LeetCode 2151: Maximum Good People Based on Statements Solution

Master LeetCode problem 2151 (Maximum Good People Based on Statements), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2151. Maximum Good People Based on Statements

Hard

Array Backtracking Bit Manipulation Enumeration

Problem Explanation

Explanation

To solve this problem, we can iterate through each person and simulate whether they are good or bad based on the statements made by others. We can start by assuming each person as good and bad alternatively and check for contradictions. If a contradiction is found, we backtrack and try the opposite assumption. This way, we can determine the maximum number of people that can be good based on the given statements.

Start with assuming each person as good and bad alternatively.
For each assumption, check if it leads to contradictions based on the statements made by others.
Backtrack if contradictions are found and try the opposite assumption.
Keep track of the maximum number of people that can be good without contradictions.

Time Complexity: O(2^n * n^2) where n is the number of people. Space Complexity: O(n)

Solution Code

class Solution {
    public int maximumGoodPeople(int[][] statements) {
        int n = statements.length;
        int maxGood = 0;
        
        for (int i = 0; i < (1 << n); i++) {
            boolean[] isGood = new boolean[n];
            int goodCount = 0;
            boolean valid = true;
            
            for (int j = 0; j < n; j++) {
                if ((i & (1 << j)) > 0) {
                    isGood[j] = true;
                    goodCount++;
                }
            }
            
            for (int j = 0; j < n && valid; j++) {
                for (int k = 0; k < n; k++) {
                    if (statements[j][k] == 0 && isGood[j] && isGood[k]) {
                        valid = false;
                        break;
                    }
                    if (statements[j][k] == 1 && (!isGood[j] || !isGood[k])) {
                        valid = false;
                        break;
                    }
                }
            }
            
            if (valid) {
                maxGood = Math.max(maxGood, goodCount);
            }
        }
        
        return maxGood;
    }
}

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