How do you solve Maximum Good People Based on Statements on LeetCode?

Maximum Good People Based on Statements (LeetCode #2151) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(1) space complexity. Key insight: Understanding the implications of statements is crucial for determining the validity of configurations.

What is the brute force solution for Maximum Good People Based on Statements?

The brute force approach for Maximum Good People Based on Statements is Brute Force, with O(n²) time complexity and O(1) space complexity. We will check every possible combination of good and bad people using a bitmask. This means we will explore all possible assignments of truth-tellers and liars to see which configuration allows the maximum number of good people. The optimal Optimal Solution approach improves this to O(n²).

What are the interview tips for Maximum Good People Based on Statements?

Clearly explain your thought process and how you are validating each configuration. Be prepared to discuss the time and space complexity of your approach. Practice bit manipulation techniques, as they can significantly optimize your solutions.

What are common mistakes when solving Maximum Good People Based on Statements?

Failing to correctly validate statements against the current configuration. Not considering all possible configurations, leading to missed valid combinations.

#2151

Maximum Good People Based on Statements

Hard

ArrayBacktrackingBit ManipulationEnumerationBit ManipulationEnumeration

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(1)

💡

Intuition

Time O(n²)Space O(1)

We can optimize the brute-force approach by using bit manipulation to efficiently check the validity of each configuration. This allows us to quickly determine if a certain configuration of good and bad people is valid without checking every statement repeatedly.

⚙️

Algorithm

3 steps

1Step 1: Use a bitmask to represent each person's status (good or bad).
2Step 2: For each bitmask, validate the statements based on the current assignment of good and bad people.
3Step 3: Count the number of good people in valid configurations and keep track of the maximum.

solution.py17 lines

1def maxGoodPeople(statements):
2    n = len(statements)
3    max_good = 0
4    for mask in range(1 << n):
5        valid = True
6        good_count = 0
7        for i in range(n):
8            if mask & (1 << i):
9                good_count += 1
10                for j in range(n):
11                    if statements[i][j] == 1 and not (mask & (1 << j)):
12                        valid = False
13                    if statements[i][j] == 0 and (mask & (1 << j)):
14                        valid = False
15        if valid:
16            max_good = max(max_good, good_count)
17    return max_good

ℹ

Complexity note: The time complexity remains O(n²) due to the need to check n statements for each of the 2^n configurations, but we efficiently manage the checks using bit manipulation.

1Understanding the implications of statements is crucial for determining the validity of configurations.
2Bit manipulation allows for efficient representation and checking of configurations.

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