LeetCode 825. Friends Of Appropriate Ages _leetcode

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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 { 2public int numFriendRequests(int[] ages) { 3if(ages == null || ages.length == 0){ 4return 0; 5} 6 7HashMap< Integer, Integer> hm = new HashMap< > (); 8for(int age : ages){ 9hm.put(age, hm.getOrDefault(age, 0) + 1); 10} 11 12int res = 0; 13for(int a : hm.keySet()){ 14for(int b : hm.keySet()){ 15if(couldSendRequest(a, b)){ 16res += hm.get(a) * (hm.get(b) - (a == b ? 1 : 0)); 17} 18} 19} 20 21return res; 22} 23 24private boolean couldSendRequest(int a, int b){ 25return !(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.
【LeetCode 825. Friends Of Appropriate Ages】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 { 2public int numFriendRequests(int[] ages) { 3if(ages == null || ages.length == 0){ 4return 0; 5} 6 7int [] count = new int[121]; 8for(int age : ages){ 9count[age]++; 10} 11 12int [] sum = new int[121]; 13for(int i = 1; i< 121; i++){ 14sum[i] = sum[i-1] + count[i]; 15} 16 17int res = 0; 18for(int i = 15; i< 121; i++){ 19if(count[i] == 0){ 20continue; 21} 22 23int bCount = sum[i] - sum[i/2+7]; 24res += bCount * count[i] - count[i]; 25} 26 27return res; 28} 29 }