Accounts Merge

📋

Accounts Merge

Edge cases:

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IDE

def accountsMerge(accounts: list[list[str]]) -> list[list[str]]:public List<List<String>> accountsMerge(List<List<String>> accounts)vector<vector<string>> accountsMerge(vector<vector<string>>& accounts)function accountsMerge(accounts)

def accountsMerge(accounts: list[list[str]]) -> list[list[str]]:
    # Write your solution here
    pass

class Solution {
    public List<List<String>> accountsMerge(List<List<String>> accounts) {
        // Write your solution here
        return new ArrayList<>();
    }
}

#include <vector>
#include <string>
using namespace std;

vector<vector<string>> accountsMerge(vector<vector<string>>& accounts) {
    // Write your solution here
    return {};
}

function accountsMerge(accounts) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: [["John", "[e-mail]"], ["John", "[e-mail]"]]Failing to merge accounts sharing emails; treating accounts independently.✅ Implement union-find on emails to merge connected accounts before output.

Wrong: [["Anna", "[e-mail]", "[e-mail]", "[e-mail]"]]Merging accounts based on name only, ignoring email overlap.✅ Merge accounts only if they share at least one email, not just by name.

Wrong: [["X", "[e-mail]"], ["Y", "[e-mail]"], ["Z", "[e-mail]"]]Failing to merge accounts connected by email cycles or chains.✅ Use union-find with path compression to merge all connected emails.

Wrong: [["A", "[e-mail]"], ["B", "[e-mail]"], ["C", "[e-mail]"]]Incorrectly merging accounts with unique emails or failing to sort emails.✅ Only merge accounts with shared emails and sort emails in output.

Wrong: Solution times out on large inputs due to nested loops checking all pairs.✅ Optimize to union-find with email to account mapping to achieve near-linear time.

✓

Test Cases

Try testing your solution on empty and minimal inputs to ensure base cases are handled.

Consider tricky cases with cycles and chains of shared emails to catch subtle bugs.

Focus on optimizing your solution to run efficiently on large inputs using union-find.

t1_01basic

Input[["John","[e-mail]","[e-mail]"],["John","[e-mail]"],["John","[e-mail]","[e-mail]"],["Mary","[e-mail]"]]

Expected[["John","[e-mail]","[e-mail]","[e-mail]"],["John","[e-mail]"],["Mary","[e-mail]"]]

⏱ Performance - must finish in 2000ms

The first and third accounts share '[e-mail]', so they are merged. The second and fourth accounts have unique emails and remain separate.

💡 Think about grouping emails that belong to the same person.

💡 Use union-find or graph traversal to connect accounts sharing emails.

💡 After union, collect emails per group and sort them before returning.

Why it failed: Failed to merge accounts sharing emails or incorrect grouping. Fix by implementing union-find on emails and merging accounts accordingly.

✓ Correctly merges accounts sharing emails and returns sorted unique emails.

t1_02basic

Input[["Alice","[e-mail]","[e-mail]"],["Bob","[e-mail]"],["Alice","[e-mail]","[e-mail]"],["Charlie","[e-mail]"]]

Expected[["Alice","[e-mail]","[e-mail]","[e-mail]"],["Bob","[e-mail]"],["Charlie","[e-mail]"]]

⏱ Performance - must finish in 2000ms

Accounts 1 and 3 share '[e-mail]' so they merge; others remain separate.

💡 Identify emails that appear in multiple accounts to merge them.

💡 Union emails to find connected components representing merged accounts.

💡 Sort emails in each merged account before returning.

Why it failed: Incorrect merging of accounts sharing emails or missing some emails in output. Fix union-find implementation to union all connected emails.

✓ Accounts merged correctly with all emails included and sorted.

t2_01edge

Input[]

Expected[]

⏱ Performance - must finish in 2000ms

Empty input should return empty output.

💡 Check how your code handles empty input lists.

💡 Ensure no assumptions about non-empty accounts.

💡 Return empty list immediately if input is empty.

Why it failed: Code crashes or returns non-empty output on empty input. Fix by adding a guard clause for empty input.

✓ Correctly returns empty list for empty input.

t2_02edge

Input[["Solo","[e-mail]"]]

Expected[["Solo","[e-mail]"]]

⏱ Performance - must finish in 2000ms

Single account with one email should return itself.

💡 Test handling of single account with single email.

💡 No merging needed; just return the account with sorted emails.

💡 Ensure code does not break on minimal input.

Why it failed: Fails to handle single account or returns incorrect output. Fix by handling single account case properly.

✓ Correctly returns single account with its email.

t2_03edge

Input[["A","[e-mail]"],["B","[e-mail]"],["C","[e-mail]"]]

Expected[["A","[e-mail]"],["B","[e-mail]"],["C","[e-mail]"]]

⏱ Performance - must finish in 2000ms

All accounts have unique emails; no merges; output emails sorted.

💡 Check that accounts with no shared emails remain separate.

💡 Verify that emails are sorted in each account.

💡 Avoid merging accounts without common emails.

Why it failed: Incorrectly merges accounts without shared emails or fails to sort emails. Fix by only merging accounts with shared emails and sorting emails before output.

✓ Accounts remain separate with sorted emails as expected.

t2_04edge

Input[["Same","[e-mail]"],["Same","[e-mail]"],["Same","[e-mail]"]]

Expected[["Same","[e-mail]"]]

⏱ Performance - must finish in 2000ms

All accounts share the same email, so they merge into one account.

💡 Check that accounts with identical emails merge into one.

💡 Union-find should unify all accounts sharing the same email.

💡 Return one account with sorted unique emails.

Why it failed: Fails to merge all accounts sharing the same email. Fix union-find to union all accounts with common emails.

✓ Correctly merges all accounts sharing the same email into one.

t3_01corner

Input[["X","[e-mail]","[e-mail]"],["Y","[e-mail]","[e-mail]"],["Z","[e-mail]","[e-mail]"]]

Expected[["X","[e-mail]","[e-mail]","[e-mail]"]]

⏱ Performance - must finish in 2000ms

All accounts share emails forming a cycle; all merge into one account.

💡 Beware of cycles in email connections causing multiple merges.

💡 Union-find must handle transitive merges correctly.

💡 Ensure path compression or union by rank to unify all connected emails.

Why it failed: Fails to merge accounts connected by cycles, resulting in multiple outputs. Fix union-find to handle transitive closures properly.

✓ Correctly merges all accounts connected by shared emails including cycles.

t3_02corner

Input[["Anna","[e-mail]"],["Anna","[e-mail]"],["Anna","[e-mail]"]]

Expected[["Anna","[e-mail]"],["Anna","[e-mail]"],["Anna","[e-mail]"]]

⏱ Performance - must finish in 2000ms

Multiple accounts with same name but no shared emails remain separate.

💡 Name alone does not imply merging; emails must overlap.

💡 Check that union-find merges only accounts with shared emails.

💡 Verify output keeps separate accounts if no email overlap.

Why it failed: Incorrectly merges accounts just because names match. Fix by merging only on shared emails, not names.

✓ Accounts with same name but no shared emails remain separate as expected.

t3_03corner

Input[["D","[e-mail]","[e-mail]"],["E","[e-mail]","[e-mail]"],["F","[e-mail]","[e-mail]"],["G","[e-mail]"]]

Expected[["D","[e-mail]","[e-mail]","[e-mail]","[e-mail]"],["G","[e-mail]"]]

⏱ Performance - must finish in 2000ms

Chain of shared emails merges first three accounts; last account remains separate.

💡 Check that union-find merges accounts connected by chains of emails.

💡 Ensure all connected emails are grouped regardless of chain length.

💡 Sort emails and pick the name from the first account in the group.

Why it failed: Fails to merge accounts connected by email chains, splitting them incorrectly. Fix union-find to merge all connected components fully.

✓ Correctly merges accounts connected by email chains and keeps separate accounts separate.

t4_01performance

Input{"_description":"n=100000 at constraint boundary - executor generates this input with unique and some shared emails"}

Expectednull

⏱ Performance - must finish in 2000ms

Test with 100000 accounts and up to 1000 emails each to verify O(N*M*α(N*M)) union-find solution runs within 2 seconds.

💡 Use union-find with path compression for efficient merges.

💡 Avoid nested loops checking all pairs of accounts.

💡 Pre-map emails to account indices to union efficiently.

Why it failed: Brute force or naive solutions time out due to O(n^2*m) complexity. Fix by implementing union-find with email to account mapping.

✓ Efficient union-find solution runs within time limits.

Practice

1. You are given a list of edges representing an undirected graph with n nodes. The graph started as a tree with n nodes and one additional edge was added, creating a cycle. Which algorithmic approach guarantees finding the edge that creates the cycle efficiently?

easy

A. Union-Find (Disjoint Set Union) with path compression and union by rank to detect cycles during edge insertion

B. Dynamic programming to store reachable nodes and detect cycles

C. Depth-first search (DFS) cycle detection on each edge, checking connectivity repeatedly

D. Greedy algorithm that removes edges with the highest node degree first

Solution

1. Step 1: Understand problem constraints

   The graph is initially a tree with n nodes, so it has exactly n-1 edges. Adding one edge creates exactly one cycle.

2. Step 2: Identify suitable algorithm

   Union-Find efficiently tracks connected components and detects if adding an edge connects two nodes already in the same component, indicating a cycle.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Union-Find detects cycles in near-linear time [OK]

Hint: Union-Find detects cycles during edge insertion [OK]

Common Mistakes:

• Thinking DFS on each edge is efficient enough
• Believing DP can track cycles here
• Assuming greedy edge removal finds cycles

2. Identify the bug in the following union-find code snippet used to solve the Most Stones Removed with Same Row or Column problem:

class UnionFind:
    def __init__(self):
        self.parent = {}
        self.rank = {}
        self.count = 0

    def find(self, x):
        if x not in self.parent:
            self.parent[x] = x
            self.rank[x] = 0
            self.count += 1
        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
            # BUG: missing decrement of count here

uf = UnionFind()
for x, y in stones:
    uf.union(x, ~y)

return len(stones) - uf.count

medium

A. Line where rank is incremented after union

B. Line where parent is assigned in find method

C. Line where new nodes are initialized in find

D. Line missing decrement of count after union

Solution

1. Step 1: Understand count variable role

   Count tracks number of connected components, incremented on new node creation.

2. Step 2: Identify missing update

   After union merges two components, count must decrement but code misses this step.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Without decrement, count overestimates components -> wrong answer [OK]

Hint: Count must decrease on union of distinct sets [OK]

Common Mistakes:

• Forgetting to decrement count after union
• Incorrect rank updates
• Misusing parent assignments in find

3. The following code attempts to find a redundant connection using union-find but contains a subtle bug. Identify the line causing incorrect behavior.

medium

A. Line setting parent[i] = i in the for loop

B. Line incrementing rank[rootA] after union

C. Line checking if rootA == rootB in union function

D. Line initializing parent with zeros: parent = [0]*(n+1)

Solution

1. Step 1: Analyze parent array initialization

   Initializing parent with zeros causes all nodes to initially point to 0, which is incorrect for 1-based nodes.

2. Step 2: Check for correct parent assignment

   The for loop tries to fix this, but if n=0, loop may not run properly or indexing may be off; better to initialize parent as list(range(n+1)) directly.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Incorrect parent initialization leads to wrong find results [OK]

Hint: Parent array must be initialized to self indices [OK]

Common Mistakes:

• Forgetting to initialize parent properly
• Assuming zeros are valid parents
• Misindexing parent array

4. What is the time complexity of the optimal union-find + priority queue approach to the smallest string with swaps problem, where n is the length of the string and m is the number of pairs?

medium

A. O(n log n + m α(n)) where α(n) is the inverse Ackermann function from union-find operations

B. O(n * m) due to iterating over all pairs for each character

C. O(n²) because sorting characters in each connected component can be quadratic in worst case

D. O(m log n) because union-find operations dominate the runtime

Solution

1. Step 1: Analyze union-find operations

   Union-find performs m union/find operations in O(m α(n)) time, where α(n) is very slow growing and almost constant.

2. Step 2: Analyze sorting step

   Sorting characters in all connected components combined takes O(n log n) time.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Union-find plus sorting dominates runtime as O(n log n + m α(n)) [OK]

Hint: Union-find nearly linear; sorting dominates with n log n [OK]

Common Mistakes:

• Assuming O(n*m) due to nested loops
• Confusing sorting cost as O(n²)
• Ignoring inverse Ackermann factor

5. Suppose the problem is modified so that edges can be reused multiple times (i.e., multi-edges allowed), and we want to remove the maximum number of edges while keeping the graph traversable for both Alice and Bob. Which modification to the optimal union-find approach is necessary?

hard

A. Use a multigraph union-find variant that counts edge multiplicities and only removes redundant edges after full connectivity.

B. Switch to a shortest path algorithm since union-find cannot handle multi-edges.

C. No change needed; the original union-find approach works as is.

D. Process edges in arbitrary order without sorting since multi-edges guarantee connectivity.

Solution

1. Step 1: Understand multi-edge impact

   Multi-edges mean multiple edges connect same nodes; union-find alone merges sets once, ignoring multiplicity.

2. Step 2: Modify union-find approach

   Need to track edge counts per connection and only remove edges that do not break connectivity, requiring a multigraph-aware union-find or additional bookkeeping.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Original union-find ignores multiplicity -> must track edge counts [OK]

Hint: Union-find merges sets once; multi-edges require counting edges [OK]

Common Mistakes:

• Assuming union-find handles multi-edges without change
• Ignoring edge multiplicity
• Switching to unrelated algorithms unnecessarily