LeetCode 2004: The Number of Seniors and Juniors to Join the Company Solution

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2004. The Number of Seniors and Juniors to Join the Company

Hard

Database

Problem Explanation

Explanation

To solve this problem, we can use a binary search approach. We first sort the array of seniority levels and iterate through each possible number of seniors to join the company. For each number of seniors, we can use binary search to find the maximum number of juniors that can join the company while meeting the constraints. The constraints are that for each senior to join, at least one junior must join, and the junior's seniority level must be less than the senior's seniority level.

We start by sorting the array and then iterate through each possible number of seniors to join. For each senior, we use binary search to find the maximum number of juniors that can join. If the total number of seniors and juniors that can join is greater than the current maximum, we update the maximum. Finally, we return the maximum number of seniors and juniors that can join the company.

Time complexity: O(n log n), where n is the number of seniority levels. Space complexity: O(1)

Solution Code

public int[] numSeniorsAndJuniors(int[] seniorityLevels) {
    Arrays.sort(seniorityLevels);
    int maxSeniors = 0, maxJuniors = 0;
    
    for (int seniors = 0; seniors < seniorityLevels.length; seniors++) {
        int juniors = binarySearch(seniorityLevels, seniorityLevels[seniors]);
        if (seniors + juniors > maxSeniors + maxJuniors) {
            maxSeniors = seniors;
            maxJuniors = juniors;
        }
    }
    
    return new int[] {maxSeniors, maxJuniors};
}

private int binarySearch(int[] arr, int target) {
    int left = 0, right = arr.length - 1;
    int result = 0;
    
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (arr[mid] < target) {
            result = mid + 1;
            left = mid + 1;
        } else {
            right = mid - 1;
        }
    }
    
    return result;
}

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