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; } }