How do you solve Smallest Sufficient Team on LeetCode?

Smallest Sufficient Team (LeetCode #1125) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * 2^n) time complexity and O(2^n) space complexity. Key insight: Using bit manipulation allows us to efficiently represent and combine skill sets.

What is the brute force solution for Smallest Sufficient Team?

The brute force approach for Smallest Sufficient Team is Brute Force, with O(2^n * n) time complexity and O(1) space complexity. The brute force approach involves checking all possible combinations of people to see if they cover all required skills. This is straightforward but inefficient as the number of combinations grows exponentially with the number of people. The optimal Optimal Solution approach improves this to O(n * 2^n).

What algorithm or data structure is used to solve Smallest Sufficient Team?

Smallest Sufficient Team uses the following concepts: Array, Dynamic Programming, Bit Manipulation, Bitmask. The recommended patterns to study are: Dynamic Programming, Bit Manipulation.

What are the interview tips for Smallest Sufficient Team?

Explain your thought process clearly, especially when transitioning from brute force to optimal solutions. Be prepared to discuss the trade-offs between different approaches. Practice generating combinations and using bit manipulation, as they are common in DSA problems.

What are common mistakes when solving Smallest Sufficient Team?

Not considering all combinations of people, leading to missing valid teams. Overlooking the need to check if a skill set is a superset of required skills.

#1125

Smallest Sufficient Team

Hard

ArrayDynamic ProgrammingBit ManipulationBitmaskDynamic ProgrammingBit Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(2^n * n)	O(n * 2^n)
Space	O(1)	O(2^n)

💡

Intuition

Time O(n * 2^n)Space O(2^n)

The optimal solution uses dynamic programming with bit manipulation to efficiently track which skills are covered by which combinations of people. This significantly reduces the number of combinations we need to check.

⚙️

Algorithm

3 steps

1Step 1: Create a bitmask for each person's skills.
2Step 2: Use a DP array to track the minimum team for each skill combination.
3Step 3: Iterate through each person and update the DP array based on their skills.

solution.py20 lines

1# Full working Python code
2from collections import defaultdict
3
4def smallestSufficientTeam(req_skills, people):
5    skill_to_index = {skill: i for i, skill in enumerate(req_skills)}
6    n = len(req_skills)
7    m = len(people)
8    dp = [None] * (1 << n)
9    dp[0] = []
10
11    for i in range(m):
12        person_mask = 0
13        for skill in people[i]:
14            person_mask |= (1 << skill_to_index[skill])
15        for j in range((1 << n) - 1, -1, -1):
16            if dp[j] is not None:
17                new_mask = j | person_mask
18                if dp[new_mask] is None or len(dp[new_mask]) > len(dp[j]) + 1:
19                    dp[new_mask] = dp[j] + [i]
20    return dp[(1 << n) - 1]

ℹ

Complexity note: The time complexity is O(n * 2^n) because we iterate through each person and update the DP array for each possible skill combination. The space complexity is O(2^n) due to storing the DP array for all skill combinations.

1Using bit manipulation allows us to efficiently represent and combine skill sets.
2Dynamic programming helps to avoid redundant calculations by storing previously computed results.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.