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feat: add solutions to lc problems: No.3309,3310 (#3604)
* No.3309.Maximum Possible Number by Binary Concatenation * No.3310.Remove Methods From Project
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‎solution/3300-3399/3309.Maximum Possible Number by Binary Concatenation/README.md

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@@ -61,32 +61,124 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3309.Ma
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<!-- solution:start -->
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### 方法一
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### 方法一：枚举
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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。
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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 $O(1)$。
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<!-- tabs:start -->
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#### Python3
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```python
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class Solution:
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def maxGoodNumber(self, nums: List[int]) -> int:
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ans = 0
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for arr in permutations(nums):
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num = int("".join(bin(i)[2:] for i in arr), 2)
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ans = max(ans, num)
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return ans
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```
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#### Java
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```java
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class Solution {
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private int[] nums;
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public int maxGoodNumber(int[] nums) {
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this.nums = nums;
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int ans = f(0, 1, 2);
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ans = Math.max(ans, f(0, 2, 1));
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ans = Math.max(ans, f(1, 0, 2));
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ans = Math.max(ans, f(1, 2, 0));
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ans = Math.max(ans, f(2, 0, 1));
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ans = Math.max(ans, f(2, 1, 0));
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return ans;
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}
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private int f(int i, int j, int k) {
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String a = Integer.toBinaryString(nums[i]);
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String b = Integer.toBinaryString(nums[j]);
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String c = Integer.toBinaryString(nums[k]);
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return Integer.parseInt(a + b + c, 2);
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}
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}
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```
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#### C++
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```cpp
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class Solution {
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public:
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int maxGoodNumber(vector<int>& nums) {
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int ans = 0;
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auto f = [&](vector<int>& nums) {
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int res = 0;
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vector<int> t;
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for (int x : nums) {
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for (; x; x >>= 1) {
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t.push_back(x & 1);
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}
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}
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while (t.size()) {
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res = res * 2 + t.back();
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t.pop_back();
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}
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return res;
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};
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for (int i = 0; i < 6; ++i) {
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ans = max(ans, f(nums));
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next_permutation(nums.begin(), nums.end());
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}
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return ans;
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}
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};
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```
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#### Go
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```go
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func maxGoodNumber(nums []int) int {
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f := func(i, j, k int) int {
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a := strconv.FormatInt(int64(nums[i]), 2)
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b := strconv.FormatInt(int64(nums[j]), 2)
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c := strconv.FormatInt(int64(nums[k]), 2)
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res, _ := strconv.ParseInt(a+b+c, 2, 64)
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return int(res)
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}
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ans := f(0, 1, 2)
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ans = max(ans, f(0, 2, 1))
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ans = max(ans, f(1, 0, 2))
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ans = max(ans, f(1, 2, 0))
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ans = max(ans, f(2, 0, 1))
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ans = max(ans, f(2, 1, 0))
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return ans
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}
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```
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#### TypeScript
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```ts
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function maxGoodNumber(nums: number[]): number {
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const f = (i: number, j: number, k: number): number => {
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const a = nums[i].toString(2);
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const b = nums[j].toString(2);
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const c = nums[k].toString(2);
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const res = parseInt(a + b + c, 2);
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return res;
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};
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let ans = f(0, 1, 2);
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ans = Math.max(ans, f(0, 2, 1));
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ans = Math.max(ans, f(1, 0, 2));
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ans = Math.max(ans, f(1, 2, 0));
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ans = Math.max(ans, f(2, 0, 1));
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ans = Math.max(ans, f(2, 1, 0));
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return ans;
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}
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```
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<!-- tabs:end -->

‎solution/3300-3399/3309.Maximum Possible Number by Binary Concatenation/README_EN.md

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Enumeration
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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.
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The time complexity is $O(\log M)$, where $M$ represents the maximum value of the elements in $\textit{nums}$. The space complexity is $O(1)$.
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<!-- tabs:start -->
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#### Python3
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```python
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class Solution:
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def maxGoodNumber(self, nums: List[int]) -> int:
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ans = 0
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for arr in permutations(nums):
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num = int("".join(bin(i)[2:] for i in arr), 2)
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ans = max(ans, num)
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return ans
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```
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#### Java
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```java
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class Solution {
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private int[] nums;
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public int maxGoodNumber(int[] nums) {
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this.nums = nums;
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int ans = f(0, 1, 2);
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ans = Math.max(ans, f(0, 2, 1));
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ans = Math.max(ans, f(1, 0, 2));
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ans = Math.max(ans, f(1, 2, 0));
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ans = Math.max(ans, f(2, 0, 1));
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ans = Math.max(ans, f(2, 1, 0));
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return ans;
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}
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private int f(int i, int j, int k) {
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String a = Integer.toBinaryString(nums[i]);
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String b = Integer.toBinaryString(nums[j]);
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String c = Integer.toBinaryString(nums[k]);
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return Integer.parseInt(a + b + c, 2);
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}
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}
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```
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#### C++
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```cpp
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class Solution {
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public:
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int maxGoodNumber(vector<int>& nums) {
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int ans = 0;
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auto f = [&](vector<int>& nums) {
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int res = 0;
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vector<int> t;
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for (int x : nums) {
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for (; x; x >>= 1) {
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t.push_back(x & 1);
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}
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}
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while (t.size()) {
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res = res * 2 + t.back();
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t.pop_back();
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}
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return res;
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};
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for (int i = 0; i < 6; ++i) {
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ans = max(ans, f(nums));
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next_permutation(nums.begin(), nums.end());
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}
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return ans;
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}
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};
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```
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#### Go
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```go
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func maxGoodNumber(nums []int) int {
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f := func(i, j, k int) int {
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a := strconv.FormatInt(int64(nums[i]), 2)
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b := strconv.FormatInt(int64(nums[j]), 2)
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c := strconv.FormatInt(int64(nums[k]), 2)
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res, _ := strconv.ParseInt(a+b+c, 2, 64)
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return int(res)
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}
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ans := f(0, 1, 2)
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ans = max(ans, f(0, 2, 1))
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ans = max(ans, f(1, 0, 2))
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ans = max(ans, f(1, 2, 0))
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ans = max(ans, f(2, 0, 1))
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ans = max(ans, f(2, 1, 0))
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return ans
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}
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```
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#### TypeScript
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```ts
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function maxGoodNumber(nums: number[]): number {
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const f = (i: number, j: number, k: number): number => {
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const a = nums[i].toString(2);
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const b = nums[j].toString(2);
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const c = nums[k].toString(2);
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const res = parseInt(a + b + c, 2);
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return res;
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};
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let ans = f(0, 1, 2);
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ans = Math.max(ans, f(0, 2, 1));
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ans = Math.max(ans, f(1, 0, 2));
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ans = Math.max(ans, f(1, 2, 0));
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ans = Math.max(ans, f(2, 0, 1));
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ans = Math.max(ans, f(2, 1, 0));
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return ans;
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}
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```
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<!-- tabs:end -->

‎solution/3300-3399/3309.Maximum Possible Number by Binary Concatenation/Solution.cpp

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class Solution {
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public:
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int maxGoodNumber(vector<int>& nums) {
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int ans = 0;
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auto f = [&](vector<int>& nums) {
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int res = 0;
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vector<int> t;
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for (int x : nums) {
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for (; x; x >>= 1) {
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t.push_back(x & 1);
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}
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}
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while (t.size()) {
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res = res * 2 + t.back();
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t.pop_back();
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}
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return res;
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};
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for (int i = 0; i < 6; ++i) {
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ans = max(ans, f(nums));
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next_permutation(nums.begin(), nums.end());
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}
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return ans;
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}
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};

‎solution/3300-3399/3309.Maximum Possible Number by Binary Concatenation/Solution.go

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func maxGoodNumber(nums []int) int {
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f := func(i, j, k int) int {
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a := strconv.FormatInt(int64(nums[i]), 2)
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b := strconv.FormatInt(int64(nums[j]), 2)
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c := strconv.FormatInt(int64(nums[k]), 2)
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res, _ := strconv.ParseInt(a+b+c, 2, 64)
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return int(res)
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}
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ans := f(0, 1, 2)
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ans = max(ans, f(0, 2, 1))
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ans = max(ans, f(1, 0, 2))
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ans = max(ans, f(1, 2, 0))
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ans = max(ans, f(2, 0, 1))
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ans = max(ans, f(2, 1, 0))
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return ans
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}

‎solution/3300-3399/3309.Maximum Possible Number by Binary Concatenation/Solution.java

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class Solution {
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private int[] nums;
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public int maxGoodNumber(int[] nums) {
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this.nums = nums;
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int ans = f(0, 1, 2);
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ans = Math.max(ans, f(0, 2, 1));
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ans = Math.max(ans, f(1, 0, 2));
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ans = Math.max(ans, f(1, 2, 0));
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ans = Math.max(ans, f(2, 0, 1));
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ans = Math.max(ans, f(2, 1, 0));
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return ans;
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}
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private int f(int i, int j, int k) {
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String a = Integer.toBinaryString(nums[i]);
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String b = Integer.toBinaryString(nums[j]);
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String c = Integer.toBinaryString(nums[k]);
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return Integer.parseInt(a + b + c, 2);
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}
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}

‎solution/3300-3399/3309.Maximum Possible Number by Binary Concatenation/Solution.py

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

‎solution/3300-3399/3309.Maximum Possible Number by Binary Concatenation/Solution.ts

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function maxGoodNumber(nums: number[]): number {
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const f = (i: number, j: number, k: number): number => {
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const a = nums[i].toString(2);
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const b = nums[j].toString(2);
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const c = nums[k].toString(2);
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const res = parseInt(a + b + c, 2);
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return res;
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};
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let ans = f(0, 1, 2);
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ans = Math.max(ans, f(0, 2, 1));
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ans = Math.max(ans, f(1, 0, 2));
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ans = Math.max(ans, f(1, 2, 0));
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ans = Math.max(ans, f(2, 0, 1));
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ans = Math.max(ans, f(2, 1, 0));
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return ans;
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}

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