LeetCode #825 — MEDIUM

Friends Of Appropriate Ages

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

There are n persons on a social media website. You are given an integer array ages where ages[i] is the age of the i^th 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] <= 0.5 * age[x] + 7
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

Step 01

Brute Force Baseline

Problem summary: 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] <= 0.5 * age[x] + 7 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Two Pointers · Binary Search

Example 1

[16,16]

Example 2

[16,17,18]

Example 3

[20,30,100,110,120]

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #825: Friends Of Appropriate Ages
class Solution {
    public int numFriendRequests(int[] ages) {
        final int m = 121;
        int[] cnt = new int[m];
        for (int x : ages) {
            ++cnt[x];
        }
        int ans = 0;
        for (int ax = 1; ax < m; ++ax) {
            for (int ay = 1; ay < m; ++ay) {
                if (!(ay <= 0.5 * ax + 7 || ay > ax || (ay > 100 && ax < 100))) {
                    ans += cnt[ax] * (cnt[ay] - (ax == ay ? 1 : 0));
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #825: Friends Of Appropriate Ages
func numFriendRequests(ages []int) (ans int) {
	cnt := [121]int{}
	for _, x := range ages {
		cnt[x]++
	}
	for ax, x := range cnt {
		for ay, y := range cnt {
			if ay <= ax/2+7 || ay > ax || (ay > 100 && ax < 100) {
				continue
			}
			if ax == ay {
				ans += x * (x - 1)
			} else {
				ans += x * y
			}
		}
	}

	return
}

# Accepted solution for LeetCode #825: Friends Of Appropriate Ages
class Solution:
    def numFriendRequests(self, ages: List[int]) -> int:
        cnt = [0] * 121
        for x in ages:
            cnt[x] += 1
        ans = 0
        for ax, x in enumerate(cnt):
            for ay, y in enumerate(cnt):
                if not (ay <= 0.5 * ax + 7 or ay > ax or (ay > 100 and ax < 100)):
                    ans += x * (y - int(ax == ay))
        return ans

// Accepted solution for LeetCode #825: Friends Of Appropriate Ages
struct Solution;
use std::collections::HashMap;

impl Solution {
    fn num_friend_requests(ages: Vec<i32>) -> i32 {
        let mut hm: HashMap<i32, i32> = HashMap::new();
        for age in ages {
            *hm.entry(age).or_default() += 1;
        }
        let mut res = 0;
        for (&a, v) in &hm {
            for (&b, u) in &hm {
                if !(b > a || 2 * b <= a + 14) {
                    res += v * u;
                    if a == b {
                        res -= v;
                    }
                }
            }
        }
        res
    }
}

#[test]
fn test() {
    let ages = vec![16, 16];
    let res = 2;
    assert_eq!(Solution::num_friend_requests(ages), res);
    let ages = vec![16, 17, 18];
    let res = 2;
    assert_eq!(Solution::num_friend_requests(ages), res);
    let ages = vec![20, 30, 100, 110, 120];
    let res = 3;
    assert_eq!(Solution::num_friend_requests(ages), res);
    let ages = vec![73, 106, 39, 6, 26, 15, 30, 100, 71, 35, 46, 112, 6, 60, 110];
    let res = 29;
    assert_eq!(Solution::num_friend_requests(ages), res);
}

// Accepted solution for LeetCode #825: Friends Of Appropriate Ages
function numFriendRequests(ages: number[]): number {
    const m = 121;
    const cnt = Array(m).fill(0);
    for (const x of ages) {
        cnt[x]++;
    }

    let ans = 0;
    for (let ax = 0; ax < m; ax++) {
        for (let ay = 0; ay < m; ay++) {
            if (ay <= 0.5 * ax + 7 || ay > ax || (ay > 100 && ax < 100)) {
                continue;
            }
            ans += cnt[ax] * (cnt[ay] - (ax === ay ? 1 : 0));
        }
    }

    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.