Satisfiability of Equality Equations

📋

Satisfiability of Equality Equations

Edge cases:

</>

IDE

def equationsPossible(equations: list[str]) -> bool:public boolean equationsPossible(String[] equations)bool equationsPossible(vector<string> &equations)function equationsPossible(equations)

def equationsPossible(equations):
    # Write your solution here
    pass

class Solution {
    public boolean equationsPossible(String[] equations) {
        // Write your solution here
        return false;
    }
}

#include <vector>
using namespace std;

bool equationsPossible(vector<string> &equations) {
    // Write your solution here
    return false;
}

function equationsPossible(equations) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: trueIgnoring direct contradictions where a variable is equal and unequal to the same variable.✅ Add explicit check for inequalities of form x!=x and return false immediately.

Wrong: trueNot checking inequalities after union-find merges, missing contradictions.✅ After union-find merges, iterate inequalities and return false if both variables are in the same set.

Wrong: falseIncorrectly returning false when no contradictions exist, e.g., only inequalities with no conflicts.✅ Return true if no contradictions found after checking all inequalities.

Wrong: timeout or no outputUsing naive O(n^2) approach instead of efficient union-find.✅ Implement union-find with path compression and union by rank for O(n α(n)) complexity.

✓

Test Cases

Handle base cases like empty input and self-inequalities carefully.

Think about indirect contradictions caused by chains of equalities.

Optimize your union-find with path compression and union by rank.

t1_01basic

Input["a==b","b!=a"]

Expectedfalse

⏱ Performance - must finish in 2000ms

If a == b and b != a, this is a contradiction, so return false.

💡 Think about grouping variables that must be equal.

💡 Use union-find to merge equal variables and then check inequalities.

💡 If two variables are in the same set but have an inequality, return false.

Why it failed: Returned true despite direct contradiction between 'a==b' and 'b!=a'. Fix: After union-find merges, check inequalities for conflicts.

✓ Correctly detected direct contradiction in basic example.

t1_02basic

Input["a==b","b==c","a==c"]

Expectedtrue

⏱ Performance - must finish in 2000ms

All variables are equal with no contradictions, so return true.

💡 Check if all equalities form consistent groups.

💡 Union all equal variables and verify no inequalities contradict.

💡 No inequalities here, so result is true.

Why it failed: Returned false despite all equalities being consistent. Fix: Ensure union-find merges all equal variables correctly.

✓ Correctly returned true for consistent equalities.

t2_01edge

Input[]

Expectedtrue

⏱ Performance - must finish in 2000ms

Empty input means no constraints, so trivially satisfiable.

💡 Consider what happens if there are no equations.

💡 No constraints means no contradictions.

💡 Return true when input is empty.

Why it failed: Returned false or error on empty input. Fix: Handle empty input by returning true immediately.

✓ Correctly handled empty input as satisfiable.

t2_02edge

Input["a!=a"]

Expectedfalse

⏱ Performance - must finish in 2000ms

Variable cannot be unequal to itself, so contradiction.

💡 Check for inequalities where both sides are the same variable.

💡 Such inequalities are always false.

💡 Return false immediately if found.

Why it failed: Returned true despite 'a!=a' contradiction. Fix: Add check for inequalities of form x!=x and return false.

✓ Correctly detected self-inequality contradiction.

t2_03edge

Input["a==b","b!=c"]

Expectedtrue

⏱ Performance - must finish in 2000ms

No contradiction since a==b and b!=c do not conflict.

💡 Check if inequalities contradict any equalities.

💡 Variables not connected by equality can be unequal.

💡 Return true if no contradictions found.

Why it failed: Returned false incorrectly when inequalities do not contradict equalities. Fix: Only return false if inequality variables are in same union-find set.

✓ Correctly returned true for non-conflicting inequalities.

t2_04edge

Input["a==a"]

Expectedtrue

⏱ Performance - must finish in 2000ms

Equality of a variable to itself is always true.

💡 Check trivial equalities where variable equals itself.

💡 Such equalities do not affect union-find structure.

💡 Return true if no contradictions.

Why it failed: Returned false incorrectly on trivial equality. Fix: Allow equalities of form x==x without error.

✓ Correctly handled trivial equality.

t3_01corner

Input["a==b","b==c","c!=a"]

Expectedfalse

⏱ Performance - must finish in 2000ms

Indirect contradiction: a==b==c but c!=a contradicts equality chain.

💡 Check indirect equalities by union-find merges.

💡 Inequality contradicts if variables belong to same set.

💡 Return false if any inequality conflicts with union-find sets.

Why it failed: Returned true ignoring indirect contradictions. Fix: After union-find merges, verify all inequalities for conflicts.

✓ Correctly detected indirect contradiction.

t3_02corner

Input["a!=b","b!=c","a!=c"]

Expectedtrue

⏱ Performance - must finish in 2000ms

All inequalities with no equalities, no contradictions.

💡 Inequalities alone do not cause contradictions if no equalities connect variables.

💡 Union-find sets remain separate.

💡 Return true if no conflicts.

Why it failed: Returned false incorrectly for only inequalities with no conflicts. Fix: Do not assume inequalities alone cause contradictions.

✓ Correctly returned true for only inequalities with no conflicts.

t3_03corner

Input["a==b","b!=c","c==a"]

Expectedfalse

⏱ Performance - must finish in 2000ms

Contradiction due to cycle: a==b, c==a but b!=c conflicts.

💡 Detect cycles in equality sets that conflict with inequalities.

💡 Union-find merges all equal variables.

💡 Return false if inequality contradicts merged sets.

Why it failed: Returned true ignoring cycle contradiction. Fix: Check all inequalities after union-find merges for conflicts.

✓ Correctly detected cycle contradiction.

t4_01performance

Input{"_description":"n=100000 at constraint boundary - executor generates this"}

Expectednull

⏱ Performance - must finish in 2000ms

n=100000, O(n α(n)) union-find must complete within 2 seconds.

💡 Use union-find with path compression for efficiency.

💡 Avoid nested loops or quadratic checks.

💡 Two-pass approach: first union equalities, then check inequalities.

Why it failed: TLE due to O(n^2) or inefficient union-find implementation. Fix: Implement union-find with path compression and union by rank.

✓ Efficient union-find implementation confirmed by passing performance test.

Practice

1. You are given a string and a list of index pairs where you can swap characters any number of times. The goal is to find the lexicographically smallest string possible after performing any swaps allowed by these pairs. Which algorithmic approach best guarantees an optimal solution for this problem?

easy

A. Greedy approach that swaps characters whenever a smaller character is found at a connected index

B. Union-Find to group connected indices, then sort characters within each group to reconstruct the smallest string

C. Dynamic programming to try all possible swaps and pick the lexicographically smallest result

D. Breadth-first search to explore all possible strings reachable by swaps and select the smallest

Solution

1. Step 1: Understand problem constraints

   The problem allows unlimited swaps between indices connected by pairs, so indices form connected components.

2. Step 2: Identify approach that groups indices

   Union-Find efficiently groups indices into connected components, enabling sorting characters within each group to get the smallest lexicographic order.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Union-Find groups indices correctly, sorting within groups yields optimal string [OK]

Hint: Group indices first, then sort characters within groups [OK]

Common Mistakes:

• Assuming greedy swaps suffice
• Trying DP over all permutations
• Using BFS to explore all strings exhaustively

2. 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

3. 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

4. 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

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