Friends Of Appropriate Ages

Some people will make friend requests. The list of their ages is given and ages[i] is the age of the ith person.

Person A will NOT friend request person B (B != A) if any of the following conditions are true:

• age[B] <= 0.5 * age[A] + 7

• age[B] > age[A]

• age[B] > 100 && age[A] < 100

Otherwise, A will friend request B.

Note that if A requests B, B does not necessarily request A. Also, people will not friend request themselves.

How many total friend requests are made?

Example 1:

Input: [16,16]
Output: 2
Explanation: 2 people friend request each other.

Example 2:

Input: [16,17,18]
Output: 2
Explanation: Friend requests are made 17 -> 16, 18 -> 17.

Example 3:

Input: [20,30,100,110,120]
Output: 3
Explanation: Friend requests are made 110 -> 100, 120 -> 110, 120 -> 100.

Notes:

• 1 <= ages.length <= 20000.

• 1 <= ages[i] <= 120.

class Solution {
    // Three conditions could be merged to one:
    // The Person with age A can request person with age B if
    // B is in range ( 0.5 * A + 7, A ]
    public int numFriendRequests(int[] ages) {
        int res = 0;
        int[] numInAge = new int[121], sumInAge = new int[121];
        // numInAge[i]-> number of people with age=i
        for (int i : ages)
            numInAge[i]++;
        // sumInAge[i] -> number of people with age <= i
        for (int i = 1; i <= 120; ++i)
            sumInAge[i] = numInAge[i] + sumInAge[i - 1];
        // Because 0.5 * i + 7 >= i when i < 15.
        for (int i = 15; i <= 120; ++i) {
            if (numInAge[i] == 0)
                continue;
            int count = sumInAge[i] - sumInAge[i / 2 + 7];
            res += count * numInAge[i] - numInAge[i]; // people will not friend request themselves, so - numInAge[i]
        }
        return res;
    }
}