825

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.

 

Notes:

• 1 <= ages.length <= 20000.
• 1 <= ages[i] <= 120.

Companies:
Facebook

Related Topics:
Array

Solution 1.

// OJ: https://leetcode.com/problems/friends-of-appropriate-ages/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
private:
    bool request(int A, int B) {
        return !(B <= .5 * A + 7 || B > A);
    }
public:
    int numFriendRequests(vector<int>& ages) {
        int ans = 0;
        vector<int> cnts(120, 0);
        for (int age : ages) cnts[age - 1]++;
        for (int i = 0; i < 120; ++i) {
            if (!cnts[i]) continue;
            for (int j = 0; j < 120; ++j) {
                if (!cnts[j] || !request(i + 1, j + 1)) continue;
                ans += cnts[i] * cnts[j];
                if (i == j) ans -= cnts[i];
            }
        }
        return ans;
    }
};