What is the time complexity of Sort Items by Groups Respecting Dependencies?

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

What is the space complexity of Sort Items by Groups Respecting Dependencies?

The space complexity is O(n + m + E).

What data structures are used to solve Sort Items by Groups Respecting Dependencies?

This problem uses Depth-First Search, Breadth-First Search, Graph, Topological Sort. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Sort Items by Groups Respecting Dependencies?

Insight 1: Recognize that this is a topological sorting problem on two levels: items and groups. Items within the same group have dependencies, and groups themselves can have dependencies due to item dependencies between them. Insight 2: Topological sorting is crucial. Represent dependencies as a directed graph and use topological sort to obtain a valid ordering, both for items and groups. The algorithm needs to detect cycles, as cycles indicate no possible ordering.

Which companies ask LeetCode 1203. Sort Items by Groups Respecting Dependencies in interviews?

This problem is frequently asked at Citadel.

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Sort Items by Groups Respecting Dependencies - LeetCode 1203 Solution

Sort Items by Groups Respecting Dependencies - Complete Solution Guide

Sort Items by Groups Respecting Dependencies is LeetCode problem 1203, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

There are n items each belonging to zero or one of m groups where group[i] is the group that the i -th item belongs to and it's equal to -1 if the i -th item belongs to no group. The items and the groups are zero indexed. A group can have no item belonging to it. Return a sorted list of the items such that: The items that belong to the same group are next to each other in the sorted list. There are some relations between these items where beforeItems[i] is a list containing all the items that sh

Detailed Explanation

The problem requires sorting a list of 'n' items, each belonging to a group (0 to m-1, or -1 for no group), respecting dependencies specified by 'beforeItems'. The output should be a sorted list of items such that items within the same group are adjacent and all dependencies in 'beforeItems' are satisfied. If there are multiple valid solutions, return any of them. If no solution exists, return an empty list.

Solution Approach

The solution involves performing topological sorting twice: once for items and once for groups. First, any item not belonging to a group is assigned a new, unique group ID. Then, adjacency lists and in-degree counts are constructed for both items and groups. Edges between groups are created based on dependencies between items in different groups. Topological sorting is then performed on the item graph and the group graph. Finally, the result is built by iterating through the sorted groups and appending the items within each group in the order they appear in the sorted item list.

Step-by-Step Algorithm

Step 1: Assign unique group IDs to items that don't belong to any group (group[i] == -1). This ensures every item is part of a group.
Step 2: Build the item adjacency list (item_adj) and in-degree array (item_indegree) based on the 'beforeItems' list, considering only dependencies within the same group.
Step 3: Build the group adjacency list (group_adj) and in-degree array (group_indegree) based on the 'beforeItems' list, considering dependencies between items in *different* groups. Avoid adding duplicate edges between groups by using a set or similar data structure.
Step 4: Perform topological sorting on the item graph using the item adjacency list and in-degree array. If a cycle is detected, return an empty list.
Step 5: Perform topological sorting on the group graph using the group adjacency list and in-degree array. If a cycle is detected, return an empty list.
Step 6: Create a list of items for each group (items_by_group) based on the sorted item list.
Step 7: Iterate through the sorted group list, and for each group, append the corresponding items from the items_by_group list to the result list.
Step 8: Return the resulting sorted list of items.

Key Insights

Insight 1: Recognize that this is a topological sorting problem on two levels: items and groups. Items within the same group have dependencies, and groups themselves can have dependencies due to item dependencies between them.
Insight 2: Topological sorting is crucial. Represent dependencies as a directed graph and use topological sort to obtain a valid ordering, both for items and groups. The algorithm needs to detect cycles, as cycles indicate no possible ordering.
Insight 3: Handle the case where items don't belong to any group by assigning unique group IDs. This ensures that all items are part of some group and simplifies the sorting process.

Complexity Analysis

Time Complexity: O(n + m + E)

Space Complexity: O(n + m + E)

Topics

This problem involves: Depth-First Search, Breadth-First Search, Graph, Topological Sort.

Companies

Asked at: Citadel.

Sort Items by Groups Respecting Dependencies | LeetCode 1203

Hard

Depth-First Search Breadth-First Search Graph Topological Sort

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

Solution Code

Available in:

import java.util.*;
import java.util.List;

class Solution {
    public int[] sortItems(int n, int m, int[] group, List<List<Integer>> beforeItems) {
        
        class Helper {
            List<Integer> topologicalSort(List<List<Integer>> adj, int[] inDegree, int count) {
                Deque<Integer> q = new LinkedList<>();
                for (int i = 0; i < count; i++) {
                    if (inDegree[i] == 0) {
                        q.offer(i);
                    }
                }
                
                List<Integer> sortedOrder = new ArrayList<>();
                while (!q.isEmpty()) {
                    int u = q.poll();
                    sortedOrder.add(u);
                    
                    for (int v : adj.get(u)) {
                        inDegree[v] -= 1;
                        if (inDegree[v] == 0) {
                            q.offer(v);
                        }
                    }
                }
                
                return sortedOrder.size() == count ? sortedOrder : new ArrayList<>();
            }
        }
        
        Helper helper = new Helper();

        int group_id = m;
        for (int i = 0; i < n; i++) {
            if (group[i] == -1) {
                group[i] = group_id;
                group_id += 1;
            }
        }
        
        int num_groups = group_id;

        List<List<Integer>> item_adj = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            item_adj.add(new ArrayList<>());
        }
        int[] item_indegree = new int[n];
        
        List<List<Integer>> group_adj = new ArrayList<>();
        for (int i = 0; i < num_groups; i++) {
            group_adj.add(new ArrayList<>());
        }
        int[] group_indegree = new int[num_groups];
        
        Set<Pair<Integer, Integer>> group_edges = new HashSet<>();
        
        for (int i = 0; i < n; i++) {
            for (int prereq : beforeItems.get(i)) {
                if (group[prereq] == group[i]) {
                    item_adj.get(prereq).add(i);
                    item_indegree[i] += 1;
                } else {
                    if (!group_edges.contains(new Pair<>(group[prereq], group[i]))) {
                        group_adj.get(group[prereq]).add(group[i]);
                        group_indegree[group[i]] += 1;
                        group_edges.add(new Pair<>(group[prereq], group[i]));
                    }
                }
            }
        }
        
        List<Integer> sorted_items = helper.topologicalSort(item_adj, item_indegree, n);
        List<Integer> sorted_groups = helper.topologicalSort(group_adj, group_indegree, num_groups);

        if (sorted_items.isEmpty() || sorted_groups.isEmpty()) {
            return new int[0];
        }

        List<List<Integer>> items_by_group = new ArrayList<>();
        for (int i = 0; i < num_groups; i++) {
            items_by_group.add(new ArrayList<>());
        }
        for (int item : sorted_items) {
            items_by_group.get(group[item]).add(item);
        }
        
        List<Integer> resultList = new ArrayList<>();
        for (int g_id : sorted_groups) {
            resultList.addAll(items_by_group.get(g_id));
        }
        
        int[] result = new int[resultList.size()];
        for (int i = 0; i < resultList.size(); i++) {
            result[i] = resultList.get(i);
        }
        
        return result;
    }
    
    static class Pair<U, V> {
        private final U first;
        private final V second;

        public Pair(U first, V second) {
            this.first = first;
            this.second = second;
        }

        public U getFirst() {
            return first;
        }

        public V getSecond() {
            return second;
        }

        @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (o == null || getClass() != o.getClass()) return false;
            Pair<?, ?> pair = (Pair<?, ?>) o;
            return Objects.equals(first, pair.first) && Objects.equals(second, pair.second);
        }

        @Override
        public int hashCode() {
            return Objects.hash(first, second);
        }
    }
}

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Problem Info

Description

Problem 1203: Sort Items by Groups Respecting Dependencies - Detailed explanation and solution approaches.

Topics

Depth-First Search Breadth-First Search Graph Topological Sort

Companies

Citadel

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1203

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

8 steps

1
Step 1: Assign unique group IDs to items that don't belong to any group (group[i] == -1). This ensures every item is part of a group.
2
Step 2: Build the item adjacency list (item_adj) and in-degree array (item_indegree) based on the 'beforeItems' list, considering only dependencies within the same group.
3
Step 3: Build the group adjacency list (group_adj) and in-degree array (group_indegree) based on the 'beforeItems' list, considering dependencies between items in *different* groups. Avoid adding duplicate edges between groups by using a set or similar data structure.
4
Step 4: Perform topological sorting on the item graph using the item adjacency list and in-degree array. If a cycle is detected, return an empty list.
5
Step 5: Perform topological sorting on the group graph using the group adjacency list and in-degree array. If a cycle is detected, return an empty list.
6
Step 6: Create a list of items for each group (items_by_group) based on the sorted item list.
7
Step 7: Iterate through the sorted group list, and for each group, append the corresponding items from the items_by_group list to the result list.
8
Step 8: Return the resulting sorted list of items.

💡

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 + E)

💾

Space Complexity

O(n + m + E)

📝

Why?

The time complexity is determined by the topological sorts on the item and group graphs. The space complexity is dominated by the adjacency lists representing the graphs and auxiliary data structures like the queue in the topological sort.

📖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) based Topological Sort

Instead of using a queue-based BFS approach for topological sorting, DFS can be used. In DFS, we traverse the graph recursively, marking nodes as visited and adding them to the sorted order *after* visiting all their descendants.

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

Trade-offs: DFS can be more memory-efficient in some cases, but it can also lead to stack overflow issues with very large graphs. The iterative BFS approach is generally preferred due to its predictable memory usage.

✨

Key Insights

1Insight 1: Recognize that this is a topological sorting problem on two levels: items and groups. Items within the same group have dependencies, and groups themselves can have dependencies due to item dependencies between them.
2Insight 2: Topological sorting is crucial. Represent dependencies as a directed graph and use topological sort to obtain a valid ordering, both for items and groups. The algorithm needs to detect cycles, as cycles indicate no possible ordering.
3Insight 3: Handle the case where items don't belong to any group by assigning unique group IDs. This ensures that all items are part of some group and simplifies the sorting process.

📚

Key Terms Explained

For Students

Graph

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

Edge Cases

Edge case 1: Circular dependencies within items or groups. The topological sort detects cycles and returns an empty list, indicating no solution.
Edge case 2: Empty 'beforeItems' list. The topological sort correctly handles this case, as there are no dependencies, resulting in a valid ordering.
Edge case 3: An item belongs to a group, but 'beforeItems' contains a dependency to an item that is either in a non-existent group (group index out of bounds) or a non-assigned group (-1 before processing). The provided code assigns a new group ID to items that are initially -1, therefore it can handle this case.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to handle items that don't initially belong to any group (group[i] == -1). Assigning unique group IDs is essential.

✗Common mistake 2: Not detecting cycles in the item or group dependency graphs. If cycles exist, no valid ordering is possible, and the algorithm must return an empty list.

✗Common mistake 3: Failing to eliminate duplicate group dependency edges. Without eliminating duplicates the topological sort might get a wrong indegree count and provide a wrong result or return an empty list.

✗Common mistake 4: Incorrectly constructing the adjacency lists or in-degree arrays. A single mistake in these data structures can lead to incorrect topological sorting results.

✅

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: Task scheduling in software development projects, where certain tasks depend on others and must be completed in a specific order, considering team affiliations (groups).

🏦Real-world scenario 2: Course scheduling in a university, where some courses have prerequisites (dependencies) and students can be grouped based on their majors.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

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 2, 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

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

@TechLeadJul 7

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

@DevExpert92Jun 30

❓

Frequently Asked Questions

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 "Sort Items by Groups Respecting Dependencies"?

For this hard-level Depth-First Search problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether breadth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?