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feat: add solutions to lc problem: No.0869 #4633

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124 changes: 95 additions & 29 deletions solution/0800-0899/0869.Reordered Power of 2/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -57,7 +57,15 @@ tags:

<!-- solution:start -->

### 方法一
### 方法一：枚举

我们可以在 $[1, 10^9]$ 的范围内枚举所有的 $2$ 的幂，判断它们的数字组成是否与给定的数字相同。

定义一个函数 $f(x)$，表示数字 $x$ 的数字组成。我们可以将数字 $x$ 转换为一个长度为 $10$ 的数组，或者一个按数字大小排序的字符串。

首先，我们计算给定数字 $n$ 的数字组成 $\text{target} = f(n)$。然后，我们枚举 $i$ 从 1 开始，每次将 $i$ 左移一位（相当于乘以 $2$），直到 $i$ 超过 $10^9$。对于每个 $i$，我们计算它的数字组成，并与 $\text{target}$ 进行比较。如果相同，则返回 $\text{true}$；如果枚举结束仍未找到相同的数字组成，则返回 $\text{false}$。

时间复杂度 $O(\log^2 M)$，空间复杂度 $O(\log M)$。其中 $M$ 是本题的输入范围上限 ${10}^9$。

<!-- tabs:start -->

Expand All @@ -66,16 +74,17 @@ tags:
```python
class Solution:
def reorderedPowerOf2(self, n: int) -> bool:
def convert(n):
def f(x: int) -> List[int]:
cnt = [0] * 10
while n:
n, v = divmod(n, 10)
while x:
x, v = divmod(x, 10)
cnt[v] += 1
return cnt

i, s = 1, convert(n)
target = f(n)
i = 1
while i <= 10**9:
if convert(i) == s:
if f(i) == target:
return True
i <<= 1
return False
Expand All @@ -86,19 +95,19 @@ class Solution:
```java
class Solution {
public boolean reorderedPowerOf2(int n) {
String s = convert(n);
for (int i = 1; i <= Math.pow(10, 9); i <<= 1) {
if (s.equals(convert(i))) {
String target = f(n);
for (int i = 1; i <= 1000000000; i <<= 1) {
if (target.equals(f(i))) {
return true;
}
}
return false;
}

private String convert(int n) {
private String f(int x) {
char[] cnt = new char[10];
for (; n > 0; n /= 10) {
cnt[n % 10]++;
for (; x > 0; x /= 10) {
cnt[x % 10]++;
}
return new String(cnt);
}
Expand All @@ -111,17 +120,23 @@ class Solution {
class Solution {
public:
bool reorderedPowerOf2(int n) {
vector<int> s = convert(n);
for (int i = 1; i <= pow(10, 9); i <<= 1)
if (s == convert(i))
string target = f(n);
for (int i = 1; i <= 1000000000; i <<= 1) {
if (target == f(i)) {
return true;
}
}
return false;
}

vector<int> convert(int n) {
vector<int> cnt(10);
for (; n; n /= 10) ++cnt[n % 10];
return cnt;
private:
string f(int x) {
char cnt[10] = {};
while (x > 0) {
cnt[x % 10]++;
x /= 10;
}
return string(cnt, cnt + 10);
}
};
```
Expand All @@ -130,21 +145,72 @@ public:

```go
func reorderedPowerOf2(n int) bool {
convert := func(n int) []byte {
cnt := make([]byte, 10)
for ; n > 0; n /= 10 {
cnt[n%10]++
}
return cnt
}
s := convert(n)
for i := 1; i <= 1e9; i <<= 1 {
if bytes.Equal(s, convert(i)) {
target := f(n)
for i := 1; i <= 1000000000; i <<= 1 {
if bytes.Equal(target, f(i)) {
return true
}
}
return false
}

func f(x int) []byte {
cnt := make([]byte, 10)
for x > 0 {
cnt[x%10]++
x /= 10
}
return cnt
}
```

#### TypeScript

```ts
function reorderedPowerOf2(n: number): boolean {
const f = (x: number) => {
const cnt = Array(10).fill(0);
while (x > 0) {
cnt[x % 10]++;
x = (x / 10) | 0;
}
return cnt.join(',');
};
const target = f(n);
for (let i = 1; i <= 1_000_000_000; i <<= 1) {
if (target === f(i)) {
return true;
}
}
return false;
}
```

#### Rust

```rust
impl Solution {
pub fn reordered_power_of2(n: i32) -> bool {
fn f(mut x: i32) -> [u8; 10] {
let mut cnt = [0u8; 10];
while x > 0 {
cnt[(x % 10) as usize] += 1;
x /= 10;
}
cnt
}

let target = f(n);
let mut i = 1i32;
while i <= 1_000_000_000 {
if target == f(i) {
return true;
}
i <<= 1;
}
false
}
}
```

<!-- tabs:end -->
Expand Down
124 changes: 95 additions & 29 deletions solution/0800-0899/0869.Reordered Power of 2/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -52,7 +52,15 @@ tags:

<!-- solution:start -->

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

We can enumerate all powers of 2 in the range $[1, 10^9]$ and check if their digit composition is the same as the given number.

Define a function $f(x)$ that represents the digit composition of number $x$. We can convert the number $x$ into an array of length 10, or a string sorted by digit size.

First, we calculate the digit composition of the given number $n$ as $\text{target} = f(n)$. Then, we enumerate $i$ starting from 1, shifting $i$ left by one bit each time (equivalent to multiplying by 2), until $i$ exceeds $10^9$. For each $i$, we calculate its digit composition and compare it with $\text{target}$. If they are the same, we return $\text{true}$; if the enumeration ends without finding the same digit composition, we return $\text{false}$.

Time complexity $O(\log^2 M)$, space complexity $O(\log M)$. Where $M$ is the upper limit of the input range ${10}^9$ for this problem.

<!-- tabs:start -->

Expand All @@ -61,16 +69,17 @@ tags:
```python
class Solution:
def reorderedPowerOf2(self, n: int) -> bool:
def convert(n):
def f(x: int) -> List[int]:
cnt = [0] * 10
while n:
n, v = divmod(n, 10)
while x:
x, v = divmod(x, 10)
cnt[v] += 1
return cnt

i, s = 1, convert(n)
target = f(n)
i = 1
while i <= 10**9:
if convert(i) == s:
if f(i) == target:
return True
i <<= 1
return False
Expand All @@ -81,19 +90,19 @@ class Solution:
```java
class Solution {
public boolean reorderedPowerOf2(int n) {
String s = convert(n);
for (int i = 1; i <= Math.pow(10, 9); i <<= 1) {
if (s.equals(convert(i))) {
String target = f(n);
for (int i = 1; i <= 1000000000; i <<= 1) {
if (target.equals(f(i))) {
return true;
}
}
return false;
}

private String convert(int n) {
private String f(int x) {
char[] cnt = new char[10];
for (; n > 0; n /= 10) {
cnt[n % 10]++;
for (; x > 0; x /= 10) {
cnt[x % 10]++;
}
return new String(cnt);
}
Expand All @@ -106,17 +115,23 @@ class Solution {
class Solution {
public:
bool reorderedPowerOf2(int n) {
vector<int> s = convert(n);
for (int i = 1; i <= pow(10, 9); i <<= 1)
if (s == convert(i))
string target = f(n);
for (int i = 1; i <= 1000000000; i <<= 1) {
if (target == f(i)) {
return true;
}
}
return false;
}

vector<int> convert(int n) {
vector<int> cnt(10);
for (; n; n /= 10) ++cnt[n % 10];
return cnt;
private:
string f(int x) {
char cnt[10] = {};
while (x > 0) {
cnt[x % 10]++;
x /= 10;
}
return string(cnt, cnt + 10);
}
};
```
Expand All @@ -125,21 +140,72 @@ public:

```go
func reorderedPowerOf2(n int) bool {
convert := func(n int) []byte {
cnt := make([]byte, 10)
for ; n > 0; n /= 10 {
cnt[n%10]++
}
return cnt
}
s := convert(n)
for i := 1; i <= 1e9; i <<= 1 {
if bytes.Equal(s, convert(i)) {
target := f(n)
for i := 1; i <= 1000000000; i <<= 1 {
if bytes.Equal(target, f(i)) {
return true
}
}
return false
}

func f(x int) []byte {
cnt := make([]byte, 10)
for x > 0 {
cnt[x%10]++
x /= 10
}
return cnt
}
```

#### TypeScript

```ts
function reorderedPowerOf2(n: number): boolean {
const f = (x: number) => {
const cnt = Array(10).fill(0);
while (x > 0) {
cnt[x % 10]++;
x = (x / 10) | 0;
}
return cnt.join(',');
};
const target = f(n);
for (let i = 1; i <= 1_000_000_000; i <<= 1) {
if (target === f(i)) {
return true;
}
}
return false;
}
```

#### Rust

```rust
impl Solution {
pub fn reordered_power_of2(n: i32) -> bool {
fn f(mut x: i32) -> [u8; 10] {
let mut cnt = [0u8; 10];
while x > 0 {
cnt[(x % 10) as usize] += 1;
x /= 10;
}
cnt
}

let target = f(n);
let mut i = 1i32;
while i <= 1_000_000_000 {
if target == f(i) {
return true;
}
i <<= 1;
}
false
}
}
```

<!-- tabs:end -->
Expand Down
22 changes: 14 additions & 8 deletions solution/0800-0899/0869.Reordered Power of 2/Solution.cpp
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,16 +1,22 @@
class Solution {
public:
bool reorderedPowerOf2(int n) {
vector<int> s = convert(n);
for (int i = 1; i <= pow(10, 9); i <<= 1)
if (s == convert(i))
string target = f(n);
for (int i = 1; i <= 1000000000; i <<= 1) {
if (target == f(i)) {
return true;
}
}
return false;
}

vector<int> convert(int n) {
vector<int> cnt(10);
for (; n; n /= 10) ++cnt[n % 10];
return cnt;
private:
string f(int x) {
char cnt[10] = {};
while (x > 0) {
cnt[x % 10]++;
x /= 10;
}
return string(cnt, cnt + 10);
}
};
};
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