LeetCode 3204: Bitwise User Permissions Analysis Solution

Master LeetCode problem 3204 (Bitwise User Permissions Analysis), a medium 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.

3204. Bitwise User Permissions Analysis

Medium

Database

Problem Explanation

Explanation:

To solve this problem, we can use bitwise operations to analyze user permissions. Each user has a set of permissions represented by a binary string, where each bit position corresponds to a specific permission. We are given a list of user permissions and a target permission mask. Our goal is to find the users who have all the permissions specified in the target mask.

We can iterate through each user's permission and check if the user has all the permissions specified in the target mask by performing a bitwise AND operation between the user's permission and the target mask. If the result is equal to the target mask, it means the user has all the permissions. We can keep track of these users and return them as the final result.

Algorithm:

Initialize an empty list to store the users who have all the permissions specified in the target mask.
Iterate through each user's permission:
- Perform a bitwise AND operation between the user's permission and the target mask.
- If the result is equal to the target mask, add the user to the list of users with all permissions.
Return the list of users with all permissions.

Time Complexity: O(N), where N is the number of users. Space Complexity: O(1) - excluding the space required to store the result.

Solution Code

class Solution {
    public List<String> analyzeUserPermissions(String[] users, String targetPermission) {
        List<String> result = new ArrayList<>();
        int targetMask = Integer.parseInt(targetPermission, 2);
        
        for (String user : users) {
            int userMask = Integer.parseInt(user, 2);
            if ((userMask & targetMask) == targetMask) {
                result.add(user);
            }
        }
        
        return result;
    }
}

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