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

Time Complexity: Minimum Spanning Tree Prim's Algorithm

O(V^2)

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.

How does the algorithm's work increase when the number of nodes and edges increases?

Scenario Under Consideration

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

void primMST(int graph[V][V]) {
    int key[V];
    bool mstSet[V];
    for (int i = 0; i < V; i++) {
        key[i] = INT_MAX;
        mstSet[i] = false;
    }
    key[0] = 0;
    for (int count = 0; count < V - 1; count++) {
        int u = minKey(key, mstSet);
        mstSet[u] = true;
        for (int v = 0; v < V; v++) {
            if (graph[u][v] && !mstSet[v] && graph[u][v] < key[v])
                key[v] = graph[u][v];
        }
    }
}
    

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

Identify Repeating Operations

Look at the loops and repeated steps:

• Primary operation: The outer loop runs once for each vertex (V times).
• Inner loop: For each vertex chosen, it checks all vertices to update keys (V times).
• Dominant operation: The nested loops together cause about V x V = V² checks.

How Execution Grows With Input

As the number of vertices (n) grows, the work grows roughly with the square of n.

Input Size (n)	Approx. Operations
10	About 100 checks
100	About 10,000 checks
1000	About 1,000,000 checks

Pattern observation: Doubling the number of vertices roughly quadruples the work.

Final Time Complexity

Time Complexity: O(V^2)

This means the time needed grows roughly with the square of the number of vertices in the graph.

Common Mistake

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

[OK] Correct: Because it checks all vertices repeatedly, the time grows with V squared, not just V.

Interview Connect

Understanding how Prim's algorithm scales helps you explain your choices and shows you know how algorithms behave with bigger data.

Self-Check

"What if we used a priority queue (min-heap) instead of scanning all vertices to find the minimum key? 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 shortest path between two nodes

B. To detect cycles in a graph

C. To find the minimum spanning tree of a connected weighted graph

D. To sort the vertices of a graph

Solution

1. Step 1: Understand Prim's Algorithm purpose

   Prim's Algorithm is designed to find a tree that connects all vertices with the minimum total edge weight.

2. Step 2: Differentiate from other graph algorithms

   It is not used for shortest path, cycle detection, or sorting vertices, which are different problems.

3. Final Answer:

   To find the minimum spanning tree of a connected weighted graph -> Option C

4. Quick Check:

   Prim's Algorithm = Minimum Spanning Tree [OK]

Hint: Prim's always builds a tree with minimum total edge weight [OK]

Common Mistakes:

• Confusing Prim's with shortest path algorithms
• Thinking it detects cycles
• Assuming it sorts nodes

2. Which of the following is the correct way to initialize the key array in Prim's Algorithm in C++?

easy

A. int key[V]; for(int i=0; i

B. int key[V] = {0};

C. int key[V]; for(int i=0; i

D. int key[V]; for(int i=0; i

Solution

1. Step 1: Understand key array role

   The key array stores minimum edge weights to include each vertex; initially, all are set to a very large value (infinity).

2. Step 2: Check initialization options

   Setting key[i] = INT_MAX correctly represents infinity; other options set to 0 or -1 are incorrect for this purpose.

3. Final Answer:

   int key[V]; for(int i=0; i -> Option A

4. Quick Check:

   Initialize keys with INT_MAX for minimum edge tracking [OK]

Hint: Initialize keys with INT_MAX to represent infinity [OK]

Common Mistakes:

• Initializing keys with 0 instead of INT_MAX
• Using negative values incorrectly
• Not initializing keys at all

3. Consider the following graph with 4 vertices and edges with weights:

0 - 1 : 10
0 - 2 : 6
0 - 3 : 5
1 - 3 : 15
2 - 3 : 4

What is the total weight of the minimum spanning tree found by Prim's Algorithm starting from vertex 0?

medium

A. 21

B. 19

C. 20

D. 25

Solution

1. Step 1: Start from vertex 0 and pick minimum edges

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

2. Step 2: Add next minimum edge connecting new vertices

   From vertices {0,3}, edges are (3-2)=4, (3-1)=15, (0-2)=6, (0-1)=10. Pick (3-2)=4 next.

3. Step 3: Add last vertex with minimum edge

   From {0,3,2}, edges to 1 are (0-1)=10 and (3-1)=15. Pick (0-1)=10.

4. Step 4: Calculate total weight

   Total = 5 + 4 + 10 = 19.

5. Final Answer:

   19 -> Option B

6. Quick Check:

   Sum of chosen edges = 19 [OK]

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

Common Mistakes:

• Choosing edges with larger weights first
• Not updating the set of included vertices
• Adding edges that form cycles

4. The following C++ code snippet is intended to implement Prim's Algorithm but has a bug. What is the error?

const int V = 4;
int key[V] = {INT_MAX, INT_MAX, INT_MAX, INT_MAX};
bool mstSet[V] = {false, false, false, false};
key[0] = 0;
for (int count = 0; count < V - 1; count++) {
    int u = -1;
    for (int v = 0; v < V; v++) {
        if (!mstSet[v] && (u == -1 || key[v] < key[u]))
            u = v;
    }
    mstSet[u] = true;
    for (int v = 0; v < V; v++) {
        if (graph[u][v] && !mstSet[v] && graph[u][v] < key[v])
            key[v] = graph[u][v];
    }
}

medium

A. The condition 'graph[u][v]' should check for non-zero edges

B. The loop should run from 0 to V instead of V - 1

C. The key array is not initialized with INT_MAX

D. The variable 'u' is initialized incorrectly inside the loop

Solution

1. Step 1: Analyze the neighbor update condition

   The condition if(graph[u][v] && !mstSet[v] && graph[u][v] < key[v]) uses graph[u][v] as a boolean check (true if non-zero).

2. Step 2: Identify the bug

   This implicit check assumes positive weights and treats 0 as no edge. Standard implementations explicitly use graph[u][v] != 0 to clearly indicate checking for edge existence (non-zero weight).

3. Step 3: Rule out other options

   A: 'u' is correctly initialized to -1 inside each iteration. B: V-1 iterations build the tree correctly. C: key array is initialized with INT_MAX.

4. Final Answer:

   The condition 'graph[u][v]' should check for non-zero edges -> Option A

5. Quick Check:

   Edge check: graph[u][v] != 0 [OK]

Hint: Check edge existence condition carefully for zero weights [OK]

Common Mistakes:

• Misinitializing 'u' outside the loop
• Incorrect loop range
• Not initializing key array properly

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

0-1: 3, 0-3: 1, 1-2: 1, 1-3: 3, 1-4: 10, 2-4: 5, 3-4: 6, 4-5: 2

If Prim's Algorithm starts from vertex 0, which edge will NOT be included in the minimum spanning tree?

hard

A. Edge 1-2 with weight 1

B. Edge 0-1 with weight 3

C. Edge 4-5 with weight 2

D. Edge 3-4 with weight 6

Solution

1. Step 1: Build MST step-by-step from vertex 0

   Start at 0: edges (0-3)=1, (0-1)=3. Pick (0-3)=1 first.

2. Step 2: Add next minimum edges connecting new vertices

   From {0,3}, edges: (3-1)=3, (3-4)=6, (0-1)=3. Pick (0-1)=3 or (3-1)=3 (same weight). Pick (0-1)=3.

3. Step 3: Continue adding minimum edges

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

4. Step 4: Add edge to vertex 4

   From {0,1,2,3}, edges: (2-4)=5, (3-4)=6, (1-4)=10. Pick (2-4)=5.

5. Step 5: Add edge to vertex 5

   From {0,1,2,3,4}, edge: (4-5)=2.

6. Step 6: Identify excluded edge

   Edge (3-4)=6 is not chosen because (2-4)=5 is cheaper and connects 4 earlier.

7. Final Answer:

   Edge 3-4 with weight 6 -> Option D

8. Quick Check:

   Exclude heavier edge when cheaper path exists [OK]

Hint: Always pick smallest edge connecting new vertex, exclude heavier alternatives [OK]

Common Mistakes:

• Including heavier edges when cheaper ones exist
• Not updating candidate edges after each addition
• Assuming all edges connected to MST are included