LeetCode 825. Friends Of Appropriate Ages

原题链接在这里：https://leetcode.com/problems/friends-of-appropriate-ages/

题目：

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.

题解：

Accumlate the frequency of different ages.

If age a and age b could send request, and a != b, then res += a freq * b freq.

If a == b, since no one could send friend request to themselves, the request is a freq * (a freq - 1). Send friend request to other people with same age.

Time Complexity: O(n^2). n = ages.length.

Space: O(n).

AC Java:

 class Solution {
     public int numFriendRequests(int[] ages) {
         if(ages == null || ages.length == 0){
             return 0;
         }

         HashMap<Integer, Integer> hm = new HashMap<>();
         for(int age : ages){
             hm.put(age, hm.getOrDefault(age, 0) + 1);
         }

         int res = 0;
         for(int a : hm.keySet()){
             for(int b : hm.keySet()){
                 if(couldSendRequest(a, b)){
                     res += hm.get(a) * (hm.get(b) - (a == b ? 1 : 0));
                 }
             }
         }

         return res;
     }

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

With 3 conditions, we only care the count of B in range (a/2+7, a].

Get the sum count of b and * a count - a count since people can't sent friend request to themselves.

Since A > B >= 0.5*A+7, A > 0.5*A+7. Then A>14. Thus i is started from 15.

Time Complexity: O(n).

Space: O(1).

AC Java:

 class Solution {
     public int numFriendRequests(int[] ages) {
         if(ages == null || ages.length == 0){
             return 0;
         }

         int [] count = new int[121];
         for(int age : ages){
             count[age]++;
         }

         int [] sum = new int[121];
         for(int i = 1; i<121; i++){
             sum[i] = sum[i-1] + count[i];
         }

         int res = 0;
         for(int i = 15; i<121; i++){
             if(count[i] == 0){
                 continue;
             } 

             int bCount = sum[i] - sum[i/2+7];
             res += bCount * count[i] - count[i];
         }

         return res;
     }
 }