825. Friends Of Appropriate Ages

Solution idea:

Brute force method

step 1: sub-function makeRequest(x,y) that checks if y will make a request to x by the listed conditions

step 2: use a map countMap that counts number of people at each unique age because age only ranges from $[1,120]$

step 3: nested-for loop. For any two keys x, y of countMap that satisfies makeRequest(x,y):

case 1: if x==y, then # of requests = countMap[x] * (countMap[y] - 1) ($-1$ to exclude friend request itself)

case 2: if x!=y, then # of requests = countMap[x] * countMap[y]

Time complexity = $O(n^2)$