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feat: add solutions to lc problem: No.3180 #3093

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129 changes: 125 additions & 4 deletions solution/3100-3199/3180.Maximum Total Reward Using Operations I/README.md
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Expand Up @@ -69,32 +69,153 @@ tags:

<!-- solution:start -->

### 方法一
### 方法一：排序 + 记忆化搜索 + 二分查找

我们可以对奖励值数组 `rewardValues` 进行排序，然后使用记忆化搜索的方法求解最大总奖励。

我们定义一个函数 $\text{dfs}(x)$，表示当前总奖励为 $x$ 时，能够获得的最大总奖励。那么答案为 $\text{dfs}(0)$。

函数 $\text{dfs}(x)$ 的执行过程如下：

1. 二分查找数组 `rewardValues` 中第一个大于 $x$ 的元素的下标 $i$；
2. 遍历数组 `rewardValues` 中从下标 $i$ 开始的元素，对于每个元素 $v$，计算 $v + \text{dfs}(x + v)$ 的最大值。
3. 将结果返回。

为了避免重复计算，我们使用记忆化数组 `f` 记录已经计算过的结果。

时间复杂度 $O(n \times (\log n + M))$，空间复杂度 $O(M)$。其中 $n$ 是数组 `rewardValues` 的长度，而 $M$ 是数组 `rewardValues` 中的最大值的两倍。

<!-- tabs:start -->

#### Python3

```python

class Solution:
def maxTotalReward(self, rewardValues: List[int]) -> int:
@cache
def dfs(x: int) -> int:
i = bisect_right(rewardValues, x)
ans = 0
for v in rewardValues[i:]:
ans = max(ans, v + dfs(x + v))
return ans

rewardValues.sort()
return dfs(0)
```

#### Java

```java

class Solution {
private int[] nums;
private Integer[] f;

public int maxTotalReward(int[] rewardValues) {
nums = rewardValues;
Arrays.sort(nums);
int n = nums.length;
f = new Integer[nums[n - 1] << 1];
return dfs(0);
}

private int dfs(int x) {
if (f[x] != null) {
return f[x];
}
int i = Arrays.binarySearch(nums, x + 1);
i = i < 0 ? -i - 1 : i;
int ans = 0;
for (; i < nums.length; ++i) {
ans = Math.max(ans, nums[i] + dfs(x + nums[i]));
}
return f[x] = ans;
}
}
```

#### C++

```cpp

class Solution {
public:
int maxTotalReward(vector<int>& rewardValues) {
sort(rewardValues.begin(), rewardValues.end());
int n = rewardValues.size();
int f[rewardValues.back() << 1];
memset(f, -1, sizeof(f));
function<int(int)> dfs = [&](int x) {
if (f[x] != -1) {
return f[x];
}
auto it = upper_bound(rewardValues.begin(), rewardValues.end(), x);
int ans = 0;
for (; it != rewardValues.end(); ++it) {
ans = max(ans, rewardValues[it - rewardValues.begin()] + dfs(x + *it));
}
return f[x] = ans;
};
return dfs(0);
}
};
```

#### Go

```go
func maxTotalReward(rewardValues []int) int {
sort.Ints(rewardValues)
n := len(rewardValues)
f := make([]int, rewardValues[n-1]<<1)
for i := range f {
f[i] = -1
}
var dfs func(int) int
dfs = func(x int) int {
if f[x] != -1 {
return f[x]
}
i := sort.SearchInts(rewardValues, x+1)
f[x] = 0
for _, v := range rewardValues[i:] {
f[x] = max(f[x], v+dfs(x+v))
}
return f[x]
}
return dfs(0)
}
```

#### TypeScript

```ts
function maxTotalReward(rewardValues: number[]): number {
rewardValues.sort((a, b) => a - b);
const search = (x: number): number => {
let [l, r] = [0, rewardValues.length];
while (l < r) {
const mid = (l + r) >> 1;
if (rewardValues[mid] > x) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
};
const f: number[] = Array(rewardValues.at(-1)! << 1).fill(-1);
const dfs = (x: number): number => {
if (f[x] !== -1) {
return f[x];
}
let ans = 0;
for (let i = search(x); i < rewardValues.length; ++i) {
ans = Math.max(ans, rewardValues[i] + dfs(x + rewardValues[i]));
}
return (f[x] = ans);
};
return dfs(0);
}
```

<!-- tabs:end -->
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129 changes: 125 additions & 4 deletions solution/3100-3199/3180.Maximum Total Reward Using Operations I/README_EN.md
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Expand Up @@ -67,32 +67,153 @@ tags:

<!-- solution:start -->

### Solution 1
### Solution 1: Sorting + Memoization + Binary Search

We can sort the `rewardValues` array and then use memoization to solve for the maximum total reward.

We define a function $\text{dfs}(x)$, representing the maximum total reward that can be obtained when the current total reward is $x$. Thus, the answer is $\text{dfs}(0)$.

The execution process of the function $\text{dfs}(x)$ is as follows:

1. Perform a binary search in the `rewardValues` array for the index $i$ of the first element greater than $x$;
2. Iterate over the elements in the `rewardValues` array starting from index $i$, and for each element $v$, calculate the maximum value of $v + \text{dfs}(x + v)$.
3. Return the result.

To avoid repeated calculations, we use a memoization array `f` to record the results that have already been computed.

The time complexity is $O(n \times (\log n + M))$, and the space complexity is $O(M)$. Where $n$ is the length of the `rewardValues` array, and $M$ is twice the maximum value in the `rewardValues` array.

<!-- tabs:start -->

#### Python3

```python

class Solution:
def maxTotalReward(self, rewardValues: List[int]) -> int:
@cache
def dfs(x: int) -> int:
i = bisect_right(rewardValues, x)
ans = 0
for v in rewardValues[i:]:
ans = max(ans, v + dfs(x + v))
return ans

rewardValues.sort()
return dfs(0)
```

#### Java

```java

class Solution {
private int[] nums;
private Integer[] f;

public int maxTotalReward(int[] rewardValues) {
nums = rewardValues;
Arrays.sort(nums);
int n = nums.length;
f = new Integer[nums[n - 1] << 1];
return dfs(0);
}

private int dfs(int x) {
if (f[x] != null) {
return f[x];
}
int i = Arrays.binarySearch(nums, x + 1);
i = i < 0 ? -i - 1 : i;
int ans = 0;
for (; i < nums.length; ++i) {
ans = Math.max(ans, nums[i] + dfs(x + nums[i]));
}
return f[x] = ans;
}
}
```

#### C++

```cpp

class Solution {
public:
int maxTotalReward(vector<int>& rewardValues) {
sort(rewardValues.begin(), rewardValues.end());
int n = rewardValues.size();
int f[rewardValues.back() << 1];
memset(f, -1, sizeof(f));
function<int(int)> dfs = [&](int x) {
if (f[x] != -1) {
return f[x];
}
auto it = upper_bound(rewardValues.begin(), rewardValues.end(), x);
int ans = 0;
for (; it != rewardValues.end(); ++it) {
ans = max(ans, rewardValues[it - rewardValues.begin()] + dfs(x + *it));
}
return f[x] = ans;
};
return dfs(0);
}
};
```

#### Go

```go
func maxTotalReward(rewardValues []int) int {
sort.Ints(rewardValues)
n := len(rewardValues)
f := make([]int, rewardValues[n-1]<<1)
for i := range f {
f[i] = -1
}
var dfs func(int) int
dfs = func(x int) int {
if f[x] != -1 {
return f[x]
}
i := sort.SearchInts(rewardValues, x+1)
f[x] = 0
for _, v := range rewardValues[i:] {
f[x] = max(f[x], v+dfs(x+v))
}
return f[x]
}
return dfs(0)
}
```

#### TypeScript

```ts
function maxTotalReward(rewardValues: number[]): number {
rewardValues.sort((a, b) => a - b);
const search = (x: number): number => {
let [l, r] = [0, rewardValues.length];
while (l < r) {
const mid = (l + r) >> 1;
if (rewardValues[mid] > x) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
};
const f: number[] = Array(rewardValues.at(-1)! << 1).fill(-1);
const dfs = (x: number): number => {
if (f[x] !== -1) {
return f[x];
}
let ans = 0;
for (let i = search(x); i < rewardValues.length; ++i) {
ans = Math.max(ans, rewardValues[i] + dfs(x + rewardValues[i]));
}
return (f[x] = ans);
};
return dfs(0);
}
```

<!-- tabs:end -->
Expand Down
21 changes: 21 additions & 0 deletions solution/3100-3199/3180.Maximum Total Reward Using Operations I/Solution.cpp
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@@ -0,0 +1,21 @@
class Solution {
public:
int maxTotalReward(vector<int>& rewardValues) {
sort(rewardValues.begin(), rewardValues.end());
int n = rewardValues.size();
int f[rewardValues.back() << 1];
memset(f, -1, sizeof(f));
function<int(int)> dfs = [&](int x) {
if (f[x] != -1) {
return f[x];
}
auto it = upper_bound(rewardValues.begin(), rewardValues.end(), x);
int ans = 0;
for (; it != rewardValues.end(); ++it) {
ans = max(ans, rewardValues[it - rewardValues.begin()] + dfs(x + *it));
}
return f[x] = ans;
};
return dfs(0);
}
};
21 changes: 21 additions & 0 deletions solution/3100-3199/3180.Maximum Total Reward Using Operations I/Solution.go
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Original file line number Diff line number Diff line change
@@ -0,0 +1,21 @@
func maxTotalReward(rewardValues []int) int {
sort.Ints(rewardValues)
n := len(rewardValues)
f := make([]int, rewardValues[n-1]<<1)
for i := range f {
f[i] = -1
}
var dfs func(int) int
dfs = func(x int) int {
if f[x] != -1 {
return f[x]
}
i := sort.SearchInts(rewardValues, x+1)
f[x] = 0
for _, v := range rewardValues[i:] {
f[x] = max(f[x], v+dfs(x+v))
}
return f[x]
}
return dfs(0)
}
25 changes: 25 additions & 0 deletions solution/3100-3199/3180.Maximum Total Reward Using Operations I/Solution.java
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Original file line number Diff line number Diff line change
@@ -0,0 +1,25 @@
class Solution {
private int[] nums;
private Integer[] f;

public int maxTotalReward(int[] rewardValues) {
nums = rewardValues;
Arrays.sort(nums);
int n = nums.length;
f = new Integer[nums[n - 1] << 1];
return dfs(0);
}

private int dfs(int x) {
if (f[x] != null) {
return f[x];
}
int i = Arrays.binarySearch(nums, x + 1);
i = i < 0 ? -i - 1 : i;
int ans = 0;
for (; i < nums.length; ++i) {
ans = Math.max(ans, nums[i] + dfs(x + nums[i]));
}
return f[x] = ans;
}
}
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