What is the time complexity of Maximum Good People Based on Statements?

The optimal solution has a time complexity of O(n*2^n*n).

What data structures are used to solve Maximum Good People Based on Statements?

This problem uses Array, Backtracking, Bit Manipulation, Enumeration. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Maximum Good People Based on Statements?

Insight 1: The core idea is to try all possible combinations of 'good' and 'bad' assignments to the people and check if the statements made by the 'good' people are consistent with the assumed assignments. Insight 2: Bit manipulation is a natural fit for representing the assignments, as each bit in a mask can represent whether a person is 'good' or 'bad'.

Which companies ask LeetCode 2151. Maximum Good People Based on Statements in interviews?

This problem is frequently asked at TuSimple.

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Maximum Good People Based on Statements - LeetCode 2151 Solution

Maximum Good People Based on Statements - Complete Solution Guide

Maximum Good People Based on Statements is LeetCode problem 2151, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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 i

Detailed Explanation

The problem asks us to find the maximum number of 'good' people possible given a set of statements made by 'n' people about each other. Each person is either 'good' (always tells the truth) or 'bad' (might lie). The statements are represented in an n x n matrix, where statements[i][j] can be 0 (person i says person j is bad), 1 (person i says person j is good), or 2 (person i makes no statement about person j). We need to determine an assignment of 'good' and 'bad' to each person such that the statements made by the 'good' people are consistent with the assignments. The goal is to maximize the number of 'good' people in a consistent assignment.

Solution Approach

The solution employs a brute-force approach using bit manipulation. It iterates through all possible assignments of 'good' and 'bad' to the 'n' people. Each assignment is represented by a bitmask of length 'n'. For each assignment, it checks if the statements made by the people assumed to be 'good' are consistent with the assumed assignment itself. If an assignment is consistent, it counts the number of 'good' people and updates the maximum count found so far.

Step-by-Step Algorithm

Step 1: Iterate through all possible bitmasks from 0 to 2^n - 1. Each bitmask represents a possible assignment of 'good' and 'bad' to the people.
Step 2: For each bitmask, check for consistency. Iterate through all people from 0 to n-1.
Step 3: If the i-th person is assumed to be 'good' (i-th bit in the mask is set), iterate through all people from 0 to n-1 again.
Step 4: If the i-th person made a statement about the j-th person (statements[i][j] is not 2), check if the statement is consistent with the assignment in the mask. Specifically, check if statements[i][j] is equal to the j-th bit in the mask.
Step 5: If any statement made by a 'good' person is inconsistent, mark the assignment as inconsistent and break the inner loop.
Step 6: If the assignment is consistent after checking all statements, count the number of 'good' people in the assignment.
Step 7: Update the maximum number of 'good' people found so far.
Step 8: Return the maximum number of 'good' people.

Key Insights

Insight 1: The core idea is to try all possible combinations of 'good' and 'bad' assignments to the people and check if the statements made by the 'good' people are consistent with the assumed assignments.
Insight 2: Bit manipulation is a natural fit for representing the assignments, as each bit in a mask can represent whether a person is 'good' or 'bad'.
Insight 3: The problem constraints (n <= 15) suggest an exponential time complexity solution (like trying all possible assignments) is acceptable.

Complexity Analysis

Time Complexity: O(n*2^n*n)

Space Complexity: O(1)

Topics

This problem involves: Array, Backtracking, Bit Manipulation, Enumeration.

Companies

Asked at: TuSimple.

Maximum Good People Based on Statements | LeetCode 2151

Hard

Array Backtracking Bit Manipulation Enumeration

Time: O(n*2^n*n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int maximumGood(int[][] statements) {
        int n = statements.length;
        int max_good = 0;

        for (int mask = 0; mask < (1 << n); mask++) {
            boolean is_consistent = true;
            for (int i = 0; i < n; i++) {
                // If person i is assumed to be good in this mask
                if (((mask >> i) & 1) == 1) {
                    for (int j = 0; j < n; j++) {
                        if (statements[i][j] == 2) {
                            continue;
                        }
                        
                        // What the mask assumes person j's status is
                        int j_is_good_in_mask = (mask >> j) & 1;
                        
                        // If what person i says doesn't match what the mask assumes
                        if (statements[i][j] != j_is_good_in_mask) {
                            is_consistent = false;
                            break;
                        }
                    }
                }
                if (!is_consistent) {
                    break;
                }
            }
            
            if (is_consistent) {
                int good_count = 0;
                for (int i = 0; i < n; i++) {
                    if (((mask >> i) & 1) == 1) {
                        good_count++;
                    }
                }
                max_good = Math.max(max_good, good_count);
            }
        }
        
        return max_good;
    }
}

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Problem Info

Description

Problem 2151: Maximum Good People Based on Statements - Detailed explanation and solution approaches.

Topics

Array Backtracking Bit Manipulation Enumeration

Companies

TuSimple

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2151

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Iterate through all possible bitmasks from 0 to 2^n - 1. Each bitmask represents a possible assignment of 'good' and 'bad' to the people.
2
Step 2: For each bitmask, check for consistency. Iterate through all people from 0 to n-1.
3
Step 3: If the i-th person is assumed to be 'good' (i-th bit in the mask is set), iterate through all people from 0 to n-1 again.
4
Step 4: If the i-th person made a statement about the j-th person (statements[i][j] is not 2), check if the statement is consistent with the assignment in the mask. Specifically, check if statements[i][j] is equal to the j-th bit in the mask.
5
Step 5: If any statement made by a 'good' person is inconsistent, mark the assignment as inconsistent and break the inner loop.
6
Step 6: If the assignment is consistent after checking all statements, count the number of 'good' people in the assignment.
7
Step 7: Update the maximum number of 'good' people found so far.
8
Step 8: Return the maximum number of 'good' people.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n*2^n*n)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The algorithm iterates through 2^n possible assignments (outer loop). For each assignment, it iterates through n people (first inner loop). For each 'good' person, it iterates through n statements (second inner loop). Hence, the time complexity is O(n * n * 2^n) which simplifies to O(n^2 * 2^n). The algorithm uses a constant amount of extra space for variables, so the space complexity is O(1).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Backtracking

A backtracking approach can also be used. Recursively try assigning each person as 'good' or 'bad', and prune the search tree when an inconsistency is detected. This could lead to some improvements in practical cases by avoiding checking all 2^n possibilities if inconsistencies are found early on.

Time: O(n * 2^n) in the worst case, but possibly better in practice.Space: O(n) due to the recursion depth.

Trade-offs: Backtracking might be faster in practice for some input instances where inconsistencies are frequently found early in the assignment process. However, the worst-case time complexity remains exponential. The iterative bit manipulation approach is often easier to implement.

✨

Key Insights

1Insight 1: The core idea is to try all possible combinations of 'good' and 'bad' assignments to the people and check if the statements made by the 'good' people are consistent with the assumed assignments.
2Insight 2: Bit manipulation is a natural fit for representing the assignments, as each bit in a mask can represent whether a person is 'good' or 'bad'.
3Insight 3: The problem constraints (n <= 15) suggest an exponential time complexity solution (like trying all possible assignments) is acceptable.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: All statements are 2 (no statements are made). In this case, all people can be good, so the solution correctly returns n.
Edge case 2: n = 2, and statements = [[2, 0], [0, 2]]. The code handles this correctly. If person 0 is good, person 1 is bad. If person 0 is bad, and tells the truth, person 1 is bad. If person 0 is bad, and lies, person 1 is good. This means the max good persons is 1.
Edge case 3: Contradictory statements, such as person 0 says person 1 is good, and person 1 says person 0 is bad. The algorithm correctly determines that all people cannot be good simultaneously and explores other possible assignments.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not checking consistency correctly. Forgetting to skip when statements[i][j] == 2 can cause incorrect results.

✗Common mistake 2: Incorrectly interpreting the bitmask. Remember that the i-th bit represents whether the i-th person is good (1) or bad (0).

✗Common mistake 3: Early termination of the inner loop before checking all the statements made by a good person.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Detecting inconsistencies in a network of sensors where sensors report on the state of other sensors. The problem maps to determining the maximum number of reliable sensors given potentially conflicting reports.

🏦Real-world scenario 2: Trust assessment in social networks. People make statements about each other (e.g., their trustworthiness). Identifying the maximum number of trustworthy individuals given the network of statements can be valuable for various applications such as fraud detection or information validation.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 19, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I avoid duplicate solutions in backtracking?

Sort the input first, then skip consecutive duplicates at the same recursion level: if (i > start && nums[i] == nums[i-1]) continue. For permutations, use a visited array or swap-based approach. Always consider if order matters (permutations vs combinations).

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Maximum Good People Based on Statements"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether backtracking techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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