825. Friends Of Appropriate Ages (Medium)

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: 
Explanation: Friend requests are made 110 -> 100, 120 -> 110, 120 -> 100.

這道題如果遍歷a，b，時間複雜度是O(n2)，妥妥超時

但是由於年齡《120，所以遍歷年齡的count，

因為一對a->b滿足條件的話，所有a年齡的人可以對所有b年齡的人發起朋友請求，總共數量count(a) * count(b)

而如果ab年齡相同，由於不能對自己發起朋友請求，所以總共數量count(a) * ( count(a) - 1)

class Solution(object):
    def numFriendRequests(self, ages):
        """
        :type ages: List[int]
        :rtype: int
        """
        count = collections.Counter(ages)
        num = 0
        ageset = list(set(ages))
        for i in range(len(ageset)):
            for j in range(len(ageset)):
                if not (ageset[i] <= 0.5*ageset[j] + 7 or (ageset[i] > ageset[j]) or (ageset[i] > 100 and ageset[j] < 100)):
                    if ageset[i] == ageset[j]:num += count[ageset[i]]*(count[ageset[i]]-1)
                    else:num += count[ageset[i]]*count[ageset[j]]
        return num