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Dec 23, 2023
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feat: add solutions to lc problem: No.2735 #2146

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88 changes: 68 additions & 20 deletions solution/2700-2799/2735.Collecting Chocolates/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -53,25 +53,27 @@

**方法一：枚举**

我们考虑枚举操作的次数，定义 $f[i][j]$ 表示类型为 $i$ 的巧克力进行了 $j$ 次操作后的最小成本。那么有：
我们考虑枚举操作的次数，定义 $f[i][j]$ 表示类型为 $i$ 的巧克力进行了 $j$ 次操作后的最小成本。

对于类型为 $i$ 的巧克力：

- 如果 $j = 0$，即没有进行操作，那么 $f[i][j] = nums[i]$；
- 如果 $0 \lt j \leq n-1$，那么它的最小成本就是下标范围为 $[i,.. (i + j) \bmod n]$ 的巧克力的最小成本，即 $f[i][j] = \min\{nums[i], nums[i + 1], \cdots, nums[(i + j) \bmod n]\}$，或者可以写成 $f[i][j] = \min\{f[i][j - 1], nums[(i + j) \bmod n]\}$。
- 如果 $j \ge n$，由于当 $j = n - 1$ 时，已经覆盖了所有巧克力的最小成本，如果 $j$ 继续增大，那么最小成本不会再变化，但是操作次数的增加却会导致最终的成本增加，因此，我们不需要考虑 $j \ge n$ 的情况。

综上，我们可以得到状态转移方程：

$$
f[i][j] =
\begin{cases}
nums[i] ,& j = 0 \\
\min(f[i][j - 1], nums[(i + j) \bmod n]) ,& j > 0
\min(f[i][j - 1], nums[(i + j) \bmod n]) ,& 0 \lt j \leq n - 1
\end{cases}
$$

接下来，我们枚举操作的次数 $j$，其中 $j \in [0,..n-1]$，那么进行 $j$ 次操作的最小成本为：

$$
\sum_{i=0}^{n-1} f[i][j] + j \times x
$$

我们取所有操作次数中的最小值即可。
最后，我们只需要枚举操作的次数 $j$，计算出每种操作次数下的最小成本，取最小值即可。即答案为 $\min\limits_{0 \leq j \leq n - 1} \sum\limits_{i = 0}^{n - 1} f[i][j] + x \times j$。

时间复杂度 $O(n^2)$，空间复杂度 $O(n^2)$。其中 $n$ 是数组 $nums$ 的长度。
时间复杂度 $O(n^2)$，空间复杂度 $O(n^2)$。其中 $n$ 为数组 $nums$ 的长度。

<!-- tabs:start -->

Expand All @@ -88,11 +90,7 @@ class Solution:
f[i][0] = v
for j in range(1, n):
f[i][j] = min(f[i][j - 1], nums[(i + j) % n])
ans = inf
for j in range(n):
cost = sum(f[i][j] for i in range(n)) + x * j
ans = min(ans, cost)
return ans
return min(sum(f[i][j] for i in range(n)) + x * j for j in range(n))
```

### **Java**
Expand All @@ -112,7 +110,7 @@ class Solution {
}
long ans = 1L << 60;
for (int j = 0; j < n; ++j) {
long cost = 1L * j * x;
long cost = 1L * x * j;
for (int i = 0; i < n; ++i) {
cost += f[i][j];
}
Expand All @@ -139,7 +137,7 @@ public:
}
long long ans = 1LL << 60;
for (int j = 0; j < n; ++j) {
long long cost = 1LL * j * x;
long long cost = 1LL * x * j;
for (int i = 0; i < n; ++i) {
cost += f[i][j];
}
Expand All @@ -156,17 +154,17 @@ public:
func minCost(nums []int, x int) int64 {
n := len(nums)
f := make([][]int, n)
for i := range f {
for i, v := range nums {
f[i] = make([]int, n)
f[i][0] = nums[i]
f[i][0] = v
for j := 1; j < n; j++ {
f[i][j] = min(f[i][j-1], nums[(i+j)%n])
}
}
ans := 1 << 60
for j := 0; j < n; j++ {
cost := x * j
for i := range nums {
for i := 0; i < n; i++ {
cost += f[i][j]
}
ans = min(ans, cost)
Expand All @@ -175,6 +173,56 @@ func minCost(nums []int, x int) int64 {
}
```

### **TypeScript**

```ts
function minCost(nums: number[], x: number): number {
const n = nums.length;
const f: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
for (let i = 0; i < n; ++i) {
f[i][0] = nums[i];
for (let j = 1; j < n; ++j) {
f[i][j] = Math.min(f[i][j - 1], nums[(i + j) % n]);
}
}
let ans = Infinity;
for (let j = 0; j < n; ++j) {
let cost = x * j;
for (let i = 0; i < n; ++i) {
cost += f[i][j];
}
ans = Math.min(ans, cost);
}
return ans;
}
```

### **Rust**

```rust
impl Solution {
pub fn min_cost(nums: Vec<i32>, x: i32) -> i64 {
let n = nums.len();
let mut f = vec![vec![0; n]; n];
for i in 0..n {
f[i][0] = nums[i];
for j in 1..n {
f[i][j] = f[i][j - 1].min(nums[(i + j) % n]);
}
}
let mut ans = i64::MAX;
for j in 0..n {
let mut cost = (x as i64) * (j as i64);
for i in 0..n {
cost += f[i][j] as i64;
}
ans = ans.min(cost);
}
ans
}
}
```

### **...**

```
Expand Down
90 changes: 80 additions & 10 deletions solution/2700-2799/2735.Collecting Chocolates/README_EN.md
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Expand Up @@ -45,6 +45,30 @@ Thus, the total cost will become (1 + 5 + 1 + 5 + 1) = 13. We can prove that thi

## Solutions

**Solution 1: Enumeration**

We consider enumerating the number of operations, and define $f[i][j]$ as the minimum cost after performing $j$ operations on the chocolate of type $i$.

For the chocolate of type $i$:

- If $j = 0$, i.e., no operation is performed, then $f[i][j] = nums[i]$;
- If $0 \lt j \leq n-1$, then its minimum cost is the minimum cost of the chocolates with indices in the range $[i,.. (i + j) \bmod n]$, i.e., $f[i][j] = \min\{nums[i], nums[i + 1], \cdots, nums[(i + j) \bmod n]\}$, or it can be written as $f[i][j] = \min\{f[i][j - 1], nums[(i + j) \bmod n]\}$.
- If $j \ge n$, since when $j = n - 1$, the minimum cost of all covered chocolates has been obtained. If $j$ continues to increase, the minimum cost will not change, but the increase in the number of operations will lead to an increase in the final cost. Therefore, we do not need to consider the case where $j \ge n$.

In summary, we can get the state transition equation:

$$
f[i][j] =
\begin{cases}
nums[i] ,& j = 0 \\
\min(f[i][j - 1], nums[(i + j) \bmod n]) ,& 0 \lt j \leq n - 1
\end{cases}
$$

Finally, we only need to enumerate the number of operations $j$, calculate the minimum cost under each number of operations, and take the minimum value. That is, the answer is $\min\limits_{0 \leq j \leq n - 1} \sum\limits_{i = 0}^{n - 1} f[i][j] + x \times j$.

The time complexity is $O(n^2)$, and the space complexity is $O(n^2)$. Where $n$ is the length of the array $nums$.

<!-- tabs:start -->

### **Python3**
Expand All @@ -58,11 +82,7 @@ class Solution:
f[i][0] = v
for j in range(1, n):
f[i][j] = min(f[i][j - 1], nums[(i + j) % n])
ans = inf
for j in range(n):
cost = sum(f[i][j] for i in range(n)) + x * j
ans = min(ans, cost)
return ans
return min(sum(f[i][j] for i in range(n)) + x * j for j in range(n))
```

### **Java**
Expand All @@ -80,7 +100,7 @@ class Solution {
}
long ans = 1L << 60;
for (int j = 0; j < n; ++j) {
long cost = 1L * j * x;
long cost = 1L * x * j;
for (int i = 0; i < n; ++i) {
cost += f[i][j];
}
Expand All @@ -107,7 +127,7 @@ public:
}
long long ans = 1LL << 60;
for (int j = 0; j < n; ++j) {
long long cost = 1LL * j * x;
long long cost = 1LL * x * j;
for (int i = 0; i < n; ++i) {
cost += f[i][j];
}
Expand All @@ -124,17 +144,17 @@ public:
func minCost(nums []int, x int) int64 {
n := len(nums)
f := make([][]int, n)
for i := range f {
for i, v := range nums {
f[i] = make([]int, n)
f[i][0] = nums[i]
f[i][0] = v
for j := 1; j < n; j++ {
f[i][j] = min(f[i][j-1], nums[(i+j)%n])
}
}
ans := 1 << 60
for j := 0; j < n; j++ {
cost := x * j
for i := range nums {
for i := 0; i < n; i++ {
cost += f[i][j]
}
ans = min(ans, cost)
Expand All @@ -143,6 +163,56 @@ func minCost(nums []int, x int) int64 {
}
```

### **TypeScript**

```ts
function minCost(nums: number[], x: number): number {
const n = nums.length;
const f: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
for (let i = 0; i < n; ++i) {
f[i][0] = nums[i];
for (let j = 1; j < n; ++j) {
f[i][j] = Math.min(f[i][j - 1], nums[(i + j) % n]);
}
}
let ans = Infinity;
for (let j = 0; j < n; ++j) {
let cost = x * j;
for (let i = 0; i < n; ++i) {
cost += f[i][j];
}
ans = Math.min(ans, cost);
}
return ans;
}
```

### **Rust**

```rust
impl Solution {
pub fn min_cost(nums: Vec<i32>, x: i32) -> i64 {
let n = nums.len();
let mut f = vec![vec![0; n]; n];
for i in 0..n {
f[i][0] = nums[i];
for j in 1..n {
f[i][j] = f[i][j - 1].min(nums[(i + j) % n]);
}
}
let mut ans = i64::MAX;
for j in 0..n {
let mut cost = (x as i64) * (j as i64);
for i in 0..n {
cost += f[i][j] as i64;
}
ans = ans.min(cost);
}
ans
}
}
```

### **...**

```
Expand Down
42 changes: 21 additions & 21 deletions solution/2700-2799/2735.Collecting Chocolates/Solution.cpp
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Original file line number Diff line number Diff line change
@@ -1,22 +1,22 @@
class Solution {
public:
long long minCost(vector<int>& nums, int x) {
int n = nums.size();
int f[n][n];
for (int i = 0; i < n; ++i) {
f[i][0] = nums[i];
for (int j = 1; j < n; ++j) {
f[i][j] = min(f[i][j - 1], nums[(i + j) % n]);
}
}
long long ans = 1LL << 60;
for (int j = 0; j < n; ++j) {
long long cost = 1LL * j * x;
for (int i = 0; i < n; ++i) {
cost += f[i][j];
}
ans = min(ans, cost);
}
return ans;
}
class Solution {
public:
long long minCost(vector<int>& nums, int x) {
int n = nums.size();
int f[n][n];
for (int i = 0; i < n; ++i) {
f[i][0] = nums[i];
for (int j = 1; j < n; ++j) {
f[i][j] = min(f[i][j - 1], nums[(i + j) % n]);
}
}
long long ans = 1LL << 60;
for (int j = 0; j < n; ++j) {
long long cost = 1LL * x * j;
for (int i = 0; i < n; ++i) {
cost += f[i][j];
}
ans = min(ans, cost);
}
return ans;
}
};
6 changes: 3 additions & 3 deletions solution/2700-2799/2735.Collecting Chocolates/Solution.go
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Original file line number Diff line number Diff line change
@@ -1,17 +1,17 @@
func minCost(nums []int, x int) int64 {
n := len(nums)
f := make([][]int, n)
for i := range f {
for i, v := range nums {
f[i] = make([]int, n)
f[i][0] = nums[i]
f[i][0] = v
for j := 1; j < n; j++ {
f[i][j] = min(f[i][j-1], nums[(i+j)%n])
}
}
ans := 1 << 60
for j := 0; j < n; j++ {
cost := x * j
for i := range nums {
for i := 0; i < n; i++ {
cost += f[i][j]
}
ans = min(ans, cost)
Expand Down
40 changes: 20 additions & 20 deletions solution/2700-2799/2735.Collecting Chocolates/Solution.java
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Original file line number Diff line number Diff line change
@@ -1,21 +1,21 @@
class Solution {
public long minCost(int[] nums, int x) {
int n = nums.length;
int[][] f = new int[n][n];
for (int i = 0; i < n; ++i) {
f[i][0] = nums[i];
for (int j = 1; j < n; ++j) {
f[i][j] = Math.min(f[i][j - 1], nums[(i + j) % n]);
}
}
long ans = 1L << 60;
for (int j = 0; j < n; ++j) {
long cost = 1L * j * x;
for (int i = 0; i < n; ++i) {
cost += f[i][j];
}
ans = Math.min(ans, cost);
}
return ans;
}
class Solution {
public long minCost(int[] nums, int x) {
int n = nums.length;
int[][] f = new int[n][n];
for (int i = 0; i < n; ++i) {
f[i][0] = nums[i];
for (int j = 1; j < n; ++j) {
f[i][j] = Math.min(f[i][j - 1], nums[(i + j) % n]);
}
}
long ans = 1L << 60;
for (int j = 0; j < n; ++j) {
long cost = 1L * x * j;
for (int i = 0; i < n; ++i) {
cost += f[i][j];
}
ans = Math.min(ans, cost);
}
return ans;
}
}
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