Problem

There are two types of persons:

  • The good person : The person who always tells the truth.
  • The bad person : The person who might tell the truth and might lie.

You are given a 0-indexed 2D integer array statements of size n x n that represents the statements made by n people about each other. More specifically, statements[i][j] could be one of the following:

  • 0 which represents a statement made by person i that person j is a bad person.
  • 1 which represents a statement made by person i that person j is a good person.
  • 2 represents that no statement is made by person i about person j.

Additionally, no person ever makes a statement about themselves. Formally, we have that statements[i][i] = 2 for all 0 <= i < n.

Return _themaximum number of people who can be good based on the statements made by the _n people.

Examples

Example 1

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![](https://assets.leetcode.com/uploads/2022/01/15/logic1.jpg)

Input: statements = [[2,1,2],[1,2,2],[2,0,2]]
Output: 2
Explanation: Each person makes a single statement.
- Person 0 states that person 1 is good.
- Person 1 states that person 0 is good.
- Person 2 states that person 1 is bad.
Let's take person 2 as the key.
- Assuming that person 2 is a good person:
    - Based on the statement made by person 2, person 1 is a bad person.
    - Now we know for sure that person 1 is bad and person 2 is good.
    - Based on the statement made by person 1, and since person 1 is bad, they could be:
        - telling the truth. There will be a contradiction in this case and this assumption is invalid.
        - lying. In this case, person 0 is also a bad person and lied in their statement.
    - **Following that person 2 is a good person, there will be only one good person in the group**.
- Assuming that person 2 is a bad person:
    - Based on the statement made by person 2, and since person 2 is bad, they could be:
        - telling the truth. Following this scenario, person 0 and 1 are both bad as explained before.
            - **Following that person 2 is bad but told the truth, there will be no good persons in the group**.
        - lying. In this case person 1 is a good person.
            - Since person 1 is a good person, person 0 is also a good person.
            - **Following that person 2 is bad and lied, there will be two good persons in the group**.
We can see that at most 2 persons are good in the best case, so we return 2.
Note that there is more than one way to arrive at this conclusion.

Example 2

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![](https://assets.leetcode.com/uploads/2022/01/15/logic2.jpg)

Input: statements = [[2,0],[0,2]]
Output: 1
Explanation: Each person makes a single statement.
- Person 0 states that person 1 is bad.
- Person 1 states that person 0 is bad.
Let's take person 0 as the key.
- Assuming that person 0 is a good person:
    - Based on the statement made by person 0, person 1 is a bad person and was lying.
    - **Following that person 0 is a good person, there will be only one good person in the group**.
- Assuming that person 0 is a bad person:
    - Based on the statement made by person 0, and since person 0 is bad, they could be:
        - telling the truth. Following this scenario, person 0 and 1 are both bad.
            - **Following that person 0 is bad but told the truth, there will be no good persons in the group**.
        - lying. In this case person 1 is a good person.
            - **Following that person 0 is bad and lied, there will be only one good person in the group**.
We can see that at most, one person is good in the best case, so we return 1.
Note that there is more than one way to arrive at this conclusion.

Constraints

  • n == statements.length == statements[i].length
  • 2 <= n <= 15
  • statements[i][j] is either 0, 1, or 2.
  • statements[i][i] == 2

Solution

Method 1 – Bitmask Enumeration

Intuition

Since the number of people n is small (≤15), we can try every possible combination of who is good and who is bad using bitmasking. For each combination, we check if it is valid: every good person must tell the truth, so their statements must match the assignment. We maximize the number of good people among all valid assignments.

Approach

  1. Let n be the number of people. For each bitmask from 0 to 2^n - 1:
    • The i-th bit is 1 if person i is good, 0 if bad.
  2. For each bitmask, check if it is valid:
    • For every person i marked as good, check all their statements:
      • If statements[i][j] == 2, skip.
      • If statements[i][j] == 1, person j must be good in the bitmask.
      • If statements[i][j] == 0, person j must be bad in the bitmask.
    • If all good people’s statements are consistent, update the answer with the number of good people in this bitmask.
  3. Return the maximum number of good people found.

Code

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class Solution {
public:
    int maximumGood(vector<vector<int>>& statements) {
        int n = statements.size(), ans = 0;
        for (int mask = 0; mask < (1 << n); ++mask) {
            bool valid = true;
            for (int i = 0; i < n && valid; ++i) {
                if (!((mask >> i) & 1)) continue;
                for (int j = 0; j < n; ++j) {
                    if (statements[i][j] == 2) continue;
                    if (statements[i][j] != ((mask >> j) & 1)) {
                        valid = false;
                        break;
                    }
                }
            }
            if (valid) ans = max(ans, __builtin_popcount(mask));
        }
        return ans;
    }
};
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func maximumGood(statements [][]int) int {
    n, ans := len(statements), 0
    for mask := 0; mask < 1<<n; mask++ {
        valid := true
        for i := 0; i < n && valid; i++ {
            if (mask>>i)&1 == 0 {
                continue
            }
            for j := 0; j < n; j++ {
                if statements[i][j] == 2 {
                    continue
                }
                if statements[i][j] != ((mask>>j)&1) {
                    valid = false
                    break
                }
            }
        }
        if valid {
            cnt := 0
            for x := mask; x > 0; x &= x - 1 {
                cnt++
            }
            if cnt > ans {
                ans = cnt
            }
        }
    }
    return ans
}
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class Solution {
    public int maximumGood(int[][] statements) {
        int n = statements.length, ans = 0;
        for (int mask = 0; mask < (1 << n); ++mask) {
            boolean valid = true;
            for (int i = 0; i < n && valid; ++i) {
                if (((mask >> i) & 1) == 0) continue;
                for (int j = 0; j < n; ++j) {
                    if (statements[i][j] == 2) continue;
                    if (statements[i][j] != ((mask >> j) & 1)) {
                        valid = false;
                        break;
                    }
                }
            }
            if (valid) ans = Math.max(ans, Integer.bitCount(mask));
        }
        return ans;
    }
}
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class Solution {
    fun maximumGood(statements: Array<IntArray>): Int {
        val n = statements.size
        var ans = 0
        for (mask in 0 until (1 shl n)) {
            var valid = true
            for (i in 0 until n) {
                if ((mask shr i) and 1 == 0) continue
                for (j in 0 until n) {
                    if (statements[i][j] == 2) continue
                    if (statements[i][j] != ((mask shr j) and 1)) {
                        valid = false
                        break
                    }
                }
                if (!valid) break
            }
            if (valid) ans = maxOf(ans, mask.countOneBits())
        }
        return ans
    }
}
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class Solution:
    def maximumGood(self, statements: list[list[int]]) -> int:
        n = len(statements)
        ans = 0
        for mask in range(1 << n):
            valid = True
            for i in range(n):
                if not (mask >> i) & 1:
                    continue
                for j in range(n):
                    if statements[i][j] == 2:
                        continue
                    if statements[i][j] != ((mask >> j) & 1):
                        valid = False
                        break
                if not valid:
                    break
            if valid:
                ans = max(ans, bin(mask).count('1'))
        return ans
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impl Solution {
    pub fn maximum_good(statements: Vec<Vec<i32>>) -> i32 {
        let n = statements.len();
        let mut ans = 0;
        for mask in 0..(1 << n) {
            let mut valid = true;
            for i in 0..n {
                if (mask >> i) & 1 == 0 { continue; }
                for j in 0..n {
                    if statements[i][j] == 2 { continue; }
                    if statements[i][j] != ((mask >> j) & 1) as i32 {
                        valid = false;
                        break;
                    }
                }
                if !valid { break; }
            }
            if valid {
                ans = ans.max(mask.count_ones() as i32);
            }
        }
        ans
    }
}
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class Solution {
    maximumGood(statements: number[][]): number {
        const n = statements.length;
        let ans = 0;
        for (let mask = 0; mask < (1 << n); ++mask) {
            let valid = true;
            for (let i = 0; i < n && valid; ++i) {
                if (((mask >> i) & 1) === 0) continue;
                for (let j = 0; j < n; ++j) {
                    if (statements[i][j] === 2) continue;
                    if (statements[i][j] !== ((mask >> j) & 1)) {
                        valid = false;
                        break;
                    }
                }
            }
            if (valid) ans = Math.max(ans, mask.toString(2).split('1').length - 1);
        }
        return ans;
    }
}

Complexity

  • ⏰ Time complexity: O(n * 2^n), since we check all possible assignments (2^n) and for each, check all n^2 statements.
  • 🧺 Space complexity: O(1), only a few variables are used for computation.