Minimum Spanning Tree Kruskal's Algorithm in DSA Python - Time & Space Complexity

Time Complexity: Minimum Spanning Tree Kruskal's Algorithm

O(m log m)

Understanding Time Complexity

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

How does the number of edges and vertices affect the work done?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

class DisjointSet:
    def __init__(self, n):
        self.parent = list(range(n))
        self.rank = [0] * n

    def find(self, x):
        if self.parent[x] != x:
            self.parent[x] = self.find(self.parent[x])
        return self.parent[x]

    def union(self, x, y):
        rootX = self.find(x)
        rootY = self.find(y)
        if rootX != rootY:
            if self.rank[rootX] < self.rank[rootY]:
                self.parent[rootX] = rootY
            elif self.rank[rootX] > self.rank[rootY]:
                self.parent[rootY] = rootX
            else:
                self.parent[rootY] = rootX
                self.rank[rootX] += 1


def kruskal(edges, n):
    edges.sort(key=lambda x: x[2])  # sort by weight
    ds = DisjointSet(n)
    mst = []
    for u, v, w in edges:
        if ds.find(u) != ds.find(v):
            ds.union(u, v)
            mst.append((u, v, w))
    return mst

This code finds the minimum spanning tree of a graph with n vertices and edges given as (u, v, weight).

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Sorting all edges by weight.
• How many times: Sorting runs once on all edges, which is m edges.
• Secondary operation: For each edge, find and union operations on disjoint sets.
• How many times: Up to m times, once per edge.

How Execution Grows With Input

As the number of edges (m) and vertices (n) grow, sorting edges takes more time, and union-find operations also increase.

Input Size (n vertices, m edges)	Approx. Operations
10, 15	Sorting ~15 log 15, Union-Find ~15
100, 500	Sorting ~500 log 500, Union-Find ~500
1000, 5000	Sorting ~5000 log 5000, Union-Find ~5000

Pattern observation: Sorting dominates and grows a bit faster than linearly with edges; union-find grows linearly with edges.

Final Time Complexity

Time Complexity: O(m log m)

This means the time mainly depends on sorting the edges, which grows a bit faster than the number of edges.

Common Mistake

[X] Wrong: "The union-find operations take as much time as sorting."

[OK] Correct: Union-find operations are almost constant time on average due to path compression and union by rank, so sorting edges is the real time cost.

Interview Connect

Understanding Kruskal's time complexity helps you explain how efficient your solution is and shows you can analyze algorithms beyond just coding them.

Self-Check

"What if the edges were already sorted? How would the time complexity change?"

Practice

1. What is the main goal of Kruskal's algorithm in graph theory?

easy

A. To find the shortest path between two nodes

B. To find the minimum spanning tree of a graph

C. To detect cycles in a graph

D. To find the maximum flow in a network

Solution

1. Step 1: Understand Kruskal's algorithm purpose

   Kruskal's algorithm is designed to find a subset of edges that connects all vertices with the minimum total edge weight.

2. Step 2: Compare with other graph problems

   Shortest path, cycle detection, and max flow are different problems with different algorithms.

3. Final Answer:

   To find the minimum spanning tree of a graph -> Option B

4. Quick Check:

   Kruskal's algorithm = Minimum Spanning Tree [OK]

Hint: Kruskal always builds minimum spanning tree by edge weights [OK]

Common Mistakes:

• Confusing MST with shortest path
• Thinking it detects cycles only
• Mixing with max flow problems

2. Which of the following is the correct syntax to sort edges by weight in Python for Kruskal's algorithm?

easy

A. edges.sort(key=lambda x: x[2])

B. edges.sort(key=lambda x: x[1])

C. edges.sort()

D. edges.sort(key=lambda x: x[0])

Solution

1. Step 1: Identify edge structure

   Edges are usually stored as tuples (node1, node2, weight), so weight is at index 2.

2. Step 2: Use sort with key on weight

   Sorting by weight means sorting by the third element, so key=lambda x: x[2] is correct.

3. Final Answer:

   edges.sort(key=lambda x: x[2]) -> Option A

4. Quick Check:

   Sort edges by weight = key=lambda x: x[2] [OK]

Hint: Weight is usually third element in edge tuple [OK]

Common Mistakes:

• Sorting by wrong tuple index
• Using sort() without key when edges are tuples
• Confusing node indices with weight index

3. What is the output of the following code snippet that applies Kruskal's algorithm on a graph with edges [(0, 1, 10), (0, 2, 6), (0, 3, 5), (1, 3, 15), (2, 3, 4)]?

edges = [(0, 1, 10), (0, 2, 6), (0, 3, 5), (1, 3, 15), (2, 3, 4)]
edges.sort(key=lambda x: x[2])
parent = list(range(4))
def find(x):
    while parent[x] != x:
        x = parent[x]
    return x
def union(a, b):
    rootA = find(a)
    rootB = find(b)
    if rootA != rootB:
        parent[rootB] = rootA
        return True
    return False
mst = []
for u, v, w in edges:
    if union(u, v):
        mst.append((u, v, w))
print(mst)

medium

A. [(0, 2, 6), (0, 3, 5), (2, 3, 4)]

B. [(0, 3, 5), (2, 3, 4), (1, 3, 15)]

C. [(0, 1, 10), (0, 2, 6), (1, 3, 15)]

D. [(2, 3, 4), (0, 3, 5), (0, 1, 10)]

Solution

1. Step 1: Sort edges by weight

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

2. Step 2: Apply union-find to select edges without cycles

   Pick (2,3,4), union success; pick (0,3,5), union success; pick (0,2,6), union fails (cycle); pick (0,1,10), union success; stop as MST has 3 edges for 4 nodes.

3. Final Answer:

   [(2, 3, 4), (0, 3, 5), (0, 1, 10)] -> Option D

4. Quick Check:

   Edges chosen = minimum weights without cycles [OK]

Hint: Sort edges, pick smallest without cycle until MST complete [OK]

Common Mistakes:

• Not sorting edges correctly
• Adding edges that form cycles
• Stopping too early or too late

4. Identify the bug in this Kruskal's algorithm snippet and select the fix:

edges = [(0,1,1), (1,2,2), (0,2,3)]
parent = [0, 1, 2]
def find(x):
    if parent[x] == x:
        return x
    return find(parent[x])
def union(a, b):
    rootA = find(a)
    rootB = find(b)
    if rootA != rootB:
        parent[a] = b  # Bug here
        return True
    return False
mst = []
for u,v,w in sorted(edges, key=lambda x: x[2]):
    if union(u,v):
        mst.append((u,v,w))
print(mst)

medium

A. Change 'parent[a] = b' to 'parent[rootB] = rootA'

B. Change 'parent[a] = b' to 'parent[rootA] = rootB'

C. Change 'parent[a] = b' to 'parent[b] = a'

D. No bug, code is correct

Solution

1. Step 1: Analyze union function

   Union should link roots of sets, not original nodes directly.

2. Step 2: Fix parent assignment

   Assign parent of rootB to rootA to merge sets correctly: 'parent[rootB] = rootA'.

3. Final Answer:

   Change 'parent[a] = b' to 'parent[rootB] = rootA' -> Option A

4. Quick Check:

   Union sets by root parents, not nodes [OK]

Hint: Always union by root parents, not direct nodes [OK]

Common Mistakes:

• Assigning parent of node instead of root
• Confusing rootA and rootB in assignment
• Skipping path compression in find

5. Given a graph with vertices 0 to 4 and edges [(0,1,2), (0,3,6), (1,2,3), (1,3,8), (1,4,5), (2,4,7), (3,4,9)], what is the total weight of the minimum spanning tree found by Kruskal's algorithm?

hard

A. 19

B. 20

C. 16

D. 18

Solution

1. Step 1: Sort edges by weight

   Sorted edges: (0,1,2), (1,2,3), (1,4,5), (0,3,6), (2,4,7), (1,3,8), (3,4,9)

2. Step 2: Select edges without cycles until all vertices connected

   Pick (0,1,2), (1,2,3), (1,4,5), (0,3,6). Total weight = 2+3+5+6 = 16. But check if (2,4,7) creates cycle: yes, skip. (1,3,8) and (3,4,9) create cycles, skip.

3. Step 3: Verify MST completeness

   All 5 vertices connected with 4 edges, total weight 16.

4. Final Answer:

   16 -> Option C

5. Quick Check:

   MST total weight = sum of chosen edges [OK]

Hint: Sum smallest edges connecting all nodes without cycles [OK]

Common Mistakes:

• Including edges that form cycles
• Missing edges needed to connect all vertices
• Adding heavier edges unnecessarily