Number of Operations to Make Network Connected

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Steps

setup

Initialize Union-Find

Create union-find structure with each node as its own parent and count set to 4 (number of nodes).

💡 This sets up the initial state where no computers are connected, so count equals n.

Line:

uf = UnionFind(n)

💡 Initially, each node is isolated, so count = n.

compare

Check if Enough Connections

Check if the number of connections is less than n-1. If yes, return -1 immediately.

💡 This is a quick feasibility check: at least n-1 cables are needed to connect n nodes.

Line:

if len(connections) < n - 1:
    return -1

💡 Enough cables exist to possibly connect all nodes.

traverse

Start Processing Connection [0,1]

Begin processing the first connection between nodes 0 and 1 by preparing to perform union operation.

💡 We will check if nodes 0 and 1 belong to different sets to decide if union is needed.

Line:

uf.union(0, 1)

💡 Union is only performed if nodes are in different sets.

traverse

Perform Union on Connection [0,1]

Since nodes 0 and 1 are in different sets, merge them and update the count.

💡 Union merges two sets if they are not already connected, reducing the count of disconnected components.

Line:

uf.union(0, 1)

💡 Union operation successfully connects two previously separate components.

traverse

Start Processing Connection [0,2]

Begin processing the second connection between nodes 0 and 2 by preparing to perform union operation.

💡 We will check if nodes 0 and 2 belong to different sets to decide if union is needed.

Line:

uf.union(0, 2)

💡 Union is only performed if nodes are in different sets.

traverse

Perform Union on Connection [0,2]

Since nodes 0 and 2 are in different sets, merge them and update the count.

💡 Union merges sets containing 0 and 2, further reducing disconnected components count.

Line:

uf.union(0, 2)

💡 Another successful union reduces the number of disconnected sets.

traverse

Start Processing Connection [1,2]

Begin processing the third connection between nodes 1 and 2 by preparing to perform union operation.

💡 We will check if nodes 1 and 2 belong to different sets to decide if union is needed.

Line:

uf.union(1, 2)

💡 Union is only performed if nodes are in different sets.

traverse

Check Union on Connection [1,2]

Since nodes 1 and 2 share the same root, no union is performed and count remains the same.

💡 Union checks if nodes 1 and 2 are already connected to avoid redundant unions.

Line:

uf.union(1, 2)

💡 This connection is redundant and does not reduce the number of disconnected sets.

compare

Check if Network is Fully Connected

After processing all connections, check if count equals 1 (fully connected).

💡 If count is 1, no extra operations needed; otherwise, calculate how many cables to move.

Line:

if uf.count == 1:
    return 0

💡 There are still disconnected components that need to be connected.

reconstruct

Calculate Minimum Operations

Calculate the minimum number of operations needed as count - 1, which is 1.

💡 Each operation moves a cable to connect two disconnected components, reducing count by 1.

Line:

return uf.count - 1

💡 The number of operations equals the number of extra cables needed to connect all components.

reconstruct

Final Answer

Return the final answer 1, indicating one operation is needed to connect the network.

💡 This is the minimal number of cable moves required to connect all computers.

Line:

return uf.count - 1  # returns 1

💡 The algorithm correctly identifies the minimal operations needed.

class UnionFind:
    def __init__(self, n):
        self.parent = list(range(n))  # STEP 1
        self.rank = [0] * n          # STEP 1
        self.count = n               # STEP 1

    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)         # STEP 3,4,5
        rootY = self.find(y)         # STEP 3,4,5
        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
            self.count -= 1          # STEP 3,4
            return True
        return False

def makeConnected(n, connections):
    if len(connections) < n - 1:  # STEP 2
        return -1
    uf = UnionFind(n)              # STEP 1
    for u, v in connections:
        uf.union(u, v)             # STEP 3,4,5
        if uf.count == 1:          # STEP 6
            return 0
    return uf.count - 1            # STEP 7,8

if __name__ == '__main__':
    print(makeConnected(4, [[0,1],[0,2],[1,2]]))  # 1

📊

Number of Operations to Make Network Connected - Watch the Algorithm Execute, Step by Step

Watching each union operation and the count updates reveals how connectivity is tracked efficiently, which is hard to grasp from code alone.

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Edge: 0 → 1

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Key Takeaways

✓ Union-Find efficiently tracks connected components by merging sets and updating a count.

Reading code alone hides how union operations reduce disconnected sets step-by-step.

✓ Redundant connections do not reduce the count but are crucial as spare cables to connect isolated nodes.

Visualizing redundant edges clarifies why extra cables can be reused.

✓ The minimal operations needed equals the number of disconnected components minus one.

This insight is clearer when watching count decrement after each union.

Practice

1. Consider the following Python code implementing Union-Find with path compression and union by rank to find a redundant connection. Given the input edges [[1, 2], [1, 3], [2, 3]], what is the output of the function?

easy

A. [1, 2]

B. [1, 3]

C. [2, 3]

D. None (no redundant edge)

Solution

1. Step 1: Trace union operations for edges [1,2], [1,3], [2,3]

   Union(1,2) succeeds, Union(1,3) succeeds, Union(2,3) fails because 2 and 3 are already connected.

2. Step 2: Identify returned edge

   The first edge that causes union to fail is [2,3], so it is returned as redundant.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Union-Find detects cycle at edge [2,3] [OK]

Hint: Union fails when both nodes share the same root [OK]

Common Mistakes:

• Returning first edge instead of the one causing cycle
• Confusing union success/failure
• Off-by-one indexing errors

2. Consider the following Python code implementing the optimal solution for the problem. Given the input n = 4 and edges = [[3,1,2],[3,2,3],[1,1,3],[2,3,4]], what is the final return value of max_num_edges_to_remove_sorted(n, edges)?

easy

A. 3

B. 2

C. 1

D. -1

Solution

1. Step 1: Sort edges by type descending

   Edges sorted: type 3 edges first: [3,1,2], [3,2,3], then type 1: [1,1,3], then type 2: [2,3,4].

2. Step 2: Process edges with union-find

   Type 3 edges connect nodes 0-1 and 1-2 for both Alice and Bob, used=2. Type 1 edge connects 0-2 for Alice, but already connected, so no increment. Type 2 edge connects 2-3 for Bob, used=3 total. Both have count=1 (fully connected). Edges used=3, total edges=4, so removable=1.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Used edges=3, removable=4-3=1 -> careful: used=3, removable=1 [OK]

Hint: Count used edges after union-find merges [OK]

Common Mistakes:

• Off-by-one in indexing nodes
• Ignoring type 3 edges first
• Counting used edges incorrectly

3. Consider the following code snippet implementing the optimal approach for the smallest string with swaps problem. Given input string s = "dcab" and pairs [[0,3],[1,2]], what is the final returned string?

easy

A. "bacd"

B. "bacd"

C. "acbd"

D. "dcab"

Solution

1. Step 1: Perform unions on pairs

   Union(0,3) connects indices 0 and 3; Union(1,2) connects indices 1 and 2.

2. Step 2: Group indices and sort characters

   Group 1: indices [0,3] with chars ['d','b'] sorted to ['b','d']; Group 2: indices [1,2] with chars ['c','a'] sorted to ['a','c'].

3. Step 3: Assign sorted chars back to sorted indices

   Assign 'b' to index 0, 'd' to index 3; assign 'a' to index 1, 'c' to index 2.

4. Final Answer:

   Option C -> Option C

5. Quick Check:

   Resulting string is "acbd" after swaps [OK]

Hint: Sort characters within connected index groups [OK]

Common Mistakes:

• Swapping characters without sorting indices
• Mixing up group indices and characters

4. What is the time complexity of the optimal union-find solution for the Most Stones Removed with Same Row or Column problem, given n stones?

medium

A. O(n α(n)) where α is the inverse Ackermann function, due to union-find operations

B. O(n^2) due to checking all pairs of stones

C. O(n log n) because of sorting stones by coordinates

D. O(n) since each stone is processed once without extra overhead

Solution

1. Step 1: Analyze union-find operations

   Each union/find operation runs in nearly O(1) amortized time, specifically O(α(n)) where α is inverse Ackermann.

2. Step 2: Count total operations

   We perform at most 2n find operations and n union operations, total O(n α(n)).

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Union-find with path compression and rank -> O(n α(n)) [OK]

Hint: Union-find with path compression -> near-linear time [OK]

Common Mistakes:

• Assuming O(n^2) from brute force pairwise checks
• Confusing sorting complexity with union-find
• Ignoring inverse Ackermann function in complexity

5. What is the time complexity of the optimal Union-Find algorithm with path compression and union by rank when finding the redundant connection in a graph with n edges?

medium

A. O(n²) because each union operation may traverse all nodes

B. O(n * α(n)) where α(n) is the inverse Ackermann function, nearly constant

C. O(n log n) due to sorting edges before union operations

D. O(n) because each edge is processed once without overhead

Solution

1. Step 1: Identify number of union/find operations

   There are n edges, each triggers at most one union and two finds.

2. Step 2: Analyze complexity of find with path compression and union by rank

   Each find/union operation runs in amortized O(α(n)) time, where α(n) is inverse Ackermann function, which grows extremely slowly.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Union-Find with optimizations runs in near-linear time O(n * α(n)) [OK]

Hint: Inverse Ackermann function α(n) is almost constant [OK]

Common Mistakes:

• Assuming O(n²) due to nested loops
• Confusing sorting complexity with union-find
• Ignoring path compression benefits