825. Friends Of Appropriate Ages LeetCode Solution

Last updated: February 20, 2025

In this guide, you will get 825. Friends Of Appropriate Ages LeetCode Solution with the best time and space complexity. The solution to Friends Of Appropriate Ages problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Friends Of Appropriate Ages solution in C++
Friends Of Appropriate Ages solution in Java
Friends Of Appropriate Ages solution in Python
Additional Resources

825. Friends Of Appropriate Ages LeetCode Solution image

Problem Statement of Friends Of Appropriate Ages

There are n persons on a social media website. You are given an integer array ages where ages[i] is the age of the ith person.
A Person x will not send a friend request to a person y (x != y) if any of the following conditions is true:

age[y] age[x]
age[y] > 100 && age[x] < 100

Otherwise, x will send a friend request to y.
Note that if x sends a request to y, y will not necessarily send a request to x. Also, a person will not send a friend request to themself.
Return the total number of friend requests made.

Example 1:

Input: ages = [16,16]
Output: 2
Explanation: 2 people friend request each other.

Example 2:

Input: ages = [16,17,18]
Output: 2
Explanation: Friend requests are made 17 -> 16, 18 -> 17.

Example 3:

Input: ages = [20,30,100,110,120]
Output: 3
Explanation: Friend requests are made 110 -> 100, 120 -> 110, 120 -> 100.

Constraints:

n == ages.length
1 <= n <= 2 * 104
1 <= ages[i] <= 120

Complexity Analysis

Time Complexity: O(n^2)
Space Complexity: O(n)

825. Friends Of Appropriate Ages LeetCode Solution in C++

class Solution {
 public:
  int numFriendRequests(vector<int>& ages) {
    int ans = 0;
    unordered_map<int, int> count;

    for (const int age : ages)
      ++count[age];

    for (const auto& [ageA, countA] : count)
      for (const auto& [ageB, countB] : count)
        if (request(ageA, ageB))
          if (ageA == ageB)
            ans += countA * (countB - 1);
          else
            ans += countA * countB;

    return ans;
  }

 private:
  bool request(int ageA, int ageB) {
    return !(ageB <= 0.5 * ageA + 7 || ageB > ageA || ageB > 100 && ageA < 100);
  }
};
/* code provided by PROGIEZ */

825. Friends Of Appropriate Ages LeetCode Solution in Java

class Solution {
  public int numFriendRequests(int[] ages) {
    int ans = 0;
    int[] count = new int[121];

    for (final int age : ages)
      ++count[age];

    for (int ageA = 1; ageA <= 120; ++ageA)
      for (int ageB = 1; ageB <= 120; ++ageB) {
        final int countA = count[ageA];
        final int countB = count[ageB];
        if (countA > 0 && countB > 0 && request(ageA, ageB))
          if (ageA == ageB)
            ans += countA * (countB - 1);
          else
            ans += countA * countB;
      }

    return ans;
  }

  private boolean request(int ageA, int ageB) {
    return !(ageB <= 0.5 * ageA + 7 || ageB > ageA || ageB > 100 && ageA < 100);
  }
}
// code provided by PROGIEZ

825. Friends Of Appropriate Ages LeetCode Solution in Python

class Solution:
  def numFriendRequests(self, ages: list[int]) -> int:
    ans = 0
    count = [0] * 121

    for age in ages:
      count[age] += 1

    for i in range(15, 121):
      ans += count[i] * (count[i] - 1)

    for i in range(15, 121):
      for j in range(i // 2 + 8, i):
        ans += count[i] * count[j]

    return ans
 # code by PROGIEZ

Additional Resources

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