LeetCode 825 - Friends Of Appropriate Ages

https://leetcode.com/problems/friends-of-appropriate-ages/description/

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.

Instead of processing all 20000 people, we can process pairs of (age, count) representing how many people are that age. Since there are only 120 possible ages, this is a much faster loop.

Algorithm

For each pair (ageA, countA), (ageB, countB), if the conditions are satisfied with respect to age, then countA * countB pairs of people made friend requests.

If ageA == ageB, then we overcounted: we should have countA * (countA - 1) pairs of people making friend requests instead, as you cannot friend request yourself.

• Time Complexity: $O(\mathcal{A}^2 + N)$, where $N$ is the number of people, and $\mathcal{A}$ is the number of ages.
• Space Complexity: $O(\mathcal{A})$, the space used to store count.

    public int numFriendRequests(int[] ages) {

        int[] count = new int[121];

        for (int age: ages) count[age]++;

        int ans = 0;

        for (int ageA = 0; ageA <= 120; ageA++) {

            int countA = count[ageA];

            for (int ageB = 0; ageB <= 120; ageB++) {

                int countB = count[ageB];

                if (ageA * 0.5 + 7 >= ageB) continue;

                if (ageA < ageB) continue;

                if (ageA < 100 && 100 < ageB) continue;

                ans += countA * countB;

                if (ageA == ageB) ans -= countA;

            }

        }

        return ans;

    }

https://leetcode.com/problems/friends-of-appropriate-ages/discuss/127341/10ms-concise-Java-solution-O(n)-time-and-O(1)-space

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];
        
        for(int i : ages) 
            numInAge[i] ++;
        
        for(int i = 1; i <= 120; ++i) 
            sumInAge[i] = numInAge[i] + sumInAge[i - 1];
        
        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;}

https://leetcode.com/problems/friends-of-appropriate-ages/discuss/127029/C++JavaPython-Easy-and-Straight-Forward

    public int numFriendRequests(int[] ages) {
        Map<Integer, Integer> count = new HashMap<>();
        for (int age : ages) count.put(age, count.getOrDefault(age, 0) + 1);
        int res = 0;
        for (Integer a : count.keySet()) for (Integer b : count.keySet())
            if (request(a, b)) res += count.get(a) * (count.get(b) - (a == b ? 1 : 0));
        return res;
    }

    private boolean request(int a, int b) {
        return !(b <= 0.5 * a + 7 || b > a || (b > 100 && a < 100));
    }




https://leetcode.com/problems/friends-of-appropriate-ages/discuss/127303/Java-two-pointer-O(1)-space-solution