适龄的朋友

/*
 * 思路：
 * 使用Java Streams来简化迭代过程。我们可以利用流的特性来实现对每一对用户进行检查。
 */

import java.util.Arrays;

public class FriendRequestsStream {
    public long numFriendRequests(int[] ages) {
        return Arrays.stream(ages)
                .boxed()
                .flatMap(ageX -> Arrays.stream(ages)
                        .filter(ageY -> ageX != ageY && canSendRequest(ageX, ageY)))
                .count();
    }

    private boolean canSendRequest(int ageX, int ageY) {
        if (ageY <= 0.5 * ageX + 7) return false;
        if (ageY > ageX) return false;
        if (ageY > 100 && ageX < 100) return false;
        return true;
    }
}

This solution utilizes a frequency array to count the occurrences of each age from 1 to 120. It creates a prefix sum array to quickly calculate the sum of frequencies up to any given age. The logic involves iterating over each age and determining the range of ages that can receive friend requests from that age based on the given conditions. For each valid age, the solution calculates the number of potential friend requests by multiplying the frequency of the current age by the number of users in the valid age range (using the prefix sum to find this number quickly). This method avoids the need for nested iterations over the `ages` array, thus improving efficiency.

O(n)

O(1)