825. Friends Of Appropriate Ages

time: O(121*121), O(1)

space:O(121), O(1)

class Solution {
    public int numFriendRequests(int[] ages) {
        int count[] = new int[121];
        for (int age : ages) {
            count[age]++;
        }
        
        int res = 0;
        for (int ageA = 0; ageA <= 120; ageA++) {
            int countA = count[ageA];
            for (int ageB = 0; ageB <= 120; ageB++) {
                
                int countB = count[ageB];
                if (notAbleToSendRequest(ageA, ageB)) {
                    continue;
                }
                
                countB = (ageA == ageB) ? countB-1 : countB; // 兩年齡相同時, should 扣掉自己的一個個數, (不能對自己發邀請)
                res += countA*countB;

            }
        }
        return res;
    }
    
    private boolean notAbleToSendRequest(int ageX, int ageY) {
        return ageY <= 0.5*ageX+7 || ageY > ageX || (ageY > 100 && ageX < 100);
    }
}

/*

16, 17, 17

1*2 = 2
1


16 17 18  => in no limitation: 6 why?

16 -> 17 18

17 -> 16,18

18 -> 16,17

so exclude yourself
(age, count)

countA * countB
ageA == ageB, countA * (countA - 1)


*/