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

Time Complexity: Minimum Spanning Tree Prim's Algorithm

O((n + m) log n)

Understanding Time Complexity

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

How does the algorithm's work increase when we add more nodes and edges?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn prim_mst(graph: &Vec>) -> i32 {
    let n = graph.len();
    let mut visited = vec![false; n];
    let mut min_heap = std::collections::BinaryHeap::new();
    min_heap.push((0, 0));
    let mut total_cost = 0;

    while let Some((cost, u)) = min_heap.pop() {
        if visited[u] { continue; }
        visited[u] = true;
        total_cost += -cost;
        for &(v, w) in &graph[u] {
            if !visited[v] {
                min_heap.push((-w, v));
            }
        }
    }
    total_cost
}
    

This code finds the minimum spanning tree cost of a graph using Prim's algorithm with a priority queue.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop runs until all nodes are visited, and inside it, edges connected to the current node are checked.
• How many times: The loop runs up to n times for nodes, and the inner for-loop runs for all edges connected to those nodes, totaling m edges.

How Execution Grows With Input

As the number of nodes (n) and edges (m) increase, the algorithm processes each edge and node to build the tree.

Input Size (n, m)	Approx. Operations
10 nodes, 15 edges	About 10 pops and 15 edge checks
100 nodes, 200 edges	About 100 pops and 200 edge checks
1000 nodes, 5000 edges	About 1000 pops and 5000 edge checks

Pattern observation: The work grows roughly with the number of nodes plus edges, so more nodes and edges mean more operations.

Final Time Complexity

Time Complexity: O((n + m) log n)

This means the time grows with the number of nodes and edges, multiplied by the cost of managing the priority queue.

Common Mistake

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

[OK] Correct: The algorithm uses a priority queue that requires extra time to add and remove edges, so the log factor from the queue operations matters.

Interview Connect

Understanding this time complexity helps you explain how efficient your solution is when working with graphs, a common topic in interviews.

Self-Check

"What if we used a simple array instead of a priority queue to pick the next edge? How would the time complexity change?"

Practice

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

easy

A. To detect cycles in a graph

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

C. To find the shortest path between two vertices

D. To sort vertices in topological order

Solution

1. Step 1: Understand Prim's Algorithm purpose

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

2. Step 2: Compare with other graph algorithms

   Shortest path algorithms find paths, cycle detection finds loops, and topological sort orders vertices, which are different goals.

3. Final Answer:

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

4. Quick Check:

   Prim's Algorithm = Minimum Spanning Tree [OK]

Hint: Prim's builds minimum spanning tree by adding cheapest edges [OK]

Common Mistakes:

• Confusing Prim's with shortest path algorithms
• Thinking it detects cycles
• Mixing it with topological sorting

2. Which Rust data structure is best suited to efficiently pick the next minimum edge in Prim's Algorithm?

easy

A. Vec (Vector)

B. LinkedList

C. HashMap

D. BinaryHeap (Min-Heap)

Solution

1. Step 1: Identify operation needed in Prim's Algorithm

   Prim's needs to repeatedly select the edge with the smallest weight efficiently.

2. Step 2: Choose data structure for minimum extraction

   A BinaryHeap (min-heap) allows quick access to the smallest element, unlike Vec or LinkedList which require scanning.

3. Final Answer:

   BinaryHeap (Min-Heap) -> Option D

4. Quick Check:

   Min-Heap = Efficient minimum edge selection [OK]

Hint: Use min-heap to get smallest edge fast in Prim's [OK]

Common Mistakes:

• Using Vec which is slow for minimum extraction
• Choosing HashMap which is for key-value, not min extraction
• Using LinkedList which is inefficient for this

3. What is the output of the following Rust code snippet implementing Prim's Algorithm on a graph with edges [(0,1,2), (1,2,3), (0,2,1)] starting from vertex 0?

let edges = vec![(0,1,2), (1,2,3), (0,2,1)];
// Prim's MST total weight calculation
// Output the total weight of MST

medium

A. 3

B. 5

C. 4

D. 6

Solution

1. Step 1: List edges and weights

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

2. Step 2: Build MST using Prim's Algorithm

   Start at 0, pick smallest edge (0-2)=1, then pick next smallest connecting edge (0-1)=2. Total weight = 1 + 2 = 3.

3. Step 3: Check if all vertices connected

   Vertices 0,1,2 connected with edges (0-2) and (0-1). Edge (1-2) with weight 3 is not needed.

4. Final Answer:

   3 -> Option A

5. Quick Check:

   MST weight = 1 + 2 = 3 [OK]

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

Common Mistakes:

• Adding all edges instead of MST edges
• Choosing heavier edges first
• Not starting from given start vertex

4. Identify the bug in this Rust snippet implementing Prim's Algorithm:

let mut visited = vec![false; n];
let mut heap = BinaryHeap::new();
heap.push(Reverse((0, 0)));
while let Some(Reverse((cost, u))) = heap.pop() {
    if visited[u] { continue; }
    visited[u] = true;
    for &(v, w) in &graph[u] {
        if !visited[v] {
            heap.push(Reverse((w, v)));
        }
    }
}

medium

A. The heap should be a max-heap, not min-heap

B. The heap should store (cost, vertex) as (u, cost) instead

C. The visited check is missing before pushing neighbors

D. The graph adjacency list should store (vertex, weight) not (weight, vertex)

Solution

1. Step 1: Analyze visited check placement

   Currently, neighbors are pushed without checking if visited, causing duplicates in heap.

2. Step 2: Fix by checking visited before pushing neighbors

   Adding a visited check before pushing neighbors avoids unnecessary heap entries and improves efficiency.

3. Final Answer:

   The visited check is missing before pushing neighbors -> Option C

4. Quick Check:

   Check visited before pushing neighbors to heap [OK]

Hint: Check visited before pushing neighbors to avoid duplicates [OK]

Common Mistakes:

• Ignoring visited check before pushing neighbors
• Confusing min-heap with max-heap
• Swapping vertex and weight in tuples

5. Given a weighted undirected graph with vertices 0 to 4 and edges:

(0,1,10), (0,2,6), (0,3,5), (1,3,15), (2,3,4), (3,4,2)

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

hard

A. 21

B. 17

C. 27

D. 22

Solution

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

   Edges from 0: (0-3)=5, (0-2)=6, (0-1)=10. Pick (0-3)=5 first.

2. Step 2: Add edges connecting new vertices with minimum weight

   From visited {0,3}, edges: (3-4)=2, (3-2)=4, (0-2)=6, (0-1)=10. Pick (3-4)=2 next.

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

   Visited {0,3,4}, edges: (3-2)=4, (0-2)=6, (0-1)=10. Pick (3-2)=4 next.

4. Step 4: Add last vertex 1 with minimum edge

   Visited {0,2,3,4}, edges: (0-1)=10, (1-3)=15. Pick (0-1)=10.

5. Step 5: Calculate total weight

   Total = 5 + 2 + 4 + 10 = 21.

6. Final Answer:

   21 -> Option A

7. Quick Check:

   Sum of MST edges = 21 [OK]

Hint: Add smallest edge connecting new vertex each step [OK]

Common Mistakes:

• Choosing heavier edges before lighter ones
• Not updating visited set correctly
• Adding edges that create cycles