How do you solve Friends Of Appropriate Ages on LeetCode?

Friends Of Appropriate Ages (LeetCode #825) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: Understanding the conditions for sending friend requests is crucial.

What is the brute force solution for Friends Of Appropriate Ages?

The brute force approach for Friends Of Appropriate Ages is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we check every possible pair of people to see if one can send a friend request to the other. This is straightforward but inefficient as it involves checking all combinations. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Friends Of Appropriate Ages?

Friends Of Appropriate Ages uses the following concepts: Array, Two Pointers, Binary Search, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Friends Of Appropriate Ages?

Always clarify the problem constraints and edge cases with the interviewer. Discuss your thought process and approach before jumping into coding. Optimize your solution step-by-step, explaining why each change is made.

What are common mistakes when solving Friends Of Appropriate Ages?

Not considering the case where ages are equal when counting requests. Failing to handle edge cases like ages above 100 correctly.

#825

Friends Of Appropriate Ages

Medium

ArrayTwo PointersBinary SearchSortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(1)

💡

Intuition

Time O(n log n)Space O(1)

We can optimize the process by sorting the ages and using a single loop with a two-pointer technique to count valid friend requests, which reduces unnecessary comparisons.

⚙️

Algorithm

3 steps

1Step 1: Sort the ages array.
2Step 2: Initialize a counter for friend requests.
3Step 3: Use a loop to iterate through each age and use a two-pointer technique to find the range of valid ages that can receive requests.

solution.py11 lines

1def numFriendRequests(ages):
2    ages.sort()
3    count = 0
4    n = len(ages)
5    for i in range(n):
6        for j in range(i + 1, n):
7            if ages[j] > ages[i]:
8                if not (ages[j] <= 0.5 * ages[i] + 7 or (ages[j] > 100 and ages[i] < 100)):
9                    count += 1
10    return count + (ages.count(ages[i]) - 1) * (ages.count(ages[i])) // 2
11

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step, while the nested loop runs in O(n²) in the worst case, but we avoid many unnecessary checks. The space complexity is O(1) as we are using a constant amount of extra space.

1Understanding the conditions for sending friend requests is crucial.
2Sorting the ages can help reduce the number of comparisons needed.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.