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feat: add solutions to lc problems: No.2928,2929 #1956

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Original file line number Diff line number Diff line change
Expand Up @@ -41,34 +41,123 @@

<!-- 这里可写通用的实现逻辑 -->

**方法一:组合数学 + 容斥原理**

根据题目描述,我们需要将 $n$ 个糖果分给 $3$ 个小孩,每个小孩分到的糖果数在 $[0, limit]$ 之间。

这实际上等价于把 $n$ 个球放入 $3$ 个盒子中。由于盒子可以为空,我们可以再增加 $3$ 个虚拟球,然后再利用隔板法,即一共有 $n + 3$ 个球,我们在其中 $n + 3 - 1$ 个位置插入 $2$ 个隔板,从而将实际的 $n$ 个球分成 $3$ 组,并且允许盒子为空,因此初始方案数为 $C_{n + 2}^2$。

我们需要在这些方案中,排除掉存在盒子分到的小球数超过 $limit$ 的方案。考虑其中有一个盒子分到的小球数超过 $limit$,那么剩下的球(包括虚拟球)最多有 $n + 3 - (limit + 1) = n - limit + 2$ 个,位置数为 $n - limit + 1$,因此方案数为 $C_{n - limit + 1}^2$。由于存在 $3$ 个盒子,因此这样的方案数为 $3 \times C_{n - limit + 1}^2$。这样子算,我们会多排除掉同时存在两个盒子分到的小球数超过 $limit$ 的方案,因此我们需要再加上这样的方案数,即 $3 \times C_{n - 2 \times limit}^2$。

时间复杂度 $O(1)$,空间复杂度 $O(1)$。

<!-- tabs:start -->

### **Python3**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```python

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
```

### **Java**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```java

class Solution {
public int 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 (int) ans;
}

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

### **C++**

```cpp

class Solution {
public:
int 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;
}
};
```

### **Go**

```go
func distributeCandies(n int, limit int) int {
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 ans
}
```

### **TypeScript**

```ts
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;
}
```

### **...**
Expand Down
Original file line number Diff line number Diff line change
Expand Up @@ -35,30 +35,119 @@

## Solutions

**Solution 1: Combinatorial Mathematics + Principle of Inclusion-Exclusion**

According to the problem description, we need to distribute $n$ candies to $3$ children, with each child receiving between $[0, limit]$ candies.

This is equivalent to placing $n$ balls into $3$ boxes. Since the boxes can be empty, we can add $3$ virtual balls, and then use the method of inserting partitions, i.e., there are a total of $n + 3$ balls, and we insert $2$ partitions among the $n + 3 - 1$ positions, thus dividing the actual $n$ balls into $3$ groups, and allowing the boxes to be empty. Therefore, the initial number of schemes is $C_{n + 2}^2$.

We need to exclude the schemes where the number of balls in a box exceeds $limit$. Consider that there is a box where the number of balls exceeds $limit$, then the remaining balls (including virtual balls) have at most $n + 3 - (limit + 1) = n - limit + 2$, and the number of positions is $n - limit + 1$, so the number of schemes is $C_{n - limit + 1}^2$. Since there are $3$ boxes, the number of such schemes is $3 \times C_{n - limit + 1}^2$. In this way, we will exclude too many schemes where the number of balls in two boxes exceeds $limit$ at the same time, so we need to add the number of such schemes, i.e., $3 \times C_{n - 2 \times limit}^2$.

The time complexity is $O(1)$, and the space complexity is $O(1)$.

<!-- tabs:start -->

### **Python3**

```python

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
```

### **Java**

```java

class Solution {
public int 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 (int) ans;
}

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

### **C++**

```cpp

class Solution {
public:
int 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;
}
};
```

### **Go**

```go
func distributeCandies(n int, limit int) int {
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 ans
}
```

### **TypeScript**

```ts
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;
}
```

### **...**
Expand Down
Original file line number Diff line number Diff line change
@@ -0,0 +1,19 @@
class Solution {
public:
int 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;
}
};
Original file line number Diff line number Diff line change
@@ -0,0 +1,16 @@
func distributeCandies(n int, limit int) int {
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 ans
}
Original file line number Diff line number Diff line change
@@ -0,0 +1,19 @@
class Solution {
public int 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 (int) ans;
}

private long comb2(int n) {
return 1L * n * (n - 1) / 2;
}
}
Original file line number Diff line number Diff line change
@@ -0,0 +1,10 @@
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
Original file line number Diff line number Diff line change
@@ -0,0 +1,14 @@
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;
}
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