Smallest Sufficient Team

10 Prerequisite LeetCode Problems

Here are 10 problems which involve similar concepts useful for preparation:

78. Subsets: Understand how to generate all subsets of a set.
90. Subsets II: An extension to the problem above with repeated elements in the array.
39. Combination Sum: This problem focuses on finding all possible combinations that sum up to a target.
40. Combination Sum II: Extension to the problem above, but in this case, each number can only be used once.
46. Permutations: Understand how to generate all possible permutations.
47. Permutations II: Similar to above but dealing with duplicate values.
491. Increasing Subsequences: Find all possible increasing subsequences in an array.
17. Letter Combinations of a Phone Number: Combining possibilities of different characters.
77. Combinations: Generate all combinations of k elements out of n.
22. Generate Parentheses: Generate all valid combinations of parentheses.

These problems cover important concepts such as combinations, permutations, and subsets, which are very relevant to the problem at hand.

This problem involves concepts related to graph theory, bit manipulation, and dynamic programming. Here are 10 problems to prepare for this problem:

46. Permutations: This problem asks you to return all permutations of a set of distinct numbers.
78. Subsets: This problem asks you to return all possible subsets of a distinct set of numbers.
90. Subsets II: Similar to Subsets, but the input set may contain duplicates.
139. Word Break: This problem asks you to determine if a string can be segmented into a space-separated sequence of one or more dictionary words.
322. Coin Change: This problem involves finding the fewest number of coins needed to make up a certain amount, which is a dynamic programming problem.
377. Combination Sum IV: This problem asks to find the number of possible combinations that add up to a target number.
1286. Iterator for Combination: This problem is related to generating combinations of a string.
136. Single Number: This problem involves bit manipulation, where you have to find the number that appears only once in an array.
191. Number of 1 Bits: This problem asks you to count the number of ‘1’ bits in an unsigned integer, which involves bit manipulation.
338. Counting Bits: This problem asks you to count the number of 1’s in the binary representation of all numbers from 0 to n.

These cover dynamic programming, bit manipulation, and permutations and combinations, which are useful to tackle the problem you’re aiming for.

Clarification Questions

Is it possible for a person to have no skills? The problem statement mentions that every skill in people[i] is a skill in req_skills, but it doesn’t explicitly state if a person can have no skills.
Can there be multiple smallest sufficient teams? If yes, should we return all of them or just one of them? The problem statement says “Return any sufficient team of the smallest possible size,” which implies there might be multiple valid answers.
What should be the output if the list of required skills is empty? Should we return an empty list or include all people?
Are the skills of a person unique or can they be repeated in their list of skills?
Is the order of people in the output important? The problem says that the answer can be returned in any order, but it would be good to clarify this.
If a single person has all the required skills, should the output be an array with just their index?

These questions can help to clarify edge cases and the expected behavior of the solution in those scenarios.

Problem Analysis and Key Insights

The key insights from analyzing the problem statement of the “Smallest Sufficient Team” problem are:

The problem involves dealing with sets of skills and people. We need to find a team with the smallest size such that all required skills are covered by at least one person in the team.
Each person can have multiple skills, and each skill can be covered by multiple people.
The output is not a specific team, but any team that meets the requirements. This indicates that there could be multiple valid answers, and we just need to find one of them.
The problem guarantees that an answer exists. This means we don’t have to handle cases where no sufficient team can be found.
The number of required skills is relatively small (<= 16), suggesting that approaches involving iterating over all subsets of skills could be feasible.
The problem asks for the indices of the people in the solution, not the skills or people themselves. This indicates that the order of the people list and the indexing are important.
From the constraints, we see that each skill is a unique string, and each person has a unique set of skills. This could allow us to use data structures that require uniqueness, such as sets or dictionaries.
We can use a dynamic programming approach where the state is the set of skills, and the value is the smallest team that can cover these skills.
We can also leverage bit manipulation. As the length of required skills is not more than 16, we can represent the skill set of a person as a mask of 16 bits. This way, adding a person to a team or checking if a person’s skills are needed can be done using bitwise operations.
As we need to return the smallest team, we may need to keep track of the current smallest team during the process and update it whenever we find a smaller team.

Problem Boundary

The scope of the problem “Smallest Sufficient Team” is focused on efficient algorithms, data structures, and bit manipulation techniques for handling sets, especially when dealing with problems related to combinatorics and set cover.

The problem is about finding the minimum set of people who together possess all the skills required for a project. It is essentially a variant of the Set Cover problem, which is a well-known problem in computer science that falls into the category of combinatorics and is known to be NP-hard.

In practical terms, this problem could be encountered in project management or team formation scenarios where a project manager needs to assemble a team with a diverse set of skills to accomplish a specific project or task.

To solve this problem, it is necessary to understand concepts such as:

Set operations: As the problem deals with sets of skills and people, understanding how to manipulate sets, such as checking for membership, adding elements, or taking unions, is necessary.
Dynamic Programming: A dynamic programming approach can be used to keep track of the smallest team that can cover a particular set of skills.
Bit Manipulation: As the length of required skills is not more than 16, the problem can be optimized using bit manipulation. Each skill set can be represented as a 16-bit mask, and operations such as adding a person to a team or checking if a person’s skills are needed can be done using bitwise operations.
Data Structures: Efficiently storing and retrieving information about the people and their skills may require using advanced data structures, such as lists, sets, dictionaries, or priority queues.
Search Techniques: Finding the smallest sufficient team may involve searching through the space of all possible teams, which could be done using techniques like breadth-first search, depth-first search, or iterative deepening.
Combinatorics: Understanding the principles of combinatorics may be helpful for analyzing the complexity of the problem and thinking about possible solutions.

The boundary of the problem “Smallest Sufficient Team” can be established by understanding the constraints and requirements of the problem. Here is how you can define the boundaries:

Input:
- The input consists of a list of required skills req_skills, which can have up to 16 unique skills. Each skill is a string of lowercase English letters with a length of up to 16.
- The input also includes a list of people people, with up to 60 people. Each person is represented as a list of their skills. Each skill is a string of lowercase English letters with a length of up to 16.
- Every skill in people is a skill in req_skills.
- A sufficient team exists, meaning that there is at least one combination of people that covers all the required skills.
Output:
- The output is a list of indices representing the smallest sufficient team, i.e., the smallest set of people that covers all required skills. The indices can be returned in any order.
- The output indices refer to the people list in the input.
Problem Constraints:
- The problem involves set operations, string handling, list operations, and potentially bit manipulation and dynamic programming.
- It’s a combinatorial problem, which suggests that a brute force approach (trying all combinations of teams) may not be efficient for larger inputs due to the exponential increase in possibilities.
- The problem guarantees that a solution exists, so you don’t need to handle cases where no team covers all the required skills.
Objective:
- The main goal is to find the minimum set of people who together possess all the skills required for a project, represented by their indices.
- While doing so, the solution should be optimized to handle the maximum constraints of the problem efficiently.

By understanding these points, you can establish the boundaries of the problem. The solution should operate within these boundaries to be considered correct and efficient.

Problem Classification

Domain Classification: This problem falls within the domains of Combinatorics and Optimization.

‘What’ Components:

List of Required Skills (req_skills): A list of distinct strings representing unique skills required for a project.
List of People (people): A list of people, with each person represented as a list of their skills. Each person’s skills are a subset of the required skills.
Sufficient Team: A set of people from the given list such that for every required skill in req_skills, there is at least one person in the team who has that skill.
Objective: Find the smallest sufficient team, represented by the indices of each person in the ‘people’ list. The solution can be in any order. It is guaranteed that a solution exists.

Further Classification:

Based on the ‘What’ components and constraints of the problem, this problem can be further categorized into the following:

Combinatorial Problem: The problem involves finding a valid combination of people to form a team that satisfies the given requirements. This falls under the umbrella of combinatorics as there are multiple ways to choose a team from the given list of people, and we need to find a specific combination (smallest team that covers all required skills).
Optimization Problem: The problem is not just about finding a valid team but finding the ‘smallest’ valid team. Therefore, it’s an optimization problem where we want to minimize the size of the team.
Subset Problem: The problem involves finding a subset of people who possess all the required skills. This categorization comes from the requirement that each person’s skills are a subset of the required skills and we need to find a sufficient team (subset of ‘people’) that includes all the required skills.
Bit Manipulation/Dynamic Programming (Potential): Given the constraints (especially the limited number of skills), one may use Bit Manipulation and/or Dynamic Programming to efficiently solve this problem. This, however, is a ‘how’ aspect that depends on the solution approach, but worth mentioning in classification due to the problem’s nature.

Distilling the Problem to Its Core Elements

The fundamental concept this problem is based upon is combinatorial search. This problem is about finding the smallest subset from a larger set (in this case, people with their unique skills) which satisfies a given condition (in this case, covering all required skills).

I would describe this problem as follows: Imagine you are managing a project and you have a list of required skills for that project. You also have a list of potential team members, and for each person, you know what skills they have. Your task is to find the smallest possible team where all the required skills are covered by at least one team member.

The core problem we are trying to solve here is identifying the smallest group of people that together have all the required skills. In other words, we are trying to find a minimum covering set from the people set such that all required skills are covered.

Breaking down the problem into its key components:

Skills required for the project
People and the skills they have
Need to ensure that all required skills are covered by at least one person in the team
The team size should be as small as possible

The minimal set of operations we need to perform to solve this problem:

For each person, keep track of which skills they have
Start with an empty team
Search through all the possible combinations of people to find a team that covers all required skills
Out of all such teams, choose the one with the smallest size.

Visual Model of the Problem

Visualizing this problem might best be done by using a matrix or a graph.

Matrix: We can create a matrix where the rows represent people and the columns represent skills. Each cell in the matrix represents whether a person (row) has a particular skill (column). A “1” indicates the person has that skill, while a “0” indicates they do not. This would allow us to quickly see the distribution of skills among people.

    java  nodejs  reactjs
0     1       0        0
1     0       1        0
2     0       1        1

Graph: Another way to visualize this problem is with a bipartite graph, where one set of nodes represents people and the other set represents skills. Edges connect people to the skills they possess. The problem then becomes one of finding the smallest subset of people-nodes such that every skill-node is connected to at least one node in that subset.

Person Nodes  Skills Nodes

0 ------------ java

1 ------------ nodejs

2 ------------ nodejs, reactjs

These visualizations can help us understand the problem better and potentially guide us in identifying a suitable approach to solve it.

Problem Restatement

In a project, there’s a list of mandatory skills, and we have a roster of people, each having a specific set of skills. We need to form a team in such a way that every required skill is covered by at least one team member. A person’s participation in the team can be represented by their index in the list. We need to find the smallest possible team that meets the project’s requirements.

We are allowed to provide the answer in any order, and we are guaranteed that at least one solution will exist. A skill or a person’s skillset is represented as a list of lowercase English words, and all skills are unique.

Constraints to consider:

The list of required skills can have 1 to 16 unique skills, and each skill is a string of 1 to 16 lowercase English characters.
The list of people can have 1 to 60 unique individuals, and each person can have 0 to 16 unique skills.
Every skill possessed by a person is guaranteed to be a required skill.

Our goal, therefore, is to find any smallest sufficient team that can cover all required skills.

Abstract Representation of the Problem

Given a set of required elements (R) and a collection of sub-sets (P), each containing some elements of R, find the smallest composite set (T) that includes all the elements of R at least once. This composite set T is a sub-collection of P. The output should be the indices of the sub-sets in P that form the smallest composite set T. Note that there are several constraints on the sizes of the set R and the sub-sets in P.

The key elements in this problem are:

The set of required elements (R)
The collection of subsets (P)
The smallest composite set that includes all elements of R (T)

The task is to map elements from R to P, form T, and find the smallest possible T. The real-world details such as “skills”, “people”, “project”, etc., are not necessary in the abstract representation.

This problem belongs to the category of combinatorial optimization and set cover problems, which are common in computer science and operations research. It requires the identification of the smallest number of sub-sets from a collection that contain all the elements of a required set.

Terminology

There are a few key terms and concepts that play a critical role in understanding and solving this problem:

Set Cover Problem: This is a classical question in computer science, combinatorics, and complexity theory. It asks whether there is a sub-collection of subsets, which could cover all elements in the universal set. This problem is a variant of the set cover problem, where we need to find a minimal subset of ‘people’ that can cover all the ‘required skills’. Understanding this problem class helps to provide a high-level approach to solving the problem.
Combinatorial Optimization: This is a topic in mathematics and computer science that consists of finding the best, optimal solution from a finite set of possibilities. The “Smallest Sufficient Team” problem is a kind of combinatorial optimization problem where we are looking for the smallest team (set of indices) which covers all the required skills.
Bitmasking and Dynamic Programming: These are potential methods for solving this problem. Bitmasking is a technique used to store more than one piece of information in a single integer. Dynamic programming is a method for solving complex problems by breaking them down into simpler subproblems. It is useful in this problem for remembering solutions to subproblems to avoid redundant computation.
Skills and People (in the context of this problem): “Skills” refers to the unique abilities that are required for the project (elements of the universal set). Each “person” has a subset of these skills. The goal is to find the smallest set of people (indices) that cover all the required skills.

Understanding these terms and concepts is crucial to identifying the type of problem at hand, which in turn aids in determining potential solution strategies.

Problem Simplification and Explanation

Let’s think about the problem in terms of a party planning. Suppose you are organizing a party, and you have a specific list of tasks that need to be accomplished for the party to be successful. These tasks are analogous to the “required skills” in the problem.

Now, you have a group of friends, and each friend is capable of doing a certain set of tasks. Some friends might be able to handle multiple tasks (cook, DJ, etc.), while others may be good at only one task (maybe someone is an excellent DJ but can’t cook). These friends are analogous to the “people” in the problem, and the tasks each friend can do are the “skills” they possess.

The goal of the problem is to invite the minimum number of friends to your party such that all tasks will be taken care of. That is, every task (skill) on your list should be covered by at least one friend (person).

This is equivalent to the problem where we need to find the smallest team such that every required skill is possessed by at least one team member.

Key concepts here include:

Sets and Subsets: Each person can be thought of as a set of skills, and the required skills form the universal set. We need to find the minimum number of these subsets (people) that cover the universal set (required skills).
Combinatorial Optimization: We are looking for the best solution (smallest team) from all possible combinations of people.

This analogy should help you understand the problem better. As with party planning, it’s not just about getting all the tasks covered, but also doing so with the minimum number of friends (to maintain social distancing, for instance!). Similarly, in the problem, we are trying to cover all skills using the minimum number of people.

Constraints

Let’s consider the problem statement and constraints to identify specific characteristics or conditions we can use:

Limited Number of Required Skills: The problem guarantees that the length of req_skills is between 1 and 16. This is a small range and we can exploit it by using techniques like bitmasking, which is a way to represent subsets of a small set using binary numbers. For instance, if there are 3 required skills, we can represent any subset of these skills as a number between 0 and 2^3 - 1 (where each bit represents the presence or absence of a skill).
All Skills Are Unique: All the strings in req_skills are unique, which means that we can create a one-to-one mapping of the skills to indices in an array or bits in a bitmask. This simplifies the problem and ensures that each skill is represented only once.
Guaranteed Sufficient Team Exists: It’s guaranteed that a sufficient team exists. This removes the need to check for cases where it’s impossible to form a team with the required skills, simplifying the problem.
Skills to People Mapping: Every skill in people[i] is a skill in req_skills. This means that we can ignore any skills that a person may have that are not in the req_skills list. This might reduce the complexity of people[i] and simplify our problem.

These are specific characteristics and conditions that can be exploited for an efficient solution.

Analyzing the constraints provides several key insights that can help guide our approach to solving this problem:

Small Number of Skills: The maximum length of the req_skills array is 16. This is a relatively small number that allows us to consider approaches like bitmasking which is more suitable for smaller sets.
Unique Skills: The uniqueness of each skill in the req_skills array simplifies our problem as we do not need to consider duplicate skills.
People-Skills Mapping: The fact that every skill in people[i] is a skill in req_skills means that we don’t have to deal with irrelevant skills. This helps reduce the complexity of our problem.
Guaranteed Solution: The fact that a solution is guaranteed to exist eliminates the need to check for impossible cases. We know that there will always be a way to form a team that possesses all the required skills.
Array Index Representation: The solution needs to represent the people by the indices. This hints that we might need to keep track of indices during our computation, and hints towards the use of dynamic programming, where we usually build up a solution using the results of smaller subproblems.
Return Any Solution: The problem does not demand a specific team but any team of the smallest possible size. This gives us the flexibility to stop once we find a valid solution, and we do not need to traverse all possible solutions.

These insights can help us to form an efficient strategy to solve the problem.

Case Analysis

Let’s consider additional examples:

Minimal Case:
```
Input: req_skills = ["java"], people = [["java"]]
Output: [0]
```
In this case, we have only one person and one required skill, and the person possesses the required skill. The minimal sufficient team is just this person.
Multiple People with Same Skill:
```
Input: req_skills = ["java"], people = [["java"], ["java"]]
Output: [0] or [1]
```
In this case, we have two people and both possess the required skill. The solution can be either [0] or [1]. This case shows that multiple correct outputs are possible, and we need to ensure our algorithm can handle such situations.
One Skill Multiple Times in People List:
```
Input: req_skills = ["java", "python"], people = [["java"], ["java", "python"]]
Output: [1]
```
In this case, the first person only has the “java” skill, but the second person has both the “java” and “python” skills. Therefore, we only need the second person for the smallest sufficient team.
All People Required:
```
Input: req_skills = ["java", "python", "javascript"], people = [["java"], ["python"], ["javascript"]]
Output: [0, 1, 2]
```
Here, each person has a unique skill, so we need all of them to form a sufficient team. This highlights the case where all people are required.
Skills Distributed Across People:
```
Input: req_skills = ["java", "python", "javascript"], people = [["java", "python"], ["python", "javascript"], ["java", "javascript"]]
Output: [0, 1] or [0, 2] or [1, 2]
```
In this case, each person has two skills, and every required skill is covered by at least two people. There are multiple valid smallest teams in this case, any of which can be an acceptable output.

Edge Cases:

No Skills Required:
```
Input: req_skills = [], people = [["java"], ["python"], ["javascript"]]
Output: []
```
In this edge case, no skills are required so no people are needed. The output should be an empty list.
No People Available:
```
Input: req_skills = ["java"], people = []
Output: Error or special value
```
In this edge case, there are no people available but there are skills required. This goes against the problem’s guarantee that a sufficient team always exists. According to the problem statement, this should never happen, so if it does, we should return an error message or a special value to indicate that no sufficient team can be formed.

Analyzing different cases gives us several key insights:

Multiple Sufficient Teams: There can be more than one valid answer. As seen in our “Multiple People with Same Skill” and “Skills Distributed Across People” cases, there can be multiple ways to form a smallest sufficient team. The solution needs to account for this and doesn’t necessarily need to find all possible teams, just one of them.
Redundancy in Skills: Some people might possess the same skills as others and therefore can be redundant. As seen in our “One Skill Multiple Times in People List” case, if one person has all the skills that another person has and more, we only need the person with more skills. We need to design our solution in a way that it doesn’t include redundant people.
All People Needed: In some cases, we might need all people to form a sufficient team. Our “All People Required” case is an example of this. The solution needs to handle such cases.
Edge Cases Handling: Although not directly related to the core problem, edge cases like “No Skills Required” or “No People Available” need special handling. This ensures our solution is robust and works under all conditions. It’s important to note that “No People Available” is not supposed to happen according to the problem constraints, and if it does, it means there’s an issue with the input data.

These insights give us an idea of the types of scenarios our solution needs to handle, and what strategies might be effective. For example, we might use a backtracking algorithm to explore different combinations of people and stop once we find a sufficient team. Or we might use a dynamic programming approach to avoid redundancy and minimize the size of the team.

Identification of Applicable Theoretical Concepts

Several mathematical and algorithmic concepts can be applied to this problem to make it more manageable:

Bit Manipulation: The problem statement can be reduced to the subset sum problem, which is a classic computer science problem often solved using dynamic programming and bit manipulation. We can use a bit to represent whether a skill is included in a certain subset or not. This will significantly decrease the time complexity because we are compressing a lot of information (whether each skill is included or not) into a single integer.
Dynamic Programming (DP): Dynamic Programming is a method for solving complex problems by breaking them down into simpler subproblems. It is applicable when the subproblems are not independent, i.e., the solution to one subproblem may involve solutions to other subproblems. We can use DP to store the smallest team that can cover a certain subset of skills.
Backtracking: Backtracking is an algorithmic technique for solving problems recursively by trying to build a solution incrementally, one piece at a time. If a solution we are building turns out not to be satisfactory, backtracking allows us to stop the building process and go back to a previous step, from where we can try a different path. Backtracking could be used to go through all possible combinations of team members.
Breadth-First Search (BFS): BFS is an algorithm for traversing or searching tree or graph data structures. It starts at the root (in the case of a tree) or an arbitrary node of a graph, and explores all of the neighbor nodes at the present depth before moving on to nodes at the next depth level. BFS can be used to find the smallest team by expanding each state (set of skills) and using a queue to remember states to be processed.

By using these concepts, we can work towards an efficient solution for the problem.

Simple Explanation

Let’s think of it in terms of organizing a school project:

Imagine you are a school teacher, and you want to assign a group project to your students. The project requires certain skills. For example, you might need someone who can draw well, someone who can write well, someone who can do good research, and so forth. These are like the “required skills” in the problem.

Now, you have a class of students, and each student has their own set of skills. For instance, some students might be good at drawing, some might be good at writing, some might be good at both, and so on. These are like the “people” in the problem, and their skills are like the “skills” that each person has.

Your task is to form a team for the project such that all the required skills are covered. However, to make sure the workload is manageable, you want the team to be as small as possible. That means if some students have overlapping skills, you might not need to include all of them in the team.

For example, if two students are both good at drawing and writing, but you only need one person for drawing and one for writing, you don’t need to include both students in the team. Instead, you can just pick one of them.

So, in summary, you’re trying to form a “sufficient team” with the smallest possible number of students, where a “sufficient team” is a team that has all the required skills for the project. This is what the problem is asking you to do.

Problem Breakdown and Solution Methodology

This problem can be thought of as a variation of the classic Knapsack problem in computer science, which is often solved using dynamic programming. In this case, however, instead of trying to maximize the value of the items put into the knapsack, we’re trying to minimize the number of people in our team.

Here’s a high-level breakdown of how we might approach this problem:

Convert skills to binary representation: Since the maximum number of skills is 16, we can represent all skill combinations using 16 bits. This means each skill can be represented as a unique bit in a 16-bit integer. We can map each skill to a unique bit, and then we can represent a person’s skills as an integer. The benefit of this approach is that we can use bitwise operations to efficiently manipulate and compare skills.
Create a dynamic programming table: We can create a dynamic programming table dp where dp[i] stores the smallest team that can cover the skills represented by the integer i. The size of this table will be 2^n, where n is the number of skills.
Initialize the dynamic programming table: Initially, we can set dp[i] to an empty team for all i except for dp[0], which represents the case where no skills are needed. For dp[0], we can set it to a team with no people, which is the correct answer for this case.
Update the dynamic programming table with each person’s skills: For each person, we update the dp table. For each integer i, if a team is already found for the skills represented by i, we consider adding the current person to this team to cover more skills. If adding this person results in a smaller team for the new skills, we update the dp entry for the new skills.
Return the smallest sufficient team: Finally, the smallest team that can cover all required skills is stored in dp[req_skills_int], where req_skills_int is the integer representation of the required skills. We can return this team as the result.

This approach will work well because it systematically considers all combinations of people and skills and always keeps the smallest team for each combination. As the problem parameters change, the dynamic programming table will adapt to always find the optimal solution.

Let’s look at an example for a better understanding. Consider this case:

Input: req_skills = ["java","nodejs"], people = [["java"],["nodejs"],["java","nodejs"]]

First, we map the skills to binary representation:

"java" -> 1 (01 in binary)
"nodejs" -> 2 (10 in binary)

Then the people will be represented as:

person1 -> 1 (has "java")
person2 -> 2 (has "nodejs")
person3 -> 3 (has "java" and "nodejs", 3 is 11 in binary)

The required skills are [“java”, “nodejs”], which is represented by 3 (11 in binary).

Then we initialize our dp table:

dp[0] = []
dp[1] = Inf
dp[2] = Inf
dp[3] = Inf

Next, we update the dp table with each person’s skills:

For person1, we update dp[1] and dp[3], the result is:
dp[1] = [0] (person1 can cover "java")
dp[2] = Inf
dp[3] = [0] (we can use person1 to cover "java" and add more skills)

For person2, we update dp[2] and dp[3], the result is:
dp[1] = [0]
dp[2] = [1] (person2 can cover "nodejs")
dp[3] = [0, 1] (we can use person1 to cover "java" and person2 to cover "nodejs")

For person3, we update dp[3], the result is:
dp[1] = [0]
dp[2] = [1]
dp[3] = [2] (we can use person3 to cover both "java" and "nodejs", which results in a smaller team)

Finally, we return the smallest sufficient team, which is dp[3] = [2].

This is a complex problem, and this approach might not be immediately obvious. However, once understood, it provides a robust solution and can be adapted to handle a variety of related problems.

Inference of Problem-Solving Approach from the Problem Statement

Skills and People: The problem revolves around managing skills and people, making it a typical scenario for combinatorial optimization problems. Recognizing this, one might be led to consider classical methods used for such problems, like dynamic programming.
Smallest Sufficient Team: The goal is to find the smallest team that possesses all required skills. This is akin to minimizing cost or size in optimization problems, which suggests techniques like dynamic programming or greedy algorithms. Given the constraints, dynamic programming is more appropriate here.
Return any sufficient team: It’s noted that the solution does not need to be unique; any valid solution will suffice. This indicates that there might be multiple correct answers, which is often the case in dynamic programming problems. This suggests that an optimal substructure may exist, further supporting the use of dynamic programming.
Every skill in people[i] is a skill in req_skills: This constraint simplifies the problem by ensuring we only need to consider the skills listed in req_skills when building our team. We don’t need to account for any additional skills a person might have. This enables us to use a compact binary representation for the skills.
The given upper bounds for the array lengths (1 <= req_skills.length <= 16, 1 <= people.length <= 60): These bounds are quite small, which implies that an algorithm with exponential time complexity might still work within a reasonable amount of time. Specifically, a dynamic programming solution with time complexity O(n*2^n) (where n is the number of skills) would be feasible.
It is guaranteed a sufficient team exists: This constraint tells us that a solution is always possible. We don’t need to account for scenarios where it might be impossible to form a team with all the required skills, simplifying our problem-solving approach.

The key terms and constraints of this problem guide us towards a dynamic programming solution. By taking note of these terms and understanding their implications, we can tailor our approach to leverage the specific structure of the problem.

This problem involves finding an optimal solution (smallest team) over a potentially large search space (different combinations of people with skills), which makes it a good candidate for Dynamic Programming (DP). Here are some reasons why we can infer DP might be a good approach:

Overlapping subproblems: In the problem, we are looking for teams that satisfy the required skills. This can involve checking overlapping subsets of people, where one person might be a part of multiple potential teams. This points towards overlapping subproblems, a key property that makes a problem suitable for DP.
Optimal substructure: The problem asks for an optimal team size. If we know the optimal team for a subset of skills, we can use this information to build an optimal team when we add an extra skill. This property is known as optimal substructure, another key characteristic of DP problems.
Small bounds on input size: The bounds on the length of req_skills (up to 16) are small enough that we can consider all possible subsets of skills (which is 2^16 possibilities). This allows us to represent each combination as a binary number and apply a DP approach, where the state of our DP table represents a binary number corresponding to a subset of skills.

So, why binary representation?

When we consider each person in the team as having or not having a certain skill, we can represent the set of skills a person has as a binary number, where each bit corresponds to a skill. This is very useful given that there are up to 16 skills - a number that fits comfortably in an integer variable for most programming languages. This binary representation allows us to easily work with different combinations of skills using bitwise operations.

In conclusion, the nature of the problem (looking for an optimal solution and dealing with overlapping subsets) and the problem constraints (particularly the small number of required skills) suggest that a DP approach using binary representation could be an effective way to solve this problem.

Simple Explanation of the Proof

Let’s start by discussing the approach and then move to proving its correctness.

First, we can represent each skill as a bit in a binary number, giving each required skill a unique position within this number. This means that a subset of skills can be represented as an integer. For example, if our skills are [“java”,“nodejs”,“reactjs”], we might assign “java” to the first bit, “nodejs” to the second, and “reactjs” to the third. Then, a person with skills [“java”, “reactjs”] would be represented by the binary number 101, or 5 in decimal.

The algorithm works by building a DP table where the index is a binary number representing a subset of skills, and the value at that index is the smallest team that can cover those skills.

We start by initializing our DP table with a length of 2^n, where n is the number of skills, and setting all values to null or an equivalent placeholder. The value at index 0 (representing no skills) is set to an empty team.

Then, for each person, we generate their skill mask (a binary number representing their skills) and iterate through all existing states in our DP table. If adding this person to the team for a certain state results in a new state, we compare the size of the new team with the size of the team already stored at that new state in the DP table, and keep the smaller team.

The proof of correctness for this algorithm comes down to proving that it correctly implements a breadth-first search (BFS) over the state space of the problem, where each state is a subset of skills.

Breadth-first nature: The algorithm goes through the people list in order, considering each possible team that can be formed by including this person in all teams that could be formed by people before them. This ensures that smaller teams are considered before larger ones, fulfilling the breadth-first property.
Correctness of the transition function: The transition function, which is the operation of going from one state to another by including a person in the team, is correct by definition. If a person has a certain skill, including them in a team certainly gives that team the ability to cover that skill.
Optimality: The algorithm stores the smallest team that can cover a set of skills at each step. By considering smaller teams first (due to the breadth-first nature), we ensure that when we first find a team that can cover a set of skills, it is the smallest such team.

So, this algorithm is guaranteed to find the smallest sufficient team to cover all skills, thus proving its correctness.

Let’s break down the approach into more granular steps:

Represent Skills as Binary Bits: Map each skill in the required skills list to a unique bit in a binary number. This gives us an efficient way to represent any subset of skills.
Initialize Dynamic Programming (DP) Table: Create an array with a length of 2^n, where n is the number of required skills. This array (let’s call it dp) will store the smallest team that can cover the set of skills represented by the index in the array. Initialize all elements of this array to null or a placeholder value, except for dp[0], which should be set to an empty list, as no people are needed to cover no skills.
Populate DP Table: Iterate through each person in the people list and create a binary number (mask) that represents the set of skills they have. Then, iterate through all existing states (from 0 to the current length of the DP array). If adding this person to the team for a certain state (oldState) results in a new state (newState), compare the size of this new team with the team (if any) already stored at dp[newState]. If the new team is smaller, or dp[newState] is still null, update dp[newState] with the new team.

The process from step 3 is a repeatable pattern, where we’re trying out each person for every possible combination of people we’ve considered before. This is essentially what makes this problem suitable for a dynamic programming approach.

In terms of independent parts of the problem, step 1 (representing skills as binary bits) can be done independently from the rest of the steps. The initialization of the DP table (step 2) is also somewhat independent, though it does depend on the number of required skills determined in step 1.

Once the DP table is populated, the smallest sufficient team can be found at dp[(2^n) - 1], as this represents the state where all skills are covered. The -1 is needed because indices are 0-based.

This step-by-step process should provide a detailed roadmap to implementing the solution.

Solution Approach and Analysis

Let’s start by breaking down our approach into more manageable steps:

Create a Skills to Bit Map: Start by creating a unique mapping between the required skills and bits in a binary number. This is akin to giving each skill a unique ID, which we can use to efficiently represent any subset of skills.
Transform People’s Skills to Bit Representation: Transform each person’s skills into the binary representation we just defined. Essentially, for each person, we are replacing their list of skills with a binary number, where a bit at position i is 1 if they have the i-th skill in the required skills list.
Initialize the DP Table: Initialize a DP table, which is an array of size 2^n where n is the number of required skills. The DP table, named dp here, is used to keep track of the smallest team that can cover a specific set of skills. Initialize all elements of the array to a maximum value or placeholder, except dp[0] which is set to an empty team (since no people are required to cover no skills).
Populate the DP Table: For each person, represented by their binary skills mask, iterate through all states in the DP table. If this person can contribute a skill not yet in the current state (resulting in a new state), check if adding this person to the team for the current state would result in a smaller team for the new state. If so, update the DP table accordingly.

Let’s make this more concrete by using an example:

Say we have the following inputs:

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req_skills = ["java", "nodejs", "reactjs"]
people = [["java"], ["nodejs"], ["nodejs", "reactjs"]]

Map each skill to a unique bit:

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skills_to_bit = {"java": 1, "nodejs": 2, "reactjs": 4}

Convert people’s skills into a binary mask:

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people_skills = [1, 2, 6]  # 6 = 2 (nodejs) + 4 (reactjs)

Initialize the DP table:

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dp = [[], None, None, None, None, None, None, None]

Populate the DP table:

Iterating through each person and each state, we’d eventually get:

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dp = [[], [0], [1], [0, 1], [2], [0, 2], [1, 2], [0, 2]]

Here, dp[7] (or dp[2^3 - 1]) gives us the smallest team that can cover all skills, which is [0, 2].

This dynamic programming approach helps us manage the complexity of the problem by breaking it down into smaller subproblems, and then combining their solutions to solve the larger problem. The binary representation of skills is a crucial part of the approach, allowing us to leverage bit manipulation to efficiently handle skill sets.

Identify Invariant

In the context of this problem, the invariant would be that for any given state in our DP table, dp[i] always holds the smallest sufficient team that can cover the skills represented by i (in binary format). This invariant holds at all times during the execution of our algorithm.

The reason it’s called an invariant is because it does not change as we iterate over the people and states - for any given state i, once we set dp[i], it always represents the smallest team that can cover the skills corresponding to i, even as we continue to iterate over the remaining people and states.

This invariant is crucial to the correctness of our approach. By maintaining this invariant, we ensure that when we finally look at dp[(1<<n_skills) - 1] (where n_skills is the number of required skills), it will indeed hold the smallest team that can cover all the required skills.

Identify Loop Invariant

A loop invariant is a property or condition that is initially true before an iteration or loop begins and remains true after each iteration of the loop. It’s used to prove the correctness of a program.

In the context of the “Smallest Sufficient Team” problem, there are nested loops, with one iterating over the “people” and the inner one iterating over the “states”.

Here is the loop invariant for this problem:

For the outer loop (iterating over “people”): After processing the i-th person, for each state j, dp[j] contains the smallest sufficient team from the first i people that can cover the skills represented by j.
For the inner loop (iterating over “states”): After processing the j-th state, dp[j | cur] contains the smallest sufficient team from the first i people that can cover the skills represented by j | cur, where cur is the skill set of the i-th person represented in binary format.

These invariants help ensure that the dynamic programming table is being filled out correctly and, once all people have been considered and all states have been processed, the final solution (i.e., the smallest sufficient team that can cover all skills) can be obtained from dp[(1<<n_skills) - 1].

In the context of computer science and problem-solving, an invariant refers to a condition or property that remains unchanged throughout the execution of a program or a particular block of code. A loop invariant, as a subset of this concept, is a specific type of invariant that applies to loops.

So, all loop invariants are invariants (since they define a property that remains unchanged during the execution of a loop), but not all invariants are loop invariants (since some invariants might apply to the whole program or blocks of code that are not loops).

In the “Smallest Sufficient Team” problem, the loop invariant that we defined previously refers specifically to the loops iterating over “people” and “states”. This invariant property holds true before, during, and after each iteration of the loop. Therefore, it is a loop invariant.

The overall invariant for this problem could be that at any point in time, the dynamic programming array dp[] holds the smallest sufficient team that can cover the skills represented by the index of dp[], considering the people processed so far. This is a more general property of the program that holds beyond just the loops.

In conclusion, for this problem, the loop invariant is a type of invariant, specifically tied to the loops in the problem. The overall invariant is a broader concept and encompasses the entire execution of the algorithm.

Thought Process

The problem involves selecting a minimum set of people who can cover all the required skills. At first glance, this problem seems to be a variant of the classic NP-hard problem called “Set Cover”, which is known for its combinatorial nature and no known efficient exact solution for large inputs.

However, we can exploit some characteristics of this problem to solve it more efficiently. The constraints state that the maximum number of required skills is 16, which is a small number and can be represented by an integer (as each bit in the integer can represent whether a skill is present or not). This is a big hint that suggests that we can use dynamic programming and bitwise operations to solve this problem.

Here is a step by step breakdown:

First, create a mapping for each skill to a unique integer. Since there are at most 16 skills, we can assign a unique bit in a 16 bit integer to each skill.
Then, for each person, calculate a binary mask that represents the skills that this person has. This can be done by setting the bit corresponding to each skill that the person has.
Next, define a dynamic programming array, dp[], where dp[i] represents the minimum team that can cover the skills represented by the binary representation of i. Initially, all dp[i] for i > 0 are infinity (as no team can be formed yet), and dp[0] is 0 (no skills needed).
Now, iterate over each state from 0 to (1 << len(req_skills)) - 1 (all possible combinations of skills). For each state, also iterate over all people. If a person has a skill that is not covered by the current state, try to update the state that includes the person’s skills with a team that includes this person (if it’s smaller).
At the end of the dynamic programming process, dp[(1 << len(req_skills)) - 1] will hold the smallest team that can cover all skills.

Here is the Python code for the above steps:

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class Solution:
    def smallestSufficientTeam(self, req_skills: List[str], people: List[List[str]]) -> List[int]:
        skill_to_bit = {skill: i for i, skill in enumerate(req_skills)}
        n = len(req_skills)
        dp = {0: []}
        
        for i, person in enumerate(people):
            his_skills = 0
            for skill in person:
                if skill in skill_to_bit:
                    his_skills |= 1 << skill_to_bit[skill]
            
            for skillset, team in list(dp.items()):
                with_him = his_skills | skillset
                if with_him == skillset: continue
                if with_him not in dp or len(dp[with_him]) > len(team) + 1:
                    dp[with_him] = team + [i]
        
        return dp[(1 << n) - 1]

The key insights in this problem are recognizing the hints to use dynamic programming and binary masks to represent sets of skills, and designing the DP transition function correctly. This function essentially considers each person one by one and tries to improve on the smallest team for each skill set with and without this person.

This kind of approach, where we iterate over states and try to improve our solution by considering one more element (in this case, a person), is a common technique in dynamic programming problems.

Establishing Preconditions and Postconditions

Parameters:
- The method takes two parameters, req_skills and people.
- req_skills is a list of strings, where each string represents a skill that is required. people is a list of lists of strings, where each sub-list represents a person and the strings in it are the skills that the person has.
- These parameters represent the required skills for a project and the available people with their respective skills.
Preconditions:
- There is no specific precondition regarding the state of the program before this method is called.
- Constraints on the input parameters are:
  - The number of required skills (req_skills) is between 1 and 16.
  - The length of each skill in req_skills is also between 1 and 16.
  - All skills in req_skills are unique.
  - The number of people (people) is between 1 and 60.
  - The number of skills each person has is between 0 and 16.
  - Every skill in each person’s skill set is a skill in req_skills.
Method Functionality:
- This method is expected to find a smallest sufficient team from the given people to cover all required skills. The team is represented by the indices of people in the original list.
- It interacts with the inputs by processing the skills of each person and updating a dynamic programming table to keep track of the smallest team that can cover each possible set of skills.
Postconditions:
- After the method has been called and returned, the return value is a list of integers representing the indices of people who form a smallest sufficient team to cover all required skills.
- The return value indicates a solution to the problem - a smallest sufficient team of people that covers all required skills.
- The method does not have any side effects, it does not modify the input lists or any global state.
Error Handling:
- The method assumes that the preconditions are met, i.e., the input parameters adhere to the constraints specified in the problem statement.
- If the preconditions are not met, the behavior of the method is not defined. It might give incorrect results, run into an infinite loop, or raise an exception (like an out-of-bounds exception if a skill does not exist in req_skills). It is the responsibility of the caller to ensure that the inputs are valid.
- There’s no specific error handling or exception mechanism built into the code, as it is generally assumed for this kind of problem that the inputs are valid.

Problem Decomposition

Problem Understanding:
- The problem is about forming a smallest team from a list of people so that all the required skills for a project are covered by the team. Each person has a set of skills and we are required to return the indices of the people who form the smallest sufficient team.
Initial Breakdown:
- Major parts of the problem include:
  - Understanding the relationship between skills and people
  - Tracking all possible combinations of skills and their corresponding smallest team
  - Identifying the smallest team that covers all required skills
Subproblem Refinement:
- The tasks involved in the subproblems are:
  - Mapping each skill to a unique bit in the bitmask
  - Mapping each person to a bitmask representing the skills they have
  - Initializing the dynamic programming table to keep track of the smallest team for each combination of skills
  - Updating the dynamic programming table by iterating over all people and possible skill combinations
Task Identification:
- The repeated task here is the bit manipulation to represent the skill sets and updating the dynamic programming table.
Task Abstraction:
- Each task is abstracted to the level where it makes sense in the context of the problem and can be reused for all inputs satisfying the given constraints.
Method Naming:
- The method can be named findSmallestSufficientTeam to indicate its purpose.
Subproblem Interactions:
- First, we need to map each skill and each person to a unique bitmask. Then we initialize the dynamic programming table. After that, we iterate over all people and possible skill combinations to update the dynamic programming table. Finally, we extract the smallest team that covers all required skills from the dynamic programming table. The order of these tasks is important and each task depends on the results of the previous tasks.

From Brute Force to Optimal Solution

Brute Force Approach:

One naive approach to this problem would be to generate all possible subsets of people and check which ones cover all the required skills. The smallest subset that covers all the required skills would be our answer.

Pseudocode:

function findSmallestSufficientTeam(req_skills, people):
    min_team_size = infinity
    min_team = []
    for each subset in all possible subsets of people:
        if subset covers all req_skills:
            if size of subset < min_team_size:
                min_team_size = size of subset
                min_team = subset
    return min_team

However, this approach is highly inefficient. The number of possible subsets of people is 2^n, where n is the number of people. For each subset, we need to check if it covers all the required skills, which takes O(m) time, where m is the number of required skills. So the overall time complexity is O(m * 2^n), which is impractical for large inputs.

Optimization:

The above problem can be optimized using dynamic programming and bit manipulation.

Dynamic Programming:
Dynamic Programming (DP) is a technique used to solve problems by breaking them down into smaller subproblems and solving these subproblems in a way that avoids redundant computation.
In this problem, we can define a subproblem as finding the smallest sufficient team for a given subset of skills. We can use a DP table to store the smallest team for each possible subset of skills. The DP table is indexed by a bitmask, where each bit represents whether a skill is included in the subset or not.
Bit Manipulation:
Bit Manipulation is a technique used to perform operations at the bit level. In this problem, we can use bit manipulation to represent subsets of skills. We assign each skill a unique bit in a bitmask, and each person can be represented by a bitmask indicating the skills they have.

The combination of DP and bit manipulation significantly reduces the time and space complexity of our solution. The time complexity is reduced from O(m * 2^n) to O(n * 2^m), and the space complexity is reduced from O(2^n) to O(2^m), where n is the number of people and m is the number of skills. This is much more efficient for the given constraints, where m is at most 16 and n is at most 60.

Code Explanation and Design Decisions

Let’s dissect the approach that uses Dynamic Programming and bit manipulation.

Initial Parameters:
The initial parameters for our function are req_skills and people. req_skills is a list of the required skills and people is a list of the skills each person has. These parameters define the problem space - we want to find the smallest team of people that have all the required skills.
Primary Loop:
The primary loop iterates over each person in the people list. Each iteration represents the process of deciding whether to include this person in the team or not. Each person can potentially add new skills to our team, so each iteration contributes to the solution by exploring a new set of potential skills.
Conditions or Branches within the Loop:
Within the loop, we have a secondary loop that iterates over all possible combinations of skills that we’ve seen so far (dp array). For each combination of skills, we consider the skills that would be added by including the current person in the team. If this results in a smaller team, we update our dp array. This condition signifies the decision to include a person in the team if they add required skills.
Updates or Modifications to Parameters:
The dp array is the primary parameter that we update within the loop. Each entry in the dp array represents the smallest team that can cover a certain combination of skills. As we iterate over each person, we update the dp array to reflect the smallest team that can cover each combination of skills, considering the current person’s skills.
Invariant:
An invariant in this problem is that for any subset of skills, the size of the team in the corresponding dp entry is the smallest possible size. This invariant is maintained throughout the code by always choosing the smallest team when updating the dp array.
Final Output:
The final output is the smallest team of people that cover all required skills. This is represented by the indices of these people in the original people list. The final output satisfies the problem’s requirements as it represents the smallest team that covers all required skills.

To summarize, the solution works by systematically considering every person, and updating our current smallest team for all combinations of skills. We use bit manipulation to efficiently represent these combinations of skills, and Dynamic Programming to avoid redundant computation.

Coding Constructs

High-Level Problem-Solving Strategies:
The main strategies being used by this code are Dynamic Programming (DP) and Bit Manipulation. DP is used to avoid redundant computation by storing and reusing solutions to subproblems, while Bit Manipulation is used to efficiently represent and manage combinations of skills.
Explaining Purpose to a Non-Programmer:
This code helps to assemble the smallest possible team of people, such that the team as a whole possesses all the required skills. Imagine if you’re planning a mission to Mars and you need a crew with different abilities - some should be pilots, others engineers, doctors, botanists, etc. Given a pool of astronauts, each having a unique set of skills, this code helps you find the smallest group that covers all needed skills.
Logical Elements or Constructs:
The logical constructs used in this code include loops for iterating over data, conditions to decide whether to include a person in the team, and an array (dp) to store the smallest team for each combination of skills.
Algorithmic Approach in Plain English:
The algorithm works by considering each person one by one and deciding whether or not to include them in the team. For each person, it looks at the skills that they could add to the team and checks if adding them would result in a smaller team for any combination of skills. If so, it updates its record of the smallest team for that combination of skills. It repeats this for every person. At the end, it looks at the smallest team that covers all required skills.
Key Steps or Operations:
- Initialize an array dp to keep track of the smallest team for each combination of skills.
- For each person, calculate their skillset in binary representation.
- Loop over all combinations of skills that we’ve seen so far.
- For each combination, consider the skills that would be added by including the current person in the team. If this results in a smaller team, update dp.
- Finally, return the smallest team that covers all required skills.
Algorithmic Patterns or Strategies:
The algorithmic patterns used in this code are Dynamic Programming and Bit Manipulation. DP is used to break down the problem into smaller subproblems and store the solutions to these subproblems to avoid redundant computation. Bit Manipulation is used to efficiently represent combinations of skills as bits in an integer, allowing us to quickly compute the union of skills and compare skill sets.

Language Agnostic Coding Drills

Dissecting the Code into Distinct Concepts:
a. Arrays: Understanding how to use and manipulate arrays is fundamental to almost all programming tasks. In this problem, an array (dp) is used to store the smallest team for each combination of skills.
b. Loops: Loops are used to iterate over the list of people and combinations of skills. A good understanding of how loops work is essential for this problem.
c. Binary Representation and Bit Manipulation: Skills are represented as bits in a binary number. Bitwise operations, such as OR and AND, are used to manipulate these bits, which represent adding a person’s skills to a team or checking whether a team has all skills. Understanding binary representation and bit manipulation is crucial for this problem.
d. Conditionals: Conditional statements (if-else) are used to decide whether to include a person in a team based on whether they would result in a smaller team. This requires a firm grasp of conditionals and logical operators.
e. Dynamic Programming: The main problem-solving strategy used in this code is Dynamic Programming (DP). DP is used to avoid redundant computation by storing and reusing solutions to subproblems.
Order of Difficulty and Brief Description:
a. Arrays (Easy): Arrays are one of the basic data structures that every programmer should be familiar with. They store elements of the same type in contiguous memory locations and allow constant-time access to individual elements.
b. Loops (Easy): Loops are a fundamental control structure in programming that allow code to be executed repeatedly. Understanding how to control the flow of execution with loops is a critical skill for programming.
c. Conditionals (Easy): Conditional statements allow the execution of different code blocks depending on whether a certain condition is true or false. Understanding conditionals is important for controlling the flow of a program.
d. Binary Representation and Bit Manipulation (Medium): Understanding binary representation of numbers and bitwise operations is a more specialized skill. It’s critical for many algorithms and data structures, especially those related to low-level programming or performance optimization.
e. Dynamic Programming (Hard): Dynamic Programming is a more advanced algorithm design technique used for optimization problems. It requires careful problem analysis and design to correctly identify the subproblems and the relations between them.
Problem-Solving Approach:
a. Start by representing the skills of each person in binary form. This will allow us to represent the skills of each person as an integer and manipulate these skills using bitwise operations.
b. Initialize an array (dp) to keep track of the smallest team for each combination of skills. This is the setup for our dynamic programming approach.
c. Use a loop to iterate over each person. For each person, calculate their skillset in binary representation.
d. Inside this loop, use another loop to iterate over all combinations of skills that we’ve seen so far. This loop will help us check whether adding the current person to a team results in a smaller team.
e. Use a conditional statement inside the inner loop to check whether adding the current person results in a smaller team for a certain combination of skills. If it does, update the dp array to reflect this new smallest team.
f. Finally, once we’ve considered all people and all combinations of skills, we can find the smallest team that covers all required skills by looking at the appropriate entry in the dp array. This is the final step of our dynamic programming approach.

Targeted Drills in Python

Sure, let’s go ahead and create Python-based coding drills for the identified concepts.

Arrays

Arrays in Python can be represented using lists. Here’s a simple drill illustrating list creation and manipulation in Python:

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# Creating a list
my_list = [1, 2, 3, 4, 5]

# Accessing elements
print(my_list[0])  # Outputs: 1

# Modifying elements
my_list[1] = 10
print(my_list)  # Outputs: [1, 10, 3, 4, 5]

Loops

The following drill demonstrates how to write a basic for loop in Python:

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# Creating a list
my_list = [1, 2, 3, 4, 5]

# Using a for loop to iterate over the list
for element in my_list:
    print(element)

Conditionals

Here’s a simple drill that shows how to use if-else statements in Python:

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# Variable declaration
a = 5

# Using if-else to check a condition
if a > 10:
    print("Greater than 10")
else:
    print("Less than or equal to 10")  # This will be printed

Binary Representation and Bit Manipulation

This drill demonstrates how to convert numbers to binary and perform bitwise operations in Python:

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# Number in binary representation
num = 0b101  # 5 in decimal

# Bitwise OR
result = num | 0b011  # 7 in decimal
print(bin(result))  # Outputs: 0b111

# Bitwise AND
result = num & 0b011  # 1 in decimal
print(bin(result))  # Outputs: 0b1

Dynamic Programming

Here’s a simple drill illustrating the use of dynamic programming to compute Fibonacci numbers:

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def fib(n):
    dp = [0, 1] + [0]*(n-1)
    for i in range(2, n+1):
        dp[i] = dp[i-1] + dp[i-2]
    return dp[n]

print(fib(10))  # Outputs: 55

The problem-specific concept in our case is the bitwise manipulation of integers to represent skillsets. This is important for our problem as it allows us to represent each person’s skillset as an integer and compare and combine these skillsets using bitwise operations.

Once we’ve learned and practiced all these drills, we can start to integrate them together to solve our problem. We would start by initializing our dp array (drill 1), then iterate over each person and each combination of skills using loops (drill 2). Inside the loops, we’d perform bitwise operations on the skillsets (drill 4) and use conditionals (drill 3) to update our dp array. This process of storing and reusing solutions to subproblems is the essence of the dynamic programming approach (drill 5).

Q&A

1 <= req_skills.length <= 16 What does it mean? If using bit mask , there are 2^16 tries! 1 <= people.length <= 60 =>2^60! What a big number!

In the problem statement, the constraints 1 <= req_skills.length <= 16 and 1 <= people.length <= 60 set the limits for the number of required skills and the number of people, respectively. Yes, these can be large numbers, but the problem is still solvable within reasonable time limits due to the specific design of the problem and the optimization techniques we can apply.

Here’s why:

1 <= req_skills.length <= 16: When using bit masks, this constraint implies that each skill can be represented as a unique bit in a 16-bit integer (since 2^16 = 65536, which can accommodate all possible combinations of 16 skills). This is manageable and allows us to leverage bitwise operations for efficiency.
1 <= people.length <= 60: Yes, in theory, this can lead to 2^60 combinations of people. However, we don’t need to explore all these combinations thanks to our dynamic programming approach. Here’s a rough explanation:
We’re using the DP table dp[], where each index i represents a combination of skills. dp[i] holds the smallest team (minimum number of people) that can cover the skills represented by i. When we iterate over people, for each person, we update the DP table based on the skills they have. The key here is that we only care about the combinations of skills, not which specific people have them. So even though we have up to 60 people, we’re not directly creating combinations of these people, but rather combinations of the skills they have.

By cleverly applying bit masks and dynamic programming, we can effectively deal with the large input space and find an optimal solution within reasonable time limits.

Can I assume that the skills are sorted?

The problem statement does not mention anything about the skills being sorted. So, it’s safe to assume that the list of skills (req_skills in the problem) can be in any arbitrary order.

Also, note that the order of skills does not impact the solution, since we are interested in forming a team with a specific set of skills rather than in a specific order. In other words, the problem is a combination problem rather than a permutation problem.

During your preprocessing step, where you assign a unique binary mask to each skill, the actual order of skills in req_skills will not matter, as each skill gets its own unique binary representation. The relative ordering of skills does not affect the binary representation of different subsets of skills. This holds true for the Dynamic Programming table as well, where each index represents a unique combination of skills, not a sequence.

This is essentially knapsack.

Yes, you could frame this problem as a variation of the classic Knapsack problem. Specifically, it could be considered as an instance of the Combinatorial Optimization problem, just like the Knapsack problem.

The “knapsack” in this problem would be the set of required skills (req_skills), and the “items” would be the people, each with a certain “weight” and “value”. The weight of each person can be considered as 1 (since we want to minimize the number of people), and the value of each person would be the skills they possess. We want to “fill our knapsack” with the minimum number of people who cover all the required skills.

However, unlike the classic 0/1 Knapsack problem, which is solved typically using dynamic programming with the state defined by the current item and the remaining weight, this problem requires the use of a bit manipulation-based approach along with dynamic programming due to the constraints and the nature of the problem. In particular, using binary masks allows us to efficiently represent and manipulate sets of skills, which are crucial in this problem.

Please someone help me with the time complexity. When we are doing dp[with_him] = need + [i], We are creating new list with elements copied from need and appending [i] to it. Shouldn’t the time complexity also include this factor? It should be O(peoplepeople2^skill). One extra factor of size=people because maximum length of need can be as big as people. Are there any optimizations being done by python?

Also the space should be O(people*2^skill) because max size each elementdp array can be people

Let’s break down the time complexity. There are two main parts to consider:

The generation of all subsets of skills that each person can offer, represented as integers. The time complexity for this part is O(n*2^m), where n is the number of people and m is the number of skills. We iterate over each person and for each person, we generate all possible combinations of skills that they can offer, which in the worst case can be 2^m.
The dynamic programming (DP) portion where we fill up our dp array. For this part, the time complexity is O(2^m*n^2). The 2^m factor comes from the fact that we are iterating over all possible combinations of skills, and for each combination, we are checking each person (n) and potentially updating our dp array by copying the list of people and appending a new person (n).

So in total, the time complexity of the algorithm is O(n*2^m + 2^m*n^2).

As for the space complexity, it is O(2^m*n). We have 2^m states for different combinations of skills, and for each state, we could potentially have to store all n people.

Python internally does perform some optimizations for operations like list copying and appending, but these optimizations do not significantly change the time complexity. In Big O notation, we focus on the upper bound of the time or space complexity, and constant factors and lower order terms are ignored. So the time complexity remains O(n*2^m + 2^m*n^2) and space complexity remains O(2^m*n).

It seems only a few people care about complexity in the discussion but complexity analysis is very important in an interview, isn’t it? I think the time complexity is O(people^2 * 2^skill) as the list concat and copy operations. Space complexity is O(people * 2^skill) as the dp array stores lists. Correct me if I am wrong. A very simple optimization to make the time complexity become O(people * 2^skill) is to use another bit mask for selected people instead of list.

You’re right in saying that complexity analysis is very important during interviews and in real-world scenarios as well. It gives an estimate of the efficiency of the algorithm in terms of time and space. The original complexity analysis I provided factored in the list concatenation and copying operations which are indeed costly operations, hence the n^2 in O(n*2^m + 2^m*n^2).

Your suggested optimization using another bit mask for selected people instead of a list would indeed reduce the time complexity to O(people * 2^skills). This is because, with this approach, we wouldn’t need to deal with list copying and appending operations which can take up to O(n) time. Instead, we would be performing bitwise operations which take constant time. So we would only have to iterate over each person for each state, resulting in a time complexity of O(people * 2^skills).

As for the space complexity, it would still be O(people * 2^skills), since we would be storing one bit mask for each person for each state. Each state would be represented by a bit mask of size people and we have 2^skills states, so the space complexity would be O(people * 2^skills).

However, bear in mind that bit manipulations might make the code harder to read and understand, especially for those not familiar with such operations. It’s always a trade-off between code readability and efficiency, and the best choice depends on the specific situation.

for skill_set, need in dp.items(): this doesnt work with python 3? anyone know how to correct it for python 3?

The line of code you posted is actually compatible with Python 3. The items() method in Python 3 returns a view object that displays a list of a dictionary’s key, value tuple pairs. Here is an example of how it works:

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dictionary = {"key1": "value1", "key2": "value2"}

for key, value in dictionary.items():
    print("Key:", key)
    print("Value:", value)

In the case of the code snippet you posted, skill_set would represent the dictionary key and need would represent the corresponding value for each iteration through the loop.

Why does my solution give 812 ms. Isn’t my solutions’ time complexity similar to the above = (number of skill combinations) = 2^(number of people) ?

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class Solution {
    public:
    int N, reqSkillBitMask;
    vector <int> res;
    void dfs(int person, int skill, vector<int> &curRes, vector<int> &peopleSkills) {
        if(skill == reqSkillBitMask){
            if(curRes.size() < res.size() || res.size() == 0) {
                res = curRes;
            } 
            return;
        }
        if(res.size() > 0 && curRes.size() >= res.size()){ return;}
        for(int i = person + 1; i < N; ++i) {
            int personSkill = peopleSkills[i];
            int cumSkill = personSkill | skill;
            if(cumSkill <= skill) {continue;}
            curRes.push_back(i);
            dfs(i, cumSkill, curRes, peopleSkills);
            curRes.pop_back();
        }
    }
    
    vector<int> smallestSufficientTeam(vector<string>& req_skills, vector<vector<string>>& people) {
        unordered_map<string, int> skillIdMap;
        int count = 0;
        reqSkillBitMask = (1 << req_skills.size()) - 1;
        for(auto skill: req_skills){
            skillIdMap[skill] = count++;
        }
        N = people.size();
        vector<int> peopleSkills(N, 0);
        for(int person = 0; person < N; ++person){
            int totSkill = 0;
            for(string skill: people[person]){
                totSkill |= 1 << skillIdMap[skill];
            }
            peopleSkills[person] = totSkill;
        }
        for(int i = 0; i < N; ++i){
            for(int j = 0; j < N; ++j){
                if(i != j && (peopleSkills[i] & peopleSkills[j] == peopleSkills[i])){
                    peopleSkills[i] = 0;   
                }
            }
        }
        vector<int> curRes;
        dfs(-1, 0, curRes, peopleSkills);
        return res;
    }
};

Your approach is a backtracking approach where you’re trying all possible subsets of people to see which one forms the smallest team that has all the required skills.

The time complexity for your solution is indeed similar, and should be approximately O(2^N) in the worst case, where N is the number of people, because you’re essentially exploring all subsets of people. However, you’ve added a pruning condition if(res.size() > 0 && curRes.size() >= res.size()){ return;} which should help to reduce the actual number of recursive calls in practice.

The reason your solution takes 812 ms could be due to a number of factors:

Even within the same language, different implementations of the same algorithm can have different runtimes due to factors like constant factors, lower-order terms, and implementation details.
The pruning you’ve added to the DFS helps, but it’s still possible for the recursion to go deep before finding an optimal solution, which could slow down the execution.

So, even though your solution is theoretically efficient, there are still factors that could slow it down in practice.

Also, I noticed this part in your code:

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for(int i = 0; i < N; ++i){
    for(int j = 0; j < N; ++j){
        if(i != j && (peopleSkills[i] & peopleSkills[j] == peopleSkills[i])){
            peopleSkills[i] = 0;   
        }
    }
}

Here, you’re trying to pre-process the people to remove any people whose skills are a subset of another person’s skills. This step itself is O(N^2), which might contribute to the runtime.

If your objective is to reduce the runtime, you may want to consider optimizing the pre-processing step, or using a more efficient data structure or approach to store and update the current best team.

Remember that the reported runtime on LeetCode can also be affected by factors like server load and network latency, and small differences in runtime are usually not significant. It’s more important to understand the algorithm and its complexity.

for skill_set, need in dp.items(), I think we should loop from larger skill_set to smaller skill_set? The reason is we are using LAST round’s smaller skill_set to update THIS round’s larger skill_set. Now under space compression, if we loop from small to large, we may wrong use THIS round’s smaller skill_set to update THIS round’s larger skill_set.

This is a good observation, and it really depends on the specifics of the problem and the algorithm. In some dynamic programming problems, you do need to be careful about the order in which you iterate through the states to make sure that you’re not overwriting a state that you still need to use later on.

However, in this particular problem and solution, it doesn’t actually matter whether we iterate from smaller skill sets to larger ones, or vice versa. This is because when we’re iterating over dp.items(), we’re iterating over the states from the previous round, not the current round. In other words, we’re not modifying dp in-place while iterating over it.

When we do dp[with_him] = need + [i], this doesn’t affect the dp.items() iteration because it’s either updating an existing state (which doesn’t change the items we’re iterating over), or adding a new state (which also doesn’t affect the items, because in Python, a dictionary’s items view is dynamic and reflects changes to the dictionary, but new items added during iteration aren’t included in the iteration).

So, in this specific case, the order of iteration doesn’t matter because we’re not modifying the states we’re iterating over in a way that would affect the iteration. However, your point is well taken, and in general, it’s important to be mindful of the order of iteration in dynamic programming problems, especially when you’re updating states in-place.

I would say this is the generic solution to shortest path problem. Dijkstra’s and Bellman-Ford are based on the generic solution

While this problem does share some similarities with shortest path problems, and techniques such as Dijkstra’s or Bellman-Ford could potentially be adapted to solve it, it’s not quite a direct application of those algorithms.

Dijkstra’s algorithm and Bellman-Ford algorithm are typically used to find the shortest path between two nodes in a graph where edges have associated weights. In this problem, we’re not exactly dealing with a typical graph structure or finding a path between two nodes. Instead, we’re trying to find the smallest subset of people that cover all required skills, which is more akin to a set cover problem.

In this case, we’re using dynamic programming to essentially try all possible combinations of people and their associated skills, tracking the smallest team that can cover each combination of skills, until we find the smallest team that can cover all required skills.

So while there are some conceptual similarities with shortest path problems — in both cases, we’re trying to find an optimal way to cover certain “points” (either nodes in a graph, or skills in this problem) — the specific details and algorithmic approach differ.

That said, having a strong understanding of shortest path algorithms can certainly help when approaching this type of problem, as many of the same problem-solving strategies and principles of optimisation apply.

I think the number of people should be (for i in range(0,len(pepple))),if find the solution,over the program! It may save much time.

The solution does indeed iterate through all people in the given order using a loop. However, ending the program as soon as a solution is found might not necessarily give the smallest team possible. This problem requires finding the smallest team with all the required skills, not just any team with all the required skills. Therefore, it is important to iterate through all possible combinations of people to ensure finding the minimum size team.

It’s worth noting that ending the search early can indeed improve performance. This is a common strategy known as pruning, which can be very effective in reducing the size of the search space in certain types of problems. However, in this case, it would need to be done carefully to ensure that we don’t miss out on a potentially smaller team by ending the search prematurely.

The key in this problem is to balance the need for thoroughness (to ensure finding the smallest possible team) with the need for efficiency (to avoid unnecessary computation). This is achieved in the given solution through the use of dynamic programming, which systematically reuses results from smaller subproblems to efficiently solve larger ones.

this is my solution with cpp. there is a map copy, should we do?

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std::vector<int> smallestSufficientTeam(std::vector<std::string> const & reqSkills,
            std::vector<std::vector<std::string>> const & people) {
        // plagiarizing from https://leetcode.com/problems/smallest-sufficient-team/discuss/334572/Python-DP-Solution
        int const reqSkillsSize = reqSkills.size();
        int const reqSkillsComb = (1 << reqSkillsSize);

        std::unordered_map<std::string, int> skills2Int;
        for (int idx = 0; idx < reqSkills.size(); idx++) {
            std::string const & skill = reqSkills[idx];
            if (skills2Int.find(skill) != skills2Int.end()) {
                continue;
            }
            skills2Int[skill] = (1 << idx);
        }

        std::unordered_map<int, std::vector<int>> dpMemo;
        dpMemo[0] = {};

            std::unordered_map<int, std::vector<int>> tmpDpMemo(dpMemo);
        for (int idx = 0; idx < people.size(); idx++) {
            int curSkill = 0;
            for (std::string const & skill : people[idx]) {
                curSkill |= skills2Int[skill];
            }
            // std::cout << std::bitset<16>(curSkill) << "\tdpMemoSize: " << dpMemo.size() << "\n";
            for (std::unordered_map<int, std::vector<int>>::iterator it = dpMemo.begin();
                 it != dpMemo.end();
                 it++) {

                // std::cout << std::bitset<16>(it->first) << "\t";
                int const nextMask = it->first | curSkill;
                if (it->first == nextMask) { // there is no need to add cur person
                    continue;
                }

                // std::cout << "nextMask: " << std::bitset<16>(nextMask) << "\t";
                if (tmpDpMemo.find(nextMask) == tmpDpMemo.end() ||
                    (tmpDpMemo[nextMask].size() > dpMemo[it->first].size() + 1)) {
                    tmpDpMemo[nextMask] = it->second;
                    tmpDpMemo[nextMask].emplace_back(idx);
                    // std::cout << "nextMaskSize: " << tmpDpMemo[nextMask].size() << "\t";
                }
                // std::cout << "\n";
            }
            dpMemo = tmpDpMemo;
            // std::cout << "\n";
        }
        return dpMemo[reqSkillsComb - 1];
    }

Your solution in C++ is also using dynamic programming and bit-masking, similar to the Python solution we’ve discussed before. You’re mapping each skill to an integer and creating bitmasks to represent the skills each person has. Then, you’re iterating over the existing team combinations (represented by bitmasks) in your dpMemo map, and trying to add each person to each team, creating a new team only if it’s smaller than any existing team with the same skills.

Regarding the copy of dpMemo to tmpDpMemo, this step is necessary because you’re iterating over dpMemo and simultaneously trying to add new entries. Modifying a data structure while iterating over it can lead to unexpected behaviors and is generally avoided. By creating a temporary copy of dpMemo, you can safely add new entries without affecting the current iteration. This comes at the cost of additional space, but ensures that the logic of your program is correct.

As for whether you should do this, it depends on your specific needs and constraints. If space is not a concern, and you want to ensure the correctness of your program, then making a copy of the map is a reasonable approach. If space is a concern, you might need to consider a different approach or try to optimize your current solution.

For time complexity, your solution is O(people * 2^skills) because in the worst case, for every person, you’re checking all possible combinations of skills. For space complexity, because you’re storing all these combinations, it’s also O(people * 2^skills).

This is a valid and correct approach for solving the problem, and the trade-offs you’re making in terms of space and time complexity are necessary given the complexity of the problem.