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feat: add solutions to lc problems: No.3309,3310 (doocs#3604)
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* No.3309.Maximum Possible Number by Binary Concatenation
* No.3310.Remove Methods From Project
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@yanglbme
yanglbme authored Oct 6, 2024
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100 changes: 96 additions & 4 deletions solution/3300-3399/3309.Maximum Possible Number by Binary Concatenation/README.md
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Expand Up @@ -61,32 +61,124 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3309.Ma

<!-- solution:start -->

### 方法一
### 方法一：枚举

根据题目描述，数组 $\textit{nums}$ 的长度为 $3$，我们可以枚举 $\textit{nums}$ 的全排列，一共有 $3! = 6$ 种排列方式，然后将排列后的数组中的元素转换为二进制字符串，再将这些二进制字符串连接起来，最后将连接后的二进制字符串转换为十进制数，取最大值即可。

时间复杂度 $O(\log M)$，其中 $M$ 表示 $\textit{nums}$ 中的元素的最大值。空间复杂度 $O(1)$。

<!-- tabs:start -->

#### Python3

```python

class Solution:
def maxGoodNumber(self, nums: List[int]) -> int:
ans = 0
for arr in permutations(nums):
num = int("".join(bin(i)[2:] for i in arr), 2)
ans = max(ans, num)
return ans
```

#### Java

```java

class Solution {
private int[] nums;

public int maxGoodNumber(int[] nums) {
this.nums = nums;
int ans = f(0, 1, 2);
ans = Math.max(ans, f(0, 2, 1));
ans = Math.max(ans, f(1, 0, 2));
ans = Math.max(ans, f(1, 2, 0));
ans = Math.max(ans, f(2, 0, 1));
ans = Math.max(ans, f(2, 1, 0));
return ans;
}

private int f(int i, int j, int k) {
String a = Integer.toBinaryString(nums[i]);
String b = Integer.toBinaryString(nums[j]);
String c = Integer.toBinaryString(nums[k]);
return Integer.parseInt(a + b + c, 2);
}
}
```

#### C++

```cpp

class Solution {
public:
int maxGoodNumber(vector<int>& nums) {
int ans = 0;
auto f = [&](vector<int>& nums) {
int res = 0;
vector<int> t;
for (int x : nums) {
for (; x; x >>= 1) {
t.push_back(x & 1);
}
}
while (t.size()) {
res = res * 2 + t.back();
t.pop_back();
}
return res;
};
for (int i = 0; i < 6; ++i) {
ans = max(ans, f(nums));
next_permutation(nums.begin(), nums.end());
}
return ans;
}
};
```
#### Go
```go
func maxGoodNumber(nums []int) int {
f := func(i, j, k int) int {
a := strconv.FormatInt(int64(nums[i]), 2)
b := strconv.FormatInt(int64(nums[j]), 2)
c := strconv.FormatInt(int64(nums[k]), 2)
res, _ := strconv.ParseInt(a+b+c, 2, 64)
return int(res)
}
ans := f(0, 1, 2)
ans = max(ans, f(0, 2, 1))
ans = max(ans, f(1, 0, 2))
ans = max(ans, f(1, 2, 0))
ans = max(ans, f(2, 0, 1))
ans = max(ans, f(2, 1, 0))
return ans
}
```

#### TypeScript

```ts
function maxGoodNumber(nums: number[]): number {
const f = (i: number, j: number, k: number): number => {
const a = nums[i].toString(2);
const b = nums[j].toString(2);
const c = nums[k].toString(2);
const res = parseInt(a + b + c, 2);
return res;
};

let ans = f(0, 1, 2);
ans = Math.max(ans, f(0, 2, 1));
ans = Math.max(ans, f(1, 0, 2));
ans = Math.max(ans, f(1, 2, 0));
ans = Math.max(ans, f(2, 0, 1));
ans = Math.max(ans, f(2, 1, 0));

return ans;
}
```

<!-- tabs:end -->
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100 changes: 96 additions & 4 deletions ...ion/3300-3399/3309.Maximum Possible Number by Binary Concatenation/README_EN.md
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Expand Up @@ -59,32 +59,124 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3309.Ma

<!-- solution:start -->

### Solution 1
### Solution 1: Enumeration

According to the problem description, the length of the array $\textit{nums}$ is $3$. We can enumerate all permutations of $\textit{nums}$, which has $3! = 6$ permutations. Then, we convert the elements of the permuted array into binary strings, concatenate these binary strings, and finally convert the concatenated binary string into a decimal number to get the maximum value.

The time complexity is $O(\log M)$, where $M$ represents the maximum value of the elements in $\textit{nums}$. The space complexity is $O(1)$.

<!-- tabs:start -->

#### Python3

```python

class Solution:
def maxGoodNumber(self, nums: List[int]) -> int:
ans = 0
for arr in permutations(nums):
num = int("".join(bin(i)[2:] for i in arr), 2)
ans = max(ans, num)
return ans
```

#### Java

```java

class Solution {
private int[] nums;

public int maxGoodNumber(int[] nums) {
this.nums = nums;
int ans = f(0, 1, 2);
ans = Math.max(ans, f(0, 2, 1));
ans = Math.max(ans, f(1, 0, 2));
ans = Math.max(ans, f(1, 2, 0));
ans = Math.max(ans, f(2, 0, 1));
ans = Math.max(ans, f(2, 1, 0));
return ans;
}

private int f(int i, int j, int k) {
String a = Integer.toBinaryString(nums[i]);
String b = Integer.toBinaryString(nums[j]);
String c = Integer.toBinaryString(nums[k]);
return Integer.parseInt(a + b + c, 2);
}
}
```

#### C++

```cpp

class Solution {
public:
int maxGoodNumber(vector<int>& nums) {
int ans = 0;
auto f = [&](vector<int>& nums) {
int res = 0;
vector<int> t;
for (int x : nums) {
for (; x; x >>= 1) {
t.push_back(x & 1);
}
}
while (t.size()) {
res = res * 2 + t.back();
t.pop_back();
}
return res;
};
for (int i = 0; i < 6; ++i) {
ans = max(ans, f(nums));
next_permutation(nums.begin(), nums.end());
}
return ans;
}
};
```
#### Go
```go
func maxGoodNumber(nums []int) int {
f := func(i, j, k int) int {
a := strconv.FormatInt(int64(nums[i]), 2)
b := strconv.FormatInt(int64(nums[j]), 2)
c := strconv.FormatInt(int64(nums[k]), 2)
res, _ := strconv.ParseInt(a+b+c, 2, 64)
return int(res)
}
ans := f(0, 1, 2)
ans = max(ans, f(0, 2, 1))
ans = max(ans, f(1, 0, 2))
ans = max(ans, f(1, 2, 0))
ans = max(ans, f(2, 0, 1))
ans = max(ans, f(2, 1, 0))
return ans
}
```

#### TypeScript

```ts
function maxGoodNumber(nums: number[]): number {
const f = (i: number, j: number, k: number): number => {
const a = nums[i].toString(2);
const b = nums[j].toString(2);
const c = nums[k].toString(2);
const res = parseInt(a + b + c, 2);
return res;
};

let ans = f(0, 1, 2);
ans = Math.max(ans, f(0, 2, 1));
ans = Math.max(ans, f(1, 0, 2));
ans = Math.max(ans, f(1, 2, 0));
ans = Math.max(ans, f(2, 0, 1));
ans = Math.max(ans, f(2, 1, 0));

return ans;
}
```

<!-- tabs:end -->
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25 changes: 25 additions & 0 deletions solution/3300-3399/3309.Maximum Possible Number by Binary Concatenation/Solution.cpp
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class Solution {
public:
int maxGoodNumber(vector<int>& nums) {
int ans = 0;
auto f = [&](vector<int>& nums) {
int res = 0;
vector<int> t;
for (int x : nums) {
for (; x; x >>= 1) {
t.push_back(x & 1);
}
}
while (t.size()) {
res = res * 2 + t.back();
t.pop_back();
}
return res;
};
for (int i = 0; i < 6; ++i) {
ans = max(ans, f(nums));
next_permutation(nums.begin(), nums.end());
}
return ans;
}
};
16 changes: 16 additions & 0 deletions solution/3300-3399/3309.Maximum Possible Number by Binary Concatenation/Solution.go
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func maxGoodNumber(nums []int) int {
f := func(i, j, k int) int {
a := strconv.FormatInt(int64(nums[i]), 2)
b := strconv.FormatInt(int64(nums[j]), 2)
c := strconv.FormatInt(int64(nums[k]), 2)
res, _ := strconv.ParseInt(a+b+c, 2, 64)
return int(res)
}
ans := f(0, 1, 2)
ans = max(ans, f(0, 2, 1))
ans = max(ans, f(1, 0, 2))
ans = max(ans, f(1, 2, 0))
ans = max(ans, f(2, 0, 1))
ans = max(ans, f(2, 1, 0))
return ans
}
21 changes: 21 additions & 0 deletions solution/3300-3399/3309.Maximum Possible Number by Binary Concatenation/Solution.java
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@@ -0,0 +1,21 @@
class Solution {
private int[] nums;

public int maxGoodNumber(int[] nums) {
this.nums = nums;
int ans = f(0, 1, 2);
ans = Math.max(ans, f(0, 2, 1));
ans = Math.max(ans, f(1, 0, 2));
ans = Math.max(ans, f(1, 2, 0));
ans = Math.max(ans, f(2, 0, 1));
ans = Math.max(ans, f(2, 1, 0));
return ans;
}

private int f(int i, int j, int k) {
String a = Integer.toBinaryString(nums[i]);
String b = Integer.toBinaryString(nums[j]);
String c = Integer.toBinaryString(nums[k]);
return Integer.parseInt(a + b + c, 2);
}
}
7 changes: 7 additions & 0 deletions solution/3300-3399/3309.Maximum Possible Number by Binary Concatenation/Solution.py
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@@ -0,0 +1,7 @@
class Solution:
def maxGoodNumber(self, nums: List[int]) -> int:
ans = 0
for arr in permutations(nums):
num = int("".join(bin(i)[2:] for i in arr), 2)
ans = max(ans, num)
return ans
18 changes: 18 additions & 0 deletions solution/3300-3399/3309.Maximum Possible Number by Binary Concatenation/Solution.ts
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@@ -0,0 +1,18 @@
function maxGoodNumber(nums: number[]): number {
const f = (i: number, j: number, k: number): number => {
const a = nums[i].toString(2);
const b = nums[j].toString(2);
const c = nums[k].toString(2);
const res = parseInt(a + b + c, 2);
return res;
};

let ans = f(0, 1, 2);
ans = Math.max(ans, f(0, 2, 1));
ans = Math.max(ans, f(1, 0, 2));
ans = Math.max(ans, f(1, 2, 0));
ans = Math.max(ans, f(2, 0, 1));
ans = Math.max(ans, f(2, 1, 0));

return ans;
}
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