Problem Description

In this LeetCode problem, we're given a set of statements made by a group of n people, about each other's trustworthiness. Specifically, the input is a 2D integer array statements where:

• statements[i][j] = 0 means person i claims person j is bad (untruthful).
• statements[i][j] = 1 means person i claims person j is good (truthful).
• statements[i][j] = 2 means person i has not made any claim about person j.

Individuals are not permitted to make statements about themselves, thus statements[i][i] is always 2. The objective is to determine the maximum number of people that can be considered 'good' based on these statements. A 'good' person always tells the truth, while a 'bad' person may lie or tell the truth.

Flowchart Walkthrough

Let's analyze the problem LeetCode 2151. Maximum Good People Based on Statements using the Flowchart. Here's how we can deduce the best approach:

1. Is it a graph?

   • No: The problem description doesn't involve a typical graph structure with nodes directly connected by edges.

2. Need to solve for kth smallest/largest?

   • No: The task is not about finding a kth smallest or largest element.

3. Involves Linked Lists?

   • No: There is no manipulation or traversal of linked lists mentioned.

4. Does the problem have small constraints?

   • Yes: Each person can either be good or bad, which suggests a limited number of combinations or permutations to explore, indicative of small constraints that suit exhaustive techniques.

5. Brute force / Backtracking?

   • Yes: Given the small constraints and the nature of trying all combinations of good and bad people to validate against given statements, brute force or backtracking is most suited. We need to explore every possible combination of people's states (good or bad) and verify their consistency with provided statements, which is characteristic of backtracking.

Conclusion: By following the path detailed in the flowchart and the problem's requirements, it's clear that using a backtracking pattern to explore all possible states and validate them against provided statements is the most appropriate approach for solving LeetCode 2151. Maximum Good People Based on Statements.

Intuition

The intuition behind the solution is to utilize a brute-force approach, trying every possible combination of 'good' and 'bad' assignments for the people. Since a 'good' person always tells the truth, their statements can be used to cross-verify which people can be good and which cannot. If a 'good' person says another person is good, then both must be good; if they say someone is bad, then that person cannot be considered good in any combination that includes the first person as a good individual.

The solution uses bit masking to represent each combination of 'good' and 'bad' people. Each bit in a mask represents a person, where 1 can denote 'good' and 0 can denote 'bad'. There are 2^n possible combinations (masks) to check, where n is the number of people.

The function check(mask) takes a mask as an argument and iterates over all people. It checks the statements of those who are 'good' (bits set to 1) in the current mask. For each 'good' person's statements, it checks whether they align with the current assumption of who's good and bad. For instance, if a 'good' person claims another person is good, but the mask marks them as bad (or vice versa), the current mask is invalid, and the function returns 0. Otherwise, if no conflicts are found, it returns the number of 'good' people in the current mask (the count of 1's in the mask).

The main function calls check(mask) for all 2^n combinations and returns the maximum number of good people found among all valid masks that don't lead to contradictions.

This brute-force solution works well for smaller n, as it explores all possibilities and ensures that the maximum number of 'good' people is found according to the statements made.

Learn more about Backtracking patterns.

Solution Approach

The implementation is grounded on a bit manipulation strategy. It uses an integer to represent a set of truth assignments to individuals, where the i-th bit of the integer corresponds to the i-th person.

Here is a step-by-step walkthrough of the solution:

1. Brute Force Enumeration: We generate all possible combinations of each person being 'good' or 'bad' by iterating through integers from 1 to 2^n - 1, inclusive. Here, n is the number of people. The reason we start from 1 is because 0 would mean that nobody is good, which is not a valid scenario for us to consider.

2. Checking Each Combination: For each combination (represented by mask), we implement a function called check(mask). This function is responsible for validating the combination against the statements made.

3. Validating the 'Good' People: Within check(mask), we traverse over each person (i) by examining the current bit in the mask: (mask >> i) & 1. If this results in 1, the person is assumed 'good' in this particular combination. Then, we check each statement i has made about other people (j).

4. Statement Verification: If person i has made a definitive statement about person j (i.e., statements[i][j] is 0 or 1), we check the truthfulness of i's statement against the current combination (mask). If a 'good' i made a statement that contradicts the current assumption about j, the function returns 0, indicating this mask (or combination) is not valid.

5. Counting 'Good' People: If no contradictions are found for a 'good' person i, the count increments, tallying the number of 'good' people in this combination.

6. Finding the Maximum: Finally, the main solution function calls check(mask) for each possibility and keeps track of the maximum number of 'good' people found from all the valid combinations. It uses the built-in Python function max() to find the maximum over all return values from check(mask).

By utilizing bit masks to represent sets and checking all possible combinations, the algorithm finds the optimal solution that matches the problem's constraints. This approach, while not the most time-efficient for large n, ensures accuracy by not overlooking any potential solution when n is within a manageable size.

Example Walkthrough

Suppose we have a group of 3 people, and we are given the following statements array:

statements = [
    [2, 1, 0],  // Person 0's statements about Person 1 and 2
    [2, 2, 1],  // Person 1's statements about Person 0 and 2
    [1, 0, 2]   // Person 2's statements about Person 0 and 1
]

Here's how we would apply the solution approach:

1. Brute Force Enumeration: Since there are 3 people, we have 2^3 = 8 combinations to consider. We will ignore the mask 000 (where nobody is good) and check 001 to 111.

2. Checking Each Combination: Let's consider the mask 011, indicating Person 0 is 'bad' while Persons 1 and 2 are 'good'. We use the check(mask) function to determine if this mask is a valid representation of the 'good' people.

3. Validating the 'Good' People: In the check(mask) function, we start with Person 0 being 'bad', so we skip and move to Persons 1 and 2.

4. Statement Verification: For Person 1 (who is 'good' in this mask), their statement about Person 2 is that they are 'good' (statements[1][2] is 1), which aligns with our mask. However, we do not have a statement from Person 1 about Person 0, so we cannot verify Person 1's truthfulness this way.

   Moving to Person 2, they state that Person 0 is 'good' (statements[2][0] is 1), but our mask assumes Person 0 is 'bad', creating a contradiction. Person 2 also claims that Person 1 is 'bad' (statements[2][1] is 0), which contradicts our mask where Person 1 is 'good'.

5. Counting 'Good' People: Since there's a contradiction in Person 2's statements, we cannot count any 'good' people under this mask, so check(mask) returns 0.

6. Finding the Maximum: We perform the same check(mask) for all other combinations. Let's take another example with the mask 101, where Persons 0 and 2 are 'good', and Person 1 is 'bad'. Person 0 claims Person 1 is 'good' which is a contradiction since we assumed them to be 'bad'. Therefore, check(mask) also returns 0 for this mask.

By going through all combinations, we may find that the mask 110 yields the maximum number of 'good' people without contradiction — Person 0 claims Person 2 is 'bad', and Person 2 claims Person 0 is 'good', which both align with this mask. In this scenario, both Persons 0 and 1 can be considered good based on their statements and each other's statements.

Therefore, the final result given by the main function would be 2, as this is the maximum number of 'good' people found across all valid combinations. The function would iterate through all the masks, evaluate each one using the check(mask) function, and conclude that the combination 110 presents the maximum number of good people without causing any conflicts.

Solution Implementation

Time and Space Complexity

The provided Python code defines a function maximumGood which finds out the maximum number of people who can be considered "good" based on their statements about themselves and others. The solution employs a brute force approach to check every possible combination of "good" and "bad" people and then validate these combinations against the given statements.

Time Complexity

The time complexity of the code is determined by the number of possible combinations of people being good or bad, since we are iterating over all possible combinations using a bit mask. Given n people, there are 2^n possible combinations (since for each person, there is a binary decision to be either good or bad). For each combination, we are checking all statements made by all n persons, and each person makes n statements or observations. This results in a nested loop running n times for each of the 2^n combinations.

Thus, the time complexity is O(n * 2^n), where n is the number of people.

Space Complexity

The space complexity of the code is quite straightforward. The only additional space used is for the variable cnt, which is an integer used to keep count of the number of "good" people in the current combination, and the space used in the recursive process stack. This does not scale with n and hence is O(1).

However, it's important to note that the code uses recursive calls to the check function. The maximum depth of this recursion would be n in the worst case (although in the check function there are no recursive calls, it's important to consider for more general cases). Therefore, if we would consider the function call stack in the analysis, the space complexity would be O(n) for the recursion call stack space. However, since in the provided code check is more of an iterative function, we consider the space complexity to be O(1).

Learn more about how to find time and space complexity quickly using problem constraints.