LeetCode 2041: Accepted Candidates From the Interviews Solution

Master LeetCode problem 2041 (Accepted Candidates From the Interviews), 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.

2041. Accepted Candidates From the Interviews

Medium

Database

Problem Explanation

Explanation

To solve this problem, we can iterate through the list of candidates and interviews and keep track of the accepted candidates. We will use a hashmap to store the count of accepted candidates for each company. Then, we will iterate through the hashmap to find the company with the maximum number of accepted candidates.

Create a hashmap to store the count of accepted candidates for each company.
Iterate through the list of candidates and interviews.
For each candidate, check if they have passed the interview.
If the candidate has passed the interview, increment the count of accepted candidates for the corresponding company in the hashmap.
After iterating through all candidates, find the company with the maximum number of accepted candidates.
Return the company with the maximum number of accepted candidates.

Time Complexity

The time complexity of this approach is O(n), where n is the total number of candidates.

Space Complexity

The space complexity of this approach is O(m), where m is the number of unique companies.

Solution Code

class Solution {
    public String findAcceptedCompany(String[] candidates, String[] interviews) {
        Map<String, Integer> acceptedCandidates = new HashMap<>();
        
        for (int i = 0; i < candidates.length; i++) {
            if (interviews[i].equals("Accepted")) {
                String company = candidates[i];
                acceptedCandidates.put(company, acceptedCandidates.getOrDefault(company, 0) + 1);
            }
        }
        
        String maxCompany = "";
        int maxCount = 0;
        
        for (Map.Entry<String, Integer> entry : acceptedCandidates.entrySet()) {
            if (entry.getValue() > maxCount) {
                maxCount = entry.getValue();
                maxCompany = entry.getKey();
            }
        }
        
        return maxCompany;
    }
}

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