1045MediumArrayTwo PointersBinary SearchSorting

How Many Invitations Go Out

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A club's members have the ages ages. A member x sends an invitation to a different member y exactly when both of these hold:

y is no older than x;
y's age is more than half of x's age plus seven.

Invitations go one way only, so x inviting y and y inviting x are counted separately, and nobody invites themselves.

Return how many invitations go out altogether.

Examples

Example 1

Input: ages = [15, 15]
Output: 2

Neither member is older than the other, and fifteen is more than half of fifteen plus seven, which is fourteen and a half, so both invitations go out.

Example 2

Input: ages = [100, 100, 100]
Output: 6

A hundred is more than half of a hundred plus seven, so every member invites every other: three members with two others apiece.

Example 3

Input: ages = [1]
Output: 0

A single member has nobody else to invite, and nobody invites themselves.

Constraints

1 <= ages.length <= 2 * 10^4
1 <= ages[i] <= 120

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def num_friend_requests(ages: list[int]) -> int:

Java

public int numFriendRequests(int[] ages)

The intended approach

Time O(n + A^2) for n members and A the highest age.Space O(A) for the tally.

Whether an invitation goes out depends only on the two ages, never on which members hold them. Ages reach no higher than a hundred and twenty, so tally how many members hold each age and then work with pairs of ages. For a pair of different ages, the invitations between them are the two tallies multiplied together. For a pair of the same age, it is the tally times one less than itself, since a member cannot be their own invitee. The test on the ages is a plain comparison. Doubling rather than halving keeps everything in whole numbers: `y`'s age doubled must be more than `x`'s age plus fourteen, and `y` must not be older than `x`. Only a hundred and twenty ages exist, so a fixed number of pairs is examined however large the club is.

Pythonentry point: num_friend_requests

def num_friend_requests(ages):
    tally = [0] * 121
    for age in ages:
        tally[age] += 1
    total = 0
    for x in range(1, 121):
        if tally[x] == 0:
            continue
        for y in range(1, x + 1):
            if tally[y] == 0:
                continue
            if 2 * y <= x + 14:
                continue
            if x == y:
                total += tally[x] * (tally[x] - 1)
            else:
                total += tally[x] * tally[y]
    return total

The obvious version

Take every ordered pair of different members and test the two conditions on their ages outright, counting the invitations that go out. Costs the club's size squared rather than a tally over the ages.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_num_friend_requests(ages):
    n = len(ages)
    total = 0
    for i in range(n):
        for j in range(n):
            if i == j:
                continue
            x = ages[i]
            y = ages[j]
            if y > x:
                continue
            if 2 * y <= x + 14:
                continue
            total += 1
    return total

How this problem was checked

Reference solution executed against 15 test cases — 15 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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