What is the time complexity of Find All Possible Recipes from Given Supplies?

The optimal solution has a time complexity of O(n + m).

What is the space complexity of Find All Possible Recipes from Given Supplies?

The space complexity is O(n + m).

What data structures are used to solve Find All Possible Recipes from Given Supplies?

This problem uses Array, Hash Table, String, Graph, Topological Sort. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Find All Possible Recipes from Given Supplies?

Insight 1: Topological sorting is the core algorithmic concept. Recipes can be considered as nodes in a directed graph where an edge from ingredient 'a' to recipe 'b' exists if 'a' is an ingredient for 'b'. The problem requires finding all nodes (recipes) that can be reached from the initial supply nodes. Insight 2: Representing the dependencies between recipes and ingredients as an adjacency list or a similar graph representation is crucial for efficiently determining which recipes can be made.

Which companies ask LeetCode 2115. Find All Possible Recipes from Given Supplies in interviews?

This problem is frequently asked at PhonePe, TikTok, Verily.

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Find All Possible Recipes from Given Supplies - LeetCode 2115 Solution

Find All Possible Recipes from Given Supplies - Complete Solution Guide

Find All Possible Recipes from Given Supplies is LeetCode problem 2115, 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 have information about n different recipes. You are given a string array recipes and a 2D string array ingredients . The i th recipe has the name recipes[i] , and you can create it if you have all the needed ingredients from ingredients[i] . A recipe can also be an ingredient for other recipes, i.e., ingredients[i] may contain a string that is in recipes . You are also given a string array supplies containing all the ingredients that you initially have, and you have an infinite supply of all

Detailed Explanation

The problem asks us to find all possible recipes that can be created from a given set of initial supplies. We are provided with a list of recipes, a list of ingredients required for each recipe, and a list of initial supplies. We can create a recipe if we have all of its ingredients, and a recipe can also be an ingredient for other recipes. The goal is to return a list of all the recipes that can be created, considering the dependencies between recipes.

Solution Approach

The solution uses a topological sort-based approach. It constructs a directed graph where nodes are ingredients and recipes. An edge exists from an ingredient 'x' to a recipe 'y' if 'x' is an ingredient of 'y' and 'x' is not in the `supplies` list. It uses in-degree counting (number of unavailable ingredients) for each recipe. The algorithm initializes a queue with recipes having an in-degree of 0 (meaning all ingredients are available). It then iteratively processes the queue, making recipes and decrementing the in-degree of recipes that depend on the made recipe. If any of those dependent recipe's in-degree becomes 0, it's added to the queue. The process continues until the queue is empty.

Step-by-Step Algorithm

Step 1: Initialize an adjacency list (`adj`) to store the dependencies between ingredients and recipes. `adj[ingredient]` will contain a list of recipes that require that ingredient but which are not in the initial supplies.
Step 2: Initialize an in-degree map (`in_degree`) to store the number of unavailable ingredients for each recipe.
Step 3: Iterate through each recipe and its ingredients. If an ingredient is not in the initial `supplies`, add an edge from that ingredient to the recipe in the adjacency list and increment the recipe's in-degree.
Step 4: Create a queue and add all recipes with an in-degree of 0 to the queue.
Step 5: While the queue is not empty, dequeue a recipe. Add this recipe to the `result` list.
Step 6: Iterate through the adjacency list for the dequeued recipe. For each dependent recipe in the adjacency list, decrement its in-degree. If the in-degree becomes 0, enqueue it.
Step 7: Return the `result` list of recipes that can be created.

Key Insights

Insight 1: Topological sorting is the core algorithmic concept. Recipes can be considered as nodes in a directed graph where an edge from ingredient 'a' to recipe 'b' exists if 'a' is an ingredient for 'b'. The problem requires finding all nodes (recipes) that can be reached from the initial supply nodes.
Insight 2: Representing the dependencies between recipes and ingredients as an adjacency list or a similar graph representation is crucial for efficiently determining which recipes can be made.
Insight 3: Using in-degree counting to determine the order in which recipes can be made avoids circular dependencies and ensures that a recipe is only made after all its ingredients are available.

Complexity Analysis

Time Complexity: O(n + m)

Space Complexity: O(n + m)

Topics

This problem involves: Array, Hash Table, String, Graph, Topological Sort.

Companies

Asked at: PhonePe, TikTok, Verily.

Find All Possible Recipes from Given Supplies | LeetCode 2115

Medium

Array Hash Table String Graph Topological Sort

Time: O(n + m)Space: O(n + m)

Solution Code

Available in:

import java.util.*;

class Solution {
    public List<String> findAllRecipes(String[] recipes, List<List<String>> ingredients, String[] supplies) {
        Map<String, List<String>> adj = new HashMap<>();
        Map<String, Integer> inDegree = new HashMap<>();
        
        Set<String> suppliesSet = new HashSet<>(Arrays.asList(supplies));
        
        for (int i = 0; i < recipes.length; i++) {
            String recipe = recipes[i];
            for (String ingredient : ingredients.get(i)) {
                if (!suppliesSet.contains(ingredient)) {
                    adj.computeIfAbsent(ingredient, k -> new ArrayList<>()).add(recipe);
                    inDegree.put(recipe, inDegree.getOrDefault(recipe, 0) + 1);
                }
            }
        }
        
        Queue<String> queue = new LinkedList<>();
        for (String recipe : recipes) {
            if (!inDegree.containsKey(recipe)) {
                inDegree.put(recipe,0);
            }
            if (inDegree.get(recipe) == 0) {
                queue.offer(recipe);
            }
        }
        
        List<String> result = new ArrayList<>();
        
        while (!queue.isEmpty()) {
            String madeRecipe = queue.poll();
            result.add(madeRecipe);
            
            if (adj.containsKey(madeRecipe)) {
                for (String dependentRecipe : adj.get(madeRecipe)) {
                    inDegree.put(dependentRecipe, inDegree.get(dependentRecipe) - 1);
                    if (inDegree.get(dependentRecipe) == 0) {
                        queue.offer(dependentRecipe);
                    }
                }
            }
        }
        
        return result;
    }
}

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Problem Info

Description

Problem 2115: Find All Possible Recipes from Given Supplies - Detailed explanation and solution approaches.

Topics

Array Hash Table String Graph Topological Sort

Companies

PhonePeTikTokVerily

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:2115

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

7 steps

1
Step 1: Initialize an adjacency list (`adj`) to store the dependencies between ingredients and recipes. `adj[ingredient]` will contain a list of recipes that require that ingredient but which are not in the initial supplies.
2
Step 2: Initialize an in-degree map (`in_degree`) to store the number of unavailable ingredients for each recipe.
3
Step 3: Iterate through each recipe and its ingredients. If an ingredient is not in the initial `supplies`, add an edge from that ingredient to the recipe in the adjacency list and increment the recipe's in-degree.
4
Step 4: Create a queue and add all recipes with an in-degree of 0 to the queue.
5
Step 5: While the queue is not empty, dequeue a recipe. Add this recipe to the `result` list.
6
Step 6: Iterate through the adjacency list for the dequeued recipe. For each dependent recipe in the adjacency list, decrement its in-degree. If the in-degree becomes 0, enqueue it.
7
Step 7: Return the `result` list of recipes that can be created.

💡

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(n + m)

💾

Space Complexity

O(n + m)

📝

Why?

The time complexity is determined by the topological sort. The space complexity is dominated by the adjacency list, in-degree map, and the queue, which store the graph structure and dependencies.

📖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)

A recursive approach where we try to make a recipe, and if we are missing an ingredient, we recursively try to make that ingredient first. If all ingredients can be made or are in the supply list, the recipe is considered creatable.

Time: O(n + m), where n is the total number of ingredients and recipes and m is the number of ingredient dependencies. Although in pathological cases, this could approach O(2^n).Space: O(n + m) in the worst case due to the recursion stack and the graph representation.

Trade-offs: DFS can be more intuitive for some, but it's susceptible to stack overflow in cases with deep dependency chains. The iterative topological sort approach is generally more robust.

✨

Key Insights

1Insight 1: Topological sorting is the core algorithmic concept. Recipes can be considered as nodes in a directed graph where an edge from ingredient 'a' to recipe 'b' exists if 'a' is an ingredient for 'b'. The problem requires finding all nodes (recipes) that can be reached from the initial supply nodes.
2Insight 2: Representing the dependencies between recipes and ingredients as an adjacency list or a similar graph representation is crucial for efficiently determining which recipes can be made.
3Insight 3: Using in-degree counting to determine the order in which recipes can be made avoids circular dependencies and ensures that a recipe is only made after all its ingredients are available.

📚

Key Terms Explained

For Students

Array

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

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

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: Circular dependencies between recipes (e.g., recipe A requires recipe B, and recipe B requires recipe A). The topological sort algorithm correctly handles this by ensuring that no recipe involved in a cycle will ever have an in-degree of 0, and therefore will never be added to the queue.
Edge case 2: A recipe has no dependencies (all its ingredients are in supplies). This recipe will have an initial in-degree of 0 and will be immediately added to the queue and processed.
Edge case 3: Empty input lists for recipes, ingredients, or supplies. The solution correctly handles this by initializing the data structures appropriately and returning an empty result.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to handle circular dependencies. If the algorithm doesn't correctly identify and prevent recipes in a circular dependency from being added to the result, it will lead to an infinite loop or incorrect results.

✗Common mistake 2: Incorrectly updating the in-degree of dependent recipes. If the in-degree is not decremented properly when a recipe is made, recipes that should be creatable may not be added to the queue.

✗Common mistake 3: Not considering the case where all ingredients of a recipe are in the initial supplies and not adding it to the queue initially.

✅

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: Dependency resolution in software package management. Determining the order in which software packages need to be installed based on their dependencies.

🏦Real-world scenario 2: Task scheduling in project management. Determining the order in which tasks need to be completed based on their dependencies.

💡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: October 14, 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 13

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

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 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.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

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 "Find All Possible Recipes from Given Supplies"?

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 hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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