Formatted question description: https://leetcode.ca/all/825.html

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/
// 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;
    }
};

Java

class Solution {
    public int numFriendRequests(int[] ages) {
        Arrays.sort(ages);
        Map<Integer, Integer> ageCountMap = new HashMap<Integer, Integer>();
        for (int age : ages) {
            int count = ageCountMap.getOrDefault(age, 0) + 1;
            ageCountMap.put(age, count);
        }
        int friendRequests = 0;
        int length = ages.length;
        for (int i = 0; i < length; i++) {
            if (i > 0 && ages[i] == ages[i - 1])
                continue;
            int age = ages[i];
            int minAge = age / 2 + 8;
            if (minAge > age)
                continue;
            int minIndex = binarySearch(ages, minAge);
            int count = ageCountMap.get(age);
            friendRequests += count * (i - minIndex);
            friendRequests += count * (count - 1);
        }
        return friendRequests;
    }

    public int binarySearch(int[] ages, int minAge) {
        int low = 0, high = ages.length - 1;
        while (low < high) {
            int mid = (high - low) / 2 + low;
            int age = ages[mid];
            if (age < minAge)
                low = mid + 1;
            else
                high = mid;
        }
        return low;
    }
}

// OJ: https://leetcode.com/problems/friends-of-appropriate-ages/
// 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;
    }
};

class Solution(object):
    def numFriendRequests(self, ages):
        """
        :type ages: List[int]
        :rtype: int
        """
        count = collections.Counter(ages)
        ages = sorted(count.keys())
        N = len(ages)
        res = 0
        for A in ages:
            for B in range(int(0.5 * A) + 7 + 1, A + 1):
                res += count[A] * (count[B] - int(A == B))
        return res

825 - Friends Of Appropriate Ages

825. Friends Of Appropriate Ages (Medium)

Solution 1.

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