Thinking it finds shortest paths
• Assuming it detects cycles

2. Which of the following Go code snippets correctly initializes a priority queue for Prim's Algorithm?

easy

A. pq := new(PriorityQueue); pq = nil

B. pq := PriorityQueue{}; pq.Init()

C. pq := make(PriorityQueue, 0); heap.Init(&pq)

D. pq := []int{}; sort.Ints(pq)

Solution

1. Step 1: Check correct priority queue initialization in Go

   PriorityQueue is usually a slice; make with length 0 and call heap.Init(&pq) to initialize.

2. Step 2: Analyze other options

   pq := PriorityQueue{}; pq.Init() misses heap.Init call; C sets pq to nil; D uses slice but sorts, not heap.

3. Final Answer:

   pq := make(PriorityQueue, 0); heap.Init(&pq) -> Option C

4. Quick Check:

   Use make + heap.Init for priority queue [OK]

Hint: Use make and heap.Init to prepare priority queue [OK]

Common Mistakes:

• Forgetting to call heap.Init
• Setting priority queue to nil
• Sorting slice instead of using heap

3. Given the following Go code snippet implementing Prim's Algorithm on a graph with edges:

edges := [][]int{{0,1,2},{1,2,3},{0,2,1}}
// Graph with 3 vertices and edges with weights
// What is the total weight of the minimum spanning tree printed by this code?
func primMST() int {
  n := 3
  graph := make([][]Edge, n)
  for _, e := range edges {
    graph[e[0]] = append(graph[e[0]], Edge{to: e[1], weight: e[2]})
    graph[e[1]] = append(graph[e[1]], Edge{to: e[0], weight: e[2]})
  }
  visited := make([]bool, n)
  pq := &PriorityQueue{}
  heap.Init(pq)
  heap.Push(pq, &Item{vertex: 0, weight: 0})
  total := 0
  for pq.Len() > 0 {
    item := heap.Pop(pq).(*Item)
    if visited[item.vertex] {
      continue
    }
    visited[item.vertex] = true
    total += item.weight
    for _, e := range graph[item.vertex] {
      if !visited[e.to] {
        heap.Push(pq, &Item{vertex: e.to, weight: e.weight})
      }
    }
  }
  return total
}

medium

A. 3

B. 4

C. 5

D. 6

Solution

1. Step 1: Build the graph and list edges

   Edges: (0-1)=2, (1-2)=3, (0-2)=1. Start from vertex 0 with weight 0.

2. Step 2: Run Prim's Algorithm step-by-step

   Pick 0 (weight 0), add edges (0-1:2), (0-2:1). Next pick edge with smallest weight: (0-2:1), total=1. Then pick (0-1:2), total=3. Vertex 1 and 2 visited, done.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Sum edges 1 + 2 = 3 [OK]

Hint: Pick smallest edges connecting new vertices first [OK]

Common Mistakes:

• Adding all edges instead of minimum spanning tree
• Starting from wrong vertex
• Counting edges twice

4. The following Go code snippet for Prim's Algorithm has a bug. What is the error?

func primBuggy() int {
  n := 3
  graph := make([][]Edge, n)
  // edges omitted for brevity
  visited := make([]bool, n)
  pq := &PriorityQueue{}
  heap.Init(pq)
  heap.Push(pq, &Item{vertex: 0, weight: 0})
  total := 0
  for pq.Len() > 0 {
    item := heap.Pop(pq).(*Item)
    if visited[item.vertex] {
      total += item.weight
      continue
    }
    visited[item.vertex] = true
    for _, e := range graph[item.vertex] {
      if !visited[e.to] {
        heap.Push(pq, &Item{vertex: e.to, weight: e.weight})
      }
    }
  }
  return total
}

medium

A. Not marking vertices as visited

B. Adding edge weight after skipping visited vertex causes incorrect total

C. Not initializing the priority queue with heap.Init

D. Pushing edges with wrong weight values

Solution

1. Step 1: Identify where total is updated

   Total is increased only when visited[item.vertex] is true, which means vertex already visited.

2. Step 2: Understand why this is wrong

   Weights should be added only when vertex is visited first time, not when skipping duplicates.

3. Final Answer:

   Adding edge weight after skipping visited vertex causes incorrect total -> Option B

4. Quick Check:

   Update total only on first visit [OK]

Hint: Add weight only when vertex visited first time [OK]

Common Mistakes:

• Adding weights multiple times for same vertex
• Forgetting to call heap.Init
• Not marking visited vertices

5. You have a weighted undirected graph with 5 vertices and edges:

(0-1, 2), (0-3, 6), (1-2, 3), (1-3, 8), (1-4, 5), (2-4, 7), (3-4, 9)

Using Prim's Algorithm starting from vertex 0, what is the total weight of the minimum spanning tree?

hard

A. 21

B. 19

C. 20

D. 16

Solution

1. Step 1: Start from vertex 0 and pick smallest edges stepwise

   Start at 0: edges (0-1:2), (0-3:6). Pick (0-1:2), total=2.

2. Step 2: Add edges from visited vertices and pick minimum

   Visited: 0,1. Edges: (1-2:3), (1-3:8), (1-4:5), (0-3:6). Pick (1-2:3), total=5.

3. Step 3: Continue adding minimum edges connecting new vertices

   Visited: 0,1,2. Edges: (1-3:8), (1-4:5), (0-3:6), (2-4:7). Pick (1-4:5), total=10.

4. Step 4: Add edges from new vertex 4

   Visited: 0,1,2,4. Edges: (1-3:8), (0-3:6), (2-4:7), (3-4:9). Pick (0-3:6), total=16.

5. Step 5: Add last vertex 3

   Visited all vertices. Total weight = 16.

6. Final Answer:

   16 -> Option D

7. Quick Check:

   Sum edges 2+3+5+6 = 16 [OK]

Hint: Always pick smallest edge connecting visited to unvisited [OK]

Common Mistakes:

• Choosing edges with larger weights first
• Not updating visited vertices correctly
• Counting edges twice