What is the time complexity of Evaluate Division?

The optimal solution has a time complexity of O(Q * (V + E)).

What is the space complexity of Evaluate Division?

The space complexity is O(V + E).

What data structures are used to solve Evaluate Division?

This problem uses Array, String, Depth-First Search, Breadth-First Search, Union Find, Graph, Shortest Path. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Evaluate Division?

Insight 1: The equations can be modeled as a graph where variables are nodes, and the division values represent weighted edges between them. A/B = value is represented by an edge from A to B with weight 'value', and an edge from B to A with weight '1/value'. Insight 2: We can use Depth-First Search (DFS) or Breadth-First Search (BFS) to traverse the graph and find the path from the numerator to the denominator in each query. The product of the edge weights along the path gives the result of the division.

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Evaluate Division - LeetCode 399 Solution

Q: Which companies ask LeetCode 399. Evaluate Division in interviews?

This problem is frequently asked at BlackRock, Citadel, Coinbase, GE Healthcare, Jane Street and 8 more companies.

Evaluate Division - Complete Solution Guide

Evaluate Division is LeetCode problem 399, 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 variable pairs equations and an array of real numbers values , where equations[i] = [A i , B i ] and values[i] represent the equation A i / B i = values[i] . Each A i or B i is a string that represents a single variable. You are also given some queries , where queries[j] = [C j , D j ] represents the j th query where you must find the answer for C j / D j = ? . Return the answers to all queries . If a single answer cannot be determined, return -1.0 . Note: The input is

Detailed Explanation

The problem asks us to evaluate division equations given as pairs of variables (strings) and their corresponding division results. We're given a set of equations like 'a / b = 2.0' and a set of queries like 'a / c = ?'. The goal is to determine the result of each query based on the given equations. If a query cannot be resolved from the equations (e.g., due to disconnected variables or undefined variables), we return -1.0. The input ensures there are no contradictions or division by zero.

Solution Approach

The provided code uses Breadth-First Search (BFS) to solve this problem. First, it constructs a graph (implemented as a hash map) from the given equations and their values. Then, for each query, it performs a BFS starting from the numerator of the query, searching for the denominator. If a path is found, the product of the edge weights along the path is the answer. If no path is found, or either variable is not in the graph, the answer is -1.0.

Step-by-Step Algorithm

Step 1: Build the Graph: Create a hash map (or dictionary) to represent the graph. For each equation (A/B = value), add an edge from A to B with the given 'value', and an edge from B to A with the reciprocal value (1/value).
Step 2: Process Queries: Iterate through the given queries.
Step 3: BFS for Each Query: For each query (C/D), perform BFS starting from node C to find node D.
Step 4: Handle Base Cases in BFS: If either C or D is not in the graph, return -1.0.
Step 5: Initialize BFS: Use a queue to store nodes to visit. Store also a 'current value' associated with each node, which represents the product of edge weights from the start node (C) to the current node.
Step 6: Iterate in BFS: While the queue is not empty, dequeue a node and its corresponding 'current value'.
Step 7: Check for Target: If the dequeued node is D, return the 'current value'.
Step 8: Explore Neighbors: For each neighbor of the dequeued node, if it hasn't been visited yet, enqueue it with an updated 'current value' (current value * weight of edge to neighbor). Also mark neighbor as visited.
Step 9: No Path Found: If the BFS completes without finding D, return -1.0.
Step 10: Return Results: Return the list of results for all queries.

Key Insights

Insight 1: The equations can be modeled as a graph where variables are nodes, and the division values represent weighted edges between them. A/B = value is represented by an edge from A to B with weight 'value', and an edge from B to A with weight '1/value'.
Insight 2: We can use Depth-First Search (DFS) or Breadth-First Search (BFS) to traverse the graph and find the path from the numerator to the denominator in each query. The product of the edge weights along the path gives the result of the division.
Insight 3: Handling disconnected variables and undefined variables (those not appearing in any equation) is crucial. If either the numerator or denominator of a query is not in the graph, the result is -1.0.

Complexity Analysis

Time Complexity: O(Q * (V + E))

Space Complexity: O(V + E)

Topics

This problem involves: Array, String, Depth-First Search, Breadth-First Search, Union Find, Graph, Shortest Path.

Companies

Asked at: BlackRock, Citadel, Coinbase, GE Healthcare, Jane Street, MakeMyTrip, Nuro, PhonePe, Rippling, Snap, Snowflake, Stripe, Zeta.

Evaluate Division | LeetCode 399

Medium

Array String Depth-First Search Breadth-First Search Union Find Graph Shortest Path

Time: O(Q * (V + E))Space: O(V + E)

Solution Code

Available in:

import java.util.*;

class Solution {
    public double[] calcEquation(List<List<String>> equations, double[] values, List<List<String>> queries) {
        Map<String, Map<String, Double>> graph = new HashMap<>();
        for (int i = 0; i < equations.size(); i++) {
            List<String> equation = equations.get(i);
            String u = equation.get(0);
            String v = equation.get(1);
            double val = values[i];

            graph.computeIfAbsent(u, k -> new HashMap<>()).put(v, val);
            graph.computeIfAbsent(v, k -> new HashMap<>()).put(u, 1.0 / val);
        }

        double[] results = new double[queries.size()];
        for (int i = 0; i < queries.size(); i++) {
            List<String> query = queries.get(i);
            String startNode = query.get(0);
            String endNode = query.get(1);
            results[i] = bfs(graph, startNode, endNode);
        }

        return results;
    }

    private double bfs(Map<String, Map<String, Double>> graph, String startNode, String endNode) {
        if (!graph.containsKey(startNode) || !graph.containsKey(endNode)) {
            return -1.0;
        }

        Queue<Pair<String, Double>> queue = new LinkedList<>();
        queue.offer(new Pair<>(startNode, 1.0));
        Set<String> visited = new HashSet<>();
        visited.add(startNode);

        while (!queue.isEmpty()) {
            Pair<String, Double> current = queue.poll();
            String currentNode = current.getKey();
            double currentValue = current.getValue();

            if (currentNode.equals(endNode)) {
                return currentValue;
            }

            for (Map.Entry<String, Double> neighborEntry : graph.get(currentNode).entrySet()) {
                String neighbor = neighborEntry.getKey();
                double weight = neighborEntry.getValue();

                if (!visited.contains(neighbor)) {
                    visited.add(neighbor);
                    queue.offer(new Pair<>(neighbor, currentValue * weight));
                }
            }
        }

        return -1.0;
    }

    // Helper class to store pairs
    static class Pair<K, V> {
        private K key;
        private V value;

        public Pair(K key, V value) {
            this.key = key;
            this.value = value;
        }

        public K getKey() {
            return key;
        }

        public V getValue() {
            return value;
        }
    }
}

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Problem Info

Description

Problem 399: Evaluate Division - Detailed explanation and solution approaches.

Topics

Array String Depth-First Search Breadth-First Search Union Find Graph Shortest Path

Companies

BlackRockCitadelCoinbaseGE HealthcareJane StreetMakeMyTripNuroPhonePe

Stats

Difficulty:Medium

Languages:4

Companies:13

Problem #:399

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

10 steps

1
Step 1: Build the Graph: Create a hash map (or dictionary) to represent the graph. For each equation (A/B = value), add an edge from A to B with the given 'value', and an edge from B to A with the reciprocal value (1/value).
2
Step 2: Process Queries: Iterate through the given queries.
3
Step 3: BFS for Each Query: For each query (C/D), perform BFS starting from node C to find node D.
4
Step 4: Handle Base Cases in BFS: If either C or D is not in the graph, return -1.0.
5
Step 5: Initialize BFS: Use a queue to store nodes to visit. Store also a 'current value' associated with each node, which represents the product of edge weights from the start node (C) to the current node.
6
Step 6: Iterate in BFS: While the queue is not empty, dequeue a node and its corresponding 'current value'.
7
Step 7: Check for Target: If the dequeued node is D, return the 'current value'.
8
Step 8: Explore Neighbors: For each neighbor of the dequeued node, if it hasn't been visited yet, enqueue it with an updated 'current value' (current value * weight of edge to neighbor). Also mark neighbor as visited.
9
Step 9: No Path Found: If the BFS completes without finding D, return -1.0.
10
Step 10: Return Results: Return the list of results for all queries.

💡

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(Q * (V + E))

💾

Space Complexity

O(V + E)

📝

Why?

The time complexity is derived from processing each query using BFS, which has a time complexity of O(V+E) in the worst case. We run this BFS Q times, hence O(Q*(V+E)). The space complexity is mainly determined by the graph representation, which stores each variable and its neighbors along with the corresponding edge weights. The BFS algorithm itself uses O(V) space for the queue and the visited set, but this is dominated by the graph's space usage.

📖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

Floyd-Warshall Algorithm

The Floyd-Warshall algorithm computes the shortest paths between all pairs of nodes in a graph. We can apply it here to precompute all possible division results between all pairs of variables. Then queries can be answered in O(1) time.

Time: O(V^3) for preprocessing, O(1) per query.Space: O(V^2)

Trade-offs: Floyd-Warshall is better if you have a large number of queries because the preprocessing step is done only once. However, it has a higher initial cost and larger space complexity. BFS is better for a small number of queries because it computes the path on demand.

Union Find (Disjoint Set)

Union Find can be used to solve this problem by finding connected components and their relative values. It merges variables into sets and tracks their relative values within those sets.

Time: O(E * α(V) + Q * α(V)), where α(V) is the inverse Ackermann function, which grows very slowly, so it's practically constant. E is equations, Q is queries, V is variables.Space: O(V)

Trade-offs: Union Find provides another viable solution with a potential advantage in scenarios where the number of equations is significantly larger compared to the number of variables. It excels when efficient determination of connected components and relative values within those components is essential.

✨

Key Insights

1Insight 1: The equations can be modeled as a graph where variables are nodes, and the division values represent weighted edges between them. A/B = value is represented by an edge from A to B with weight 'value', and an edge from B to A with weight '1/value'.
2Insight 2: We can use Depth-First Search (DFS) or Breadth-First Search (BFS) to traverse the graph and find the path from the numerator to the denominator in each query. The product of the edge weights along the path gives the result of the division.
3Insight 3: Handling disconnected variables and undefined variables (those not appearing in any equation) is crucial. If either the numerator or denominator of a query is not in the graph, the result is -1.0.

📚

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: Query with undefined variable: If a query involves a variable that is not present in the given equations, the function should return -1.0. The code checks if start_node or end_node is present in the graph.
Edge case 2: Query with same numerator and denominator: If the numerator and denominator in a query are the same (e.g., a/a), and 'a' exists in the graph, the result should be 1.0. The BFS returns 'current_val' when current_node == end_node.
Edge case 3: No path between numerator and denominator: If there is no path in the graph between the numerator and denominator of a query, the function should return -1.0. The BFS algorithm will complete without finding a path.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling disconnected graphs: Failing to account for scenarios where there is no path between the numerator and denominator in a query. This results in incorrect results instead of -1.0.

✗Common mistake 2: Incorrectly calculating the division value: Errors in calculating the product of edge weights along the path from numerator to denominator. Remembering to consider the inverse values when traveling the graph backwards.

✗Common mistake 3: Memory leaks (especially in C): Forgetting to free allocated memory for the graph, nodes, and other data structures after the computation is done. The provided C code includes free operations for avoiding memory leaks.

✅

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: Currency conversion: The problem is analogous to currency conversion, where variables represent currencies and values represent exchange rates. Queries can be used to find the exchange rate between two currencies that may not have a direct exchange rate defined but can be derived from other exchange rates.

🏦Real-world scenario 2: Unit conversion: Similar to currency conversion, variables could represent different units of measurement (e.g., meters, feet, inches), and values represent conversion factors. Queries can determine conversion factors between units that are indirectly related.

💡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.

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: July 10, 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

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

@DevExpert92Jul 6

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

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Evaluate Division"?

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.

🤖 Need help with this problem?