# 2929. Distribute Candies Among Children II

## Description

You are given two positive integers n and limit.

Return the total number of ways to distribute n candies among 3 children such that no child gets more than limit candies.

Example 1:

Input: n = 5, limit = 2
Output: 3
Explanation: There are 3 ways to distribute 5 candies such that no child gets more than 2 candies: (1, 2, 2), (2, 1, 2) and (2, 2, 1).


Example 2:

Input: n = 3, limit = 3
Output: 10
Explanation: There are 10 ways to distribute 3 candies such that no child gets more than 3 candies: (0, 0, 3), (0, 1, 2), (0, 2, 1), (0, 3, 0), (1, 0, 2), (1, 1, 1), (1, 2, 0), (2, 0, 1), (2, 1, 0) and (3, 0, 0).


Constraints:

• 1 <= n <= 106
• 1 <= limit <= 106

## Solutions

• class Solution {
public long distributeCandies(int n, int limit) {
if (n > 3 * limit) {
return 0;
}
long ans = comb2(n + 2);
if (n > limit) {
ans -= 3 * comb2(n - limit + 1);
}
if (n - 2 >= 2 * limit) {
ans += 3 * comb2(n - 2 * limit);
}
return ans;
}

private long comb2(int n) {
return 1L * n * (n - 1) / 2;
}
}

• class Solution {
public:
long long distributeCandies(int n, int limit) {
auto comb2 = [](int n) {
return 1LL * n * (n - 1) / 2;
};
if (n > 3 * limit) {
return 0;
}
long long ans = comb2(n + 2);
if (n > limit) {
ans -= 3 * comb2(n - limit + 1);
}
if (n - 2 >= 2 * limit) {
ans += 3 * comb2(n - 2 * limit);
}
return ans;
}
};

• class Solution:
def distributeCandies(self, n: int, limit: int) -> int:
if n > 3 * limit:
return 0
ans = comb(n + 2, 2)
if n > limit:
ans -= 3 * comb(n - limit + 1, 2)
if n - 2 >= 2 * limit:
ans += 3 * comb(n - 2 * limit, 2)
return ans


• func distributeCandies(n int, limit int) int64 {
comb2 := func(n int) int {
return n * (n - 1) / 2
}
if n > 3*limit {
return 0
}
ans := comb2(n+2)
if n > limit {
ans -= 3 * comb2(n-limit+1)
}
if n-2 >= 2*limit {
ans += 3 * comb2(n-2*limit)
}
return int64(ans)
}

• function distributeCandies(n: number, limit: number): number {
const comb2 = (n: number) => (n * (n - 1)) / 2;
if (n > 3 * limit) {
return 0;
}
let ans = comb2(n + 2);
if (n > limit) {
ans -= 3 * comb2(n - limit + 1);
}
if (n - 2 >= 2 * limit) {
ans += 3 * comb2(n - 2 * limit);
}
return ans;
}