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feat: add solutions to lc problem: No.1363 #1637

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Sep 16, 2023
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feat: add solutions to lc problem: No.1363 #1637

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185 changes: 184 additions & 1 deletion solution/1300-1399/1363.Largest Multiple of Three/README.md
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Expand Up @@ -53,22 +53,205 @@

<!-- 这里可写通用的实现逻辑 -->

**方法一：贪心 + 动态规划 + 逆推**

我们定义 $f[i][j]$ 表示在前 $i$ 个数中选取若干个数，使得选取的数的和模 $3$ 为 $j$ 的最大长度。为了使得选取的数最大，我们需要尽可能选取更多的数，因此我们需要使得 $f[i][j]$ 尽可能大。我们初始化 $f[0][0] = 0$，其余的 $f[0][j] = -\infty$。

考虑 $f[i][j]$ 如何进行状态转移。我们可以不选取第 $i$ 个数，此时 $f[i][j] = f[i - 1][j]$；我们也可以选取第 $i$ 个数，此时 $f[i][j] = f[i - 1][(j - x_i \bmod 3 + 3) \bmod 3] + 1$，其中 $x_i$ 表示第 $i$ 个数的值。因此我们有如下的状态转移方程：

$$
f[i][j] = \max \{ f[i - 1][j], f[i - 1][(j - x_i \bmod 3 + 3) \bmod 3] + 1 \}
$$

如果 $f[n][0] \le 0$，那么我们无法选取任何数，因此答案字符串为空。否则我们可以通过 $f$ 数组进行逆推，找出选取的数。

定义 $i = n$, $j = 0$，从 $f[i][j]$ 开始逆推，记 $k = (j - x_i \bmod 3 + 3) \bmod 3$，如果 $f[i - 1][k] + 1 = f[i][j]$，那么我们选取了第 $i$ 个数，否则我们没有选取第 $i$ 个数。如果我们选取了第 $i$ 个数，那么我们将 $j$ 更新为 $k$，否则我们保持 $j$ 不变。为了使得同等长度的数最大，我们应该优先选取较大的数，因此，我们在前面首先对数组进行排序。

时间复杂度 $O(n \times \log n)$，空间复杂度 $O(n)$。其中 $n$ 为数组的长度。

<!-- tabs:start -->

### **Python3**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```python

class Solution:
def largestMultipleOfThree(self, digits: List[int]) -> str:
digits.sort()
n = len(digits)
f = [[-inf] * 3 for _ in range(n + 1)]
f[0][0] = 0
for i, x in enumerate(digits, 1):
for j in range(3):
f[i][j] = max(f[i - 1][j], f[i - 1][(j - x % 3 + 3) % 3] + 1)
if f[n][0] <= 0:
return ""
arr = []
j = 0
for i in range(n, 0, -1):
k = (j - digits[i - 1] % 3 + 3) % 3
if f[i - 1][k] + 1 == f[i][j]:
arr.append(digits[i - 1])
j = k
i = 0
while i < len(arr) - 1 and arr[i] == 0:
i += 1
return "".join(map(str, arr[i:]))
```

### **Java**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```java
class Solution {
public String largestMultipleOfThree(int[] digits) {
Arrays.sort(digits);
int n = digits.length;
int[][] f = new int[n + 1][3];
final int inf = 1 << 30;
for (var g : f) {
Arrays.fill(g, -inf);
}
f[0][0] = 0;
for (int i = 1; i <= n; ++i) {
for (int j = 0; j < 3; ++j) {
f[i][j] = Math.max(f[i - 1][j], f[i - 1][(j - digits[i - 1] % 3 + 3) % 3] + 1);
}
}
if (f[n][0] <= 0) {
return "";
}
StringBuilder sb = new StringBuilder();
for (int i = n, j = 0; i > 0; --i) {
int k = (j - digits[i - 1] % 3 + 3) % 3;
if (f[i - 1][k] + 1 == f[i][j]) {
sb.append(digits[i - 1]);
j = k;
}
}
int i = 0;
while (i < sb.length() - 1 && sb.charAt(i) == '0') {
++i;
}
return sb.substring(i);
}
}
```

### **C++**

```cpp
class Solution {
public:
string largestMultipleOfThree(vector<int>& digits) {
sort(digits.begin(), digits.end());
int n = digits.size();
int f[n + 1][3];
memset(f, -0x3f, sizeof(f));
f[0][0] = 0;
for (int i = 1; i <= n; ++i) {
for (int j = 0; j < 3; ++j) {
f[i][j] = max(f[i - 1][j], f[i - 1][(j - digits[i - 1] % 3 + 3) % 3] + 1);
}
}
if (f[n][0] <= 0) {
return "";
}
string ans;
for (int i = n, j = 0; i; --i) {
int k = (j - digits[i - 1] % 3 + 3) % 3;
if (f[i - 1][k] + 1 == f[i][j]) {
ans += digits[i - 1] + '0';
j = k;
}
}
int i = 0;
while (i < ans.size() - 1 && ans[i] == '0') {
++i;
}
return ans.substr(i);
}
};
```

### **Go**

```go
func largestMultipleOfThree(digits []int) string {
sort.Ints(digits)
n := len(digits)
const inf = 1 << 30
f := make([][]int, n+1)
for i := range f {
f[i] = make([]int, 3)
for j := range f[i] {
f[i][j] = -inf
}
}
f[0][0] = 0
for i := 1; i <= n; i++ {
for j := 0; j < 3; j++ {
f[i][j] = max(f[i-1][j], f[i-1][(j-digits[i-1]%3+3)%3]+1)
}
}
if f[n][0] <= 0 {
return ""
}
ans := []byte{}
for i, j := n, 0; i > 0; i-- {
k := (j - digits[i-1]%3 + 3) % 3
if f[i][j] == f[i-1][k]+1 {
ans = append(ans, byte('0'+digits[i-1]))
j = k
}
}
i := 0
for i < len(ans)-1 && ans[i] == '0' {
i++
}
return string(ans[i:])
}

func max(a, b int) int {
if a > b {
return a
}
return b
}
```

### **TypeScript**

```ts
function largestMultipleOfThree(digits: number[]): string {
digits.sort((a, b) => a - b);
const n = digits.length;
const f: number[][] = new Array(n + 1).fill(0).map(() => new Array(3).fill(-Infinity));
f[0][0] = 0;
for (let i = 1; i <= n; ++i) {
for (let j = 0; j < 3; ++j) {
f[i][j] = Math.max(f[i - 1][j], f[i - 1][(j - (digits[i - 1] % 3) + 3) % 3] + 1);
}
}
if (f[n][0] <= 0) {
return '';
}
const arr: number[] = [];
for (let i = n, j = 0; i; --i) {
const k = (j - (digits[i - 1] % 3) + 3) % 3;
if (f[i - 1][k] + 1 === f[i][j]) {
arr.push(digits[i - 1]);
j = k;
}
}
let i = 0;
while (i < arr.length - 1 && arr[i] === 0) {
++i;
}
return arr.slice(i).join('');
}
```

### **...**
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169 changes: 168 additions & 1 deletion solution/1300-1399/1363.Largest Multiple of Three/README_EN.md
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Expand Up @@ -45,13 +45,180 @@
### **Python3**

```python

class Solution:
def largestMultipleOfThree(self, digits: List[int]) -> str:
digits.sort()
n = len(digits)
f = [[-inf] * 3 for _ in range(n + 1)]
f[0][0] = 0
for i, x in enumerate(digits, 1):
for j in range(3):
f[i][j] = max(f[i - 1][j], f[i - 1][(j - x % 3 + 3) % 3] + 1)
if f[n][0] <= 0:
return ""
arr = []
j = 0
for i in range(n, 0, -1):
k = (j - digits[i - 1] % 3 + 3) % 3
if f[i - 1][k] + 1 == f[i][j]:
arr.append(digits[i - 1])
j = k
i = 0
while i < len(arr) - 1 and arr[i] == 0:
i += 1
return "".join(map(str, arr[i:]))
```

### **Java**

```java
class Solution {
public String largestMultipleOfThree(int[] digits) {
Arrays.sort(digits);
int n = digits.length;
int[][] f = new int[n + 1][3];
final int inf = 1 << 30;
for (var g : f) {
Arrays.fill(g, -inf);
}
f[0][0] = 0;
for (int i = 1; i <= n; ++i) {
for (int j = 0; j < 3; ++j) {
f[i][j] = Math.max(f[i - 1][j], f[i - 1][(j - digits[i - 1] % 3 + 3) % 3] + 1);
}
}
if (f[n][0] <= 0) {
return "";
}
StringBuilder sb = new StringBuilder();
for (int i = n, j = 0; i > 0; --i) {
int k = (j - digits[i - 1] % 3 + 3) % 3;
if (f[i - 1][k] + 1 == f[i][j]) {
sb.append(digits[i - 1]);
j = k;
}
}
int i = 0;
while (i < sb.length() - 1 && sb.charAt(i) == '0') {
++i;
}
return sb.substring(i);
}
}
```

### **C++**

```cpp
class Solution {
public:
string largestMultipleOfThree(vector<int>& digits) {
sort(digits.begin(), digits.end());
int n = digits.size();
int f[n + 1][3];
memset(f, -0x3f, sizeof(f));
f[0][0] = 0;
for (int i = 1; i <= n; ++i) {
for (int j = 0; j < 3; ++j) {
f[i][j] = max(f[i - 1][j], f[i - 1][(j - digits[i - 1] % 3 + 3) % 3] + 1);
}
}
if (f[n][0] <= 0) {
return "";
}
string ans;
for (int i = n, j = 0; i; --i) {
int k = (j - digits[i - 1] % 3 + 3) % 3;
if (f[i - 1][k] + 1 == f[i][j]) {
ans += digits[i - 1] + '0';
j = k;
}
}
int i = 0;
while (i < ans.size() - 1 && ans[i] == '0') {
++i;
}
return ans.substr(i);
}
};
```

### **Go**

```go
func largestMultipleOfThree(digits []int) string {
sort.Ints(digits)
n := len(digits)
const inf = 1 << 30
f := make([][]int, n+1)
for i := range f {
f[i] = make([]int, 3)
for j := range f[i] {
f[i][j] = -inf
}
}
f[0][0] = 0
for i := 1; i <= n; i++ {
for j := 0; j < 3; j++ {
f[i][j] = max(f[i-1][j], f[i-1][(j-digits[i-1]%3+3)%3]+1)
}
}
if f[n][0] <= 0 {
return ""
}
ans := []byte{}
for i, j := n, 0; i > 0; i-- {
k := (j - digits[i-1]%3 + 3) % 3
if f[i][j] == f[i-1][k]+1 {
ans = append(ans, byte('0'+digits[i-1]))
j = k
}
}
i := 0
for i < len(ans)-1 && ans[i] == '0' {
i++
}
return string(ans[i:])
}

func max(a, b int) int {
if a > b {
return a
}
return b
}
```

### **TypeScript**

```ts
function largestMultipleOfThree(digits: number[]): string {
digits.sort((a, b) => a - b);
const n = digits.length;
const f: number[][] = new Array(n + 1).fill(0).map(() => new Array(3).fill(-Infinity));
f[0][0] = 0;
for (let i = 1; i <= n; ++i) {
for (let j = 0; j < 3; ++j) {
f[i][j] = Math.max(f[i - 1][j], f[i - 1][(j - (digits[i - 1] % 3) + 3) % 3] + 1);
}
}
if (f[n][0] <= 0) {
return '';
}
const arr: number[] = [];
for (let i = n, j = 0; i; --i) {
const k = (j - (digits[i - 1] % 3) + 3) % 3;
if (f[i - 1][k] + 1 === f[i][j]) {
arr.push(digits[i - 1]);
j = k;
}
}
let i = 0;
while (i < arr.length - 1 && arr[i] === 0) {
++i;
}
return arr.slice(i).join('');
}
```

### **...**
Expand Down
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