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Commit 23f7050

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feat: add solutions to lc problem: No.3133 (doocs#2678)
No.3133.Minimum Array End
1 parent baa1174 commit 23f7050

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-28
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12 files changed

+227
-28
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‎.github/workflows/black-lint.yml‎

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name: black-linter
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push:
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- solution/**
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- lcs/**
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- lcp/**
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- lcof2/**
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- lcci/**
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- basic/**
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pull_request:
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‎.github/workflows/clang-format-lint.yml‎

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name: clang-format-linter
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- lcp/**
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- lcof2/**
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‎.github/workflows/compress.yml‎

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- cron: '00 23 * * 0'
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concurrency:
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group: ${{github.workflow}} - ${{github.ref}}
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cancel-in-progress: true
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‎.github/workflows/deploy.yml‎

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branches:
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- main
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contents: write
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‎.github/workflows/pr-add-label.yml‎

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pull_request_target:
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types: [opened, edited, reopened, synchronize]
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concurrency:
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group: ${{github.workflow}} - ${{github.event_name}}
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‎solution/3100-3199/3133.Minimum Array End/README.md‎

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## 解法
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### 方法一
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### 方法一:贪心 + 位运算
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根据题目描述,要使得数组的最后一个元素尽可能小,且数组中的元素按位与的结果为 $x,ドル那么数组的第一个元素必须为 $x$。
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假设 $x$ 的二进制表示为 $\underline{1}00\underline{1}00,ドル那么数组序列为 $\underline{1}00\underline{1}00,ドル $\underline{1}00\underline{1}01,ドル $\underline{1}00\underline{1}10,ドル $\underline{1}00\underline{1}11$...
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如果我们忽略掉下划线部分,那么数组序列为 0000ドル,ドル 0001ドル,ドル 0010ドル,ドル 0011ドル$...,第一项为 0ドル,ドル那么第 $n$ 项为 $n-1$。
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因此,答案就是在 $x$ 的基础上,将 $n-1$ 的二进制的每一位依次填入 $x$ 的二进制中的 0ドル$ 位。
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时间复杂度 $O(\log x),ドル空间复杂度 $O(1)$。
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<!-- tabs:start -->
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```python
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class Solution:
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def minEnd(self, n: int, x: int) -> int:
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n -= 1
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ans = x
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for i in range(31):
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if x >> i & 1 ^ 1:
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ans |= (n & 1) << i
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n >>= 1
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ans |= n << 31
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return ans
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```
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```java
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class Solution {
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public long minEnd(int n, int x) {
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--n;
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long ans = x;
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for (int i = 0; i < 31; ++i) {
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if ((x >> i & 1) == 0) {
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ans |= (n & 1) << i;
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n >>= 1;
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}
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}
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ans |= (long) n << 31;
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return ans;
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}
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}
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```
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```cpp
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class Solution {
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public:
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long long minEnd(int n, int x) {
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--n;
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long long ans = x;
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for (int i = 0; i < 31; ++i) {
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if (x >> i & 1 ^ 1) {
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ans |= (n & 1) << i;
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n >>= 1;
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}
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}
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ans |= (1LL * n) << 31;
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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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func minEnd(n int, x int) (ans int64) {
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n--
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ans = int64(x)
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for i := 0; i < 31; i++ {
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if x>>i&1 == 0 {
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ans |= int64((n & 1) << i)
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n >>= 1
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}
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}
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ans |= int64(n) << 31
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return
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}
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```
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```ts
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function minEnd(n: number, x: number): number {
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--n;
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let ans: bigint = BigInt(x);
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for (let i = 0; i < 31; ++i) {
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if (((x >> i) & 1) ^ 1) {
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ans |= BigInt(n & 1) << BigInt(i);
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n >>= 1;
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}
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}
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ans |= BigInt(n) << BigInt(31);
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return Number(ans);
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}
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```
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<!-- tabs:end -->

‎solution/3100-3199/3133.Minimum Array End/README_EN.md‎

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## Solutions
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### Solution 1
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### Solution 1: Greedy + Bit Manipulation
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According to the problem description, to make the last element of the array as small as possible and the bitwise AND result of the elements in the array is $x,ドル the first element of the array must be $x$.
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Assume the binary representation of $x$ is $\underline{1}00\underline{1}00,ドル then the array sequence is $\underline{1}00\underline{1}00,ドル $\underline{1}00\underline{1}01,ドル $\underline{1}00\underline{1}10,ドル $\underline{1}00\underline{1}11$...
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If we ignore the underlined part, then the array sequence is 0000ドル,ドル 0001ドル,ドル 0010ドル,ドル 0011ドル$..., the first item is 0ドル,ドル then the $n$-th item is $n-1$.
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Therefore, the answer is to fill each bit of the binary of $n-1$ into the 0ドル$ bit of the binary of $x$ based on $x$.
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The time complexity is $O(\log x),ドル and the space complexity is $O(1)$.
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<!-- tabs:start -->
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```python
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class Solution:
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def minEnd(self, n: int, x: int) -> int:
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n -= 1
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ans = x
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for i in range(31):
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if x >> i & 1 ^ 1:
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ans |= (n & 1) << i
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n >>= 1
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ans |= n << 31
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return ans
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```
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```java
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class Solution {
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public long minEnd(int n, int x) {
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--n;
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long ans = x;
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for (int i = 0; i < 31; ++i) {
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if ((x >> i & 1) == 0) {
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ans |= (n & 1) << i;
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n >>= 1;
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}
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}
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ans |= (long) n << 31;
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return ans;
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}
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}
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```
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```cpp
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class Solution {
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public:
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long long minEnd(int n, int x) {
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--n;
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long long ans = x;
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for (int i = 0; i < 31; ++i) {
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if (x >> i & 1 ^ 1) {
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ans |= (n & 1) << i;
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n >>= 1;
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}
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}
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ans |= (1LL * n) << 31;
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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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func minEnd(n int, x int) (ans int64) {
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n--
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ans = int64(x)
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for i := 0; i < 31; i++ {
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if x>>i&1 == 0 {
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ans |= int64((n & 1) << i)
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n >>= 1
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}
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}
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ans |= int64(n) << 31
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return
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}
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```
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```ts
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function minEnd(n: number, x: number): number {
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--n;
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let ans: bigint = BigInt(x);
128+
for (let i = 0; i < 31; ++i) {
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if (((x >> i) & 1) ^ 1) {
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ans |= BigInt(n & 1) << BigInt(i);
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n >>= 1;
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}
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}
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ans |= BigInt(n) << BigInt(31);
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return Number(ans);
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}
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```
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<!-- tabs:end -->
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class Solution {
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public:
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long long minEnd(int n, int x) {
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--n;
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long long ans = x;
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for (int i = 0; i < 31; ++i) {
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if (x >> i & 1 ^ 1) {
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ans |= (n & 1) << i;
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n >>= 1;
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}
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}
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ans |= (1LL * n) << 31;
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return ans;
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}
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};
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@@ -0,0 +1,12 @@
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func minEnd(n int, x int) (ans int64) {
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n--
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ans = int64(x)
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for i := 0; i < 31; i++ {
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if x>>i&1 == 0 {
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ans |= int64((n & 1) << i)
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n >>= 1
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}
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}
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ans |= int64(n) << 31
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return
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}
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class Solution {
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public long minEnd(int n, int x) {
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--n;
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long ans = x;
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for (int i = 0; i < 31; ++i) {
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if ((x >> i & 1) == 0) {
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ans |= (n & 1) << i;
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n >>= 1;
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}
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}
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ans |= (long) n << 31;
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return ans;
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}
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}

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