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:

 1 class Solution {
 2     public int numFriendRequests(int[] ages) {
 3         if(ages == null || ages.length == 0){
 4             return 0;
 5         }
 6         
 7         HashMap hm = new HashMap<>();
 8         for(int age : ages){
 9             hm.put(age, hm.getOrDefault(age, 0) + 1);
10         }
11         
12         int res = 0;
13         for(int a : hm.keySet()){
14             for(int b : hm.keySet()){
15                 if(couldSendRequest(a, b)){
16                     res += hm.get(a) * (hm.get(b) - (a == b ? 1 : 0));
17                 }
18             }
19         }
20         
21         return res;
22     }
23     
24     private boolean couldSendRequest(int a, int b){
25         return !(b <= a*0.5 + 7 || b > a || (b > 100 && a < 100));
26     }
27 }

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:

 1 class Solution {
 2     public int numFriendRequests(int[] ages) {
 3         if(ages == null || ages.length == 0){
 4             return 0;
 5         }
 6         
 7         int [] count = new int[121];
 8         for(int age : ages){
 9             count[age]++;
10         }
11         
12         int [] sum = new int[121];
13         for(int i = 1; i<121; i++){
14             sum[i] = sum[i-1] + count[i];
15         }
16         
17         int res = 0;
18         for(int i = 15; i<121; i++){
19             if(count[i] == 0){
20                 continue;
21             } 
22             
23             int bCount = sum[i] - sum[i/2+7];
24             res += bCount * count[i] - count[i];
25         }
26         
27         return res;
28     }
29 }