What is the time complexity of Satisfiability of Equality Equations?

The optimal solution has a time complexity of O(Nα(N)).

What data structures are used to solve Satisfiability of Equality Equations?

This problem uses Array, String, Union Find, Graph. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Satisfiability of Equality Equations?

Insight 1: The core idea is to use the Union Find data structure to group together variables that are equal to each other according to the '==' equations. Insight 2: After grouping equal variables, iterate through the '!=' equations. If any two variables in an inequality equation belong to the same group (i.e., their 'find' operations return the same root), then the equations cannot be satisfied, and we return `false`.

Which companies ask LeetCode 990. Satisfiability of Equality Equations in interviews?

This problem is frequently asked at Sumo Logic, UiPath.

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Satisfiability of Equality Equations - LeetCode 990 Solution

Satisfiability of Equality Equations - Complete Solution Guide

Satisfiability of Equality Equations is LeetCode problem 990, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given an array of strings equations that represent relationships between variables where each string equations[i] is of length 4 and takes one of two different forms: "x i ==y i " or "x i !=y i " .Here, x i and y i are lowercase letters (not necessarily different) that represent one-letter variable names. Return true if it is possible to assign integers to variable names so as to satisfy all the given equations, or false otherwise . Example 1: Input: equations = ["a==b","b!=a"] Output: f

Detailed Explanation

The problem asks us to determine if a given set of equality and inequality equations involving lowercase letters can be satisfied simultaneously. Each equation is in the form 'x==y' or 'x!=y', where x and y are lowercase letters representing variables. We need to return `true` if it's possible to assign integer values to the variables such that all equality and inequality constraints are met, and `false` otherwise.

Solution Approach

The provided solutions use the Union Find (also known as Disjoint Set Union) data structure to solve the problem. First, it processes all equality equations ('==') and uses the 'union' operation to merge the sets containing the variables involved. This means all variables that must be equal are grouped into the same set. Then, it processes all inequality equations ('!='). For each inequality, it checks if the variables involved belong to the same set using the 'find' operation. If they do, it means we have a contradiction (two variables that must be equal according to '==' are required to be unequal according to '!='), so the solution returns `false`. If all inequalities are checked without finding a contradiction, the solution returns `true`.

Step-by-Step Algorithm

Step 1: Initialize a `parent` array of size 26, where `parent[i]` initially equals `i`. This represents the disjoint sets for each of the 26 possible variables (a-z).
Step 2: Implement the `find` operation (with path compression) to determine the root/representative of the set a variable belongs to. Path compression optimizes subsequent 'find' operations.
Step 3: Implement the `union` operation to merge the sets of two variables, effectively indicating that they are equal.
Step 4: Iterate through the input `equations` list. If an equation is of the form 'x==y', calculate the indices of x and y (by subtracting 'a' from their ASCII values) and perform the `union` operation on these indices.
Step 5: Iterate through the input `equations` list again. If an equation is of the form 'x!=y', calculate the indices of x and y and perform the `find` operation on both. If `find(x)` equals `find(y)`, it means x and y belong to the same set (they are equal), which contradicts the '!=' constraint. In this case, return `false`.
Step 6: If all equations are processed without finding any contradictions, return `true`.

Key Insights

Insight 1: The core idea is to use the Union Find data structure to group together variables that are equal to each other according to the '==' equations.
Insight 2: After grouping equal variables, iterate through the '!=' equations. If any two variables in an inequality equation belong to the same group (i.e., their 'find' operations return the same root), then the equations cannot be satisfied, and we return `false`.
Insight 3: Representing variables with integer indices (0-25 for 'a'-'z') allows efficient implementation of the Union Find data structure using an array.

Complexity Analysis

Time Complexity: O(Nα(N))

Space Complexity: O(1)

Topics

This problem involves: Array, String, Union Find, Graph.

Companies

Asked at: Sumo Logic, UiPath.

Satisfiability of Equality Equations | LeetCode 990

Medium

Array String Union Find Graph

Time: O(Nα(N))Space: O(1)

Solution Code

Available in:

class Solution {
    public boolean equationsPossible(String[] equations) {
        int[] parent = new int[26];
        for (int i = 0; i < 26; i++) {
            parent[i] = i;
        }

        function find = (i) -> {
            if (parent[i] == i) {
                return i;
            }
            parent[i] = find.apply(parent[i]);
            return parent[i];
        };

        function union = (i, j) -> {
            int root_i = find.apply(i);
            int root_j = find.apply(j);
            if (root_i != root_j) {
                parent[root_i] = root_j;
            }
        };

        for (String eq : equations) {
            if (eq.charAt(1) == '=') {
                int x = eq.charAt(0) - 'a';
                int y = eq.charAt(3) - 'a';
                union.apply(x, y);
            }
        }

        for (String eq : equations) {
            if (eq.charAt(1) == '!') {
                int x = eq.charAt(0) - 'a';
                int y = eq.charAt(3) - 'a';
                if (find.apply(x) == find.apply(y)) {
                    return false;
                }
            }
        }

        return true;
    }

    interface function {
        int apply(int i);
        void apply(int i, int j);
    }
}

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Problem Info

Description

Problem 990: Satisfiability of Equality Equations - Detailed explanation and solution approaches.

Topics

Array String Union Find Graph

Companies

Sumo LogicUiPath

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:990

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Initialize a `parent` array of size 26, where `parent[i]` initially equals `i`. This represents the disjoint sets for each of the 26 possible variables (a-z).
2
Step 2: Implement the `find` operation (with path compression) to determine the root/representative of the set a variable belongs to. Path compression optimizes subsequent 'find' operations.
3
Step 3: Implement the `union` operation to merge the sets of two variables, effectively indicating that they are equal.
4
Step 4: Iterate through the input `equations` list. If an equation is of the form 'x==y', calculate the indices of x and y (by subtracting 'a' from their ASCII values) and perform the `union` operation on these indices.
5
Step 5: Iterate through the input `equations` list again. If an equation is of the form 'x!=y', calculate the indices of x and y and perform the `find` operation on both. If `find(x)` equals `find(y)`, it means x and y belong to the same set (they are equal), which contradicts the '!=' constraint. In this case, return `false`.
6
Step 6: If all equations are processed without finding any contradictions, return `true`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(Nα(N))

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is dominated by the iterations through the equations combined with the union-find operations. The path compression optimization in the `find` operation makes the amortized time complexity of the union-find operations almost constant. The space complexity is constant because the parent array size depends on the fixed number of lowercase alphabets (26), not the input size.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Depth-First Search (DFS)

An alternative approach involves representing the variables and their relationships as a graph. Equality equations ('==') would define edges between nodes, indicating connected components. Then, for inequality equations ('!='), we would check if the two variables are in the same connected component using DFS. If they are, it means there's a contradiction.

Time: O(N + V + E), where V is the number of vertices (26 in this case) and E is the number of edges in the graph. N is the number of equationsSpace: O(V + E), to store the graph

Trade-offs: The DFS approach might be more intuitive for some, but the Union Find approach is generally more efficient, especially with path compression. DFS can be less efficient when the graph is dense.

✨

Key Insights

1Insight 1: The core idea is to use the Union Find data structure to group together variables that are equal to each other according to the '==' equations.
2Insight 2: After grouping equal variables, iterate through the '!=' equations. If any two variables in an inequality equation belong to the same group (i.e., their 'find' operations return the same root), then the equations cannot be satisfied, and we return `false`.
3Insight 3: Representing variables with integer indices (0-25 for 'a'-'z') allows efficient implementation of the Union Find data structure using an array.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Graph

Nodes connected by edges. Like a social network or city map!

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Equations like 'a==a' or 'a!=a'. The solution handles these correctly because 'a==a' does not change anything, and 'a!=a' will immediately result in a `false` return if it is encountered.
Edge case 2: Multiple equality equations that form a chain, such as 'a==b', 'b==c', 'c==d'. The Union Find algorithm efficiently groups all these variables into the same set.
Edge case 3: Conflicting equations such as 'a==b' and 'a!=b'. The algorithm correctly identifies and returns `false` in this case.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using path compression in the `find` operation, which can lead to a higher time complexity in some cases.

✗Common mistake 2: Processing the inequality equations before the equality equations. This can lead to incorrect results because the equality relationships need to be established first.

✗Common mistake 3: Incorrectly implementing the `union` operation. It's crucial to attach the root of one set to the root of the other, not arbitrary nodes within the sets.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Determining variable dependencies in a compiler. Equality equations can represent that two variables refer to the same memory location, while inequalities represent constraints on the relationships between variables. The compiler needs to ensure these constraints are not violated.

🏦Real-world scenario 2: Analyzing network connectivity. Equality equations can represent that two computers are in the same network, while inequalities represent that they are in different networks. Network management systems need to ensure there are no connectivity conflicts.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 30, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Satisfiability of Equality Equations"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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