LeetCode 1811: Find Interview Candidates

Explanation:

To solve this problem, we can use a two-pointer approach. We sort both the scores and the lower bounds arrays in ascending order. Then, we iterate through the scores and lower bounds arrays using two pointers to find the number of candidates that meet the score requirements.

1. Sort the scores and lower_bounds arrays in ascending order.
2. Initialize a variable count to keep track of the number of candidates that meet the score requirements.
3. Iterate through both arrays using two pointers:
   • If the current score is greater than or equal to the current lower bound, increment the count.
   • Move the pointers accordingly.
4. Return the count as the result.

Time Complexity:

The time complexity of this approach is O(nlogn) where n is the length of the input arrays due to sorting the arrays before processing.

Space Complexity:

The space complexity is O(1) as we are not using any extra space other than a few variables.

:

class Solution {
    public int findInterviewCandidates(int[] scores, int[] lower_bounds) {
        Arrays.sort(scores);
        Arrays.sort(lower_bounds);
        
        int count = 0;
        int i = 0, j = 0;
        
        while (i < scores.length && j < lower_bounds.length) {
            if (scores[i] >= lower_bounds[j]) {
                count++;
                i++;
                j++;
            } else {
                i++;
            }
        }
        
        return count;
    }
}