Skip to content

feat: add solutions for lc No.0401 #5034

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Merged
yanglbme merged 1 commit into from
Feb 17, 2026
Merged

feat: add solutions for lc No.0401 #5034

Show file tree
Hide file tree
Changes from all commits
Commits
Show all changes
1 commit
Select commit
File filter

Filter by extension

Filter by extension
Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
149 changes: 81 additions & 68 deletions solution/0400-0499/0401.Binary Watch/README.md
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -71,9 +71,11 @@ tags:

### 方法一：枚举组合

题目可转换为求 i(`i∈[0,12)`) 和 j(`j∈[0,60)`) 所有可能的组合。
题目可以转换为求 $i \in [0, 12)$ 和 $j \in [0, 60)$ 的所有可能组合。

合法组合需要满足的条件是：i 的二进制形式中 1 的个数加上 j 的二进制形式中 1 的个数，结果等于 turnedOn。
合法组合需要满足的条件是 $i$ 的二进制形式中 1 的个数加上 $j$ 的二进制形式中 1 的个数，结果等于 $\textit{turnedOn}$。

时间复杂度 $O(1)$，空间复杂度 $O(1)$。

<!-- tabs:start -->

Expand Down Expand Up @@ -147,85 +149,45 @@ func readBinaryWatch(turnedOn int) []string {

```ts
function readBinaryWatch(turnedOn: number): string[] {
if (turnedOn === 0) {
return ['0:00'];
}
const n = 10;
const res = [];
const bitArr = new Array(10).fill(false);
const createTime = () => {
return [
bitArr.slice(0, 4).reduce((p, v) => (p << 1) | Number(v), 0),
bitArr.slice(4).reduce((p, v) => (p << 1) | Number(v), 0),
];
};
const helper = (i: number, count: number) => {
if (i + count > n || count === 0) {
return;
}
bitArr[i] = true;
if (count === 1) {
const [h, m] = createTime();
if (h < 12 && m < 60) {
res.push(`${h}:${m < 10 ? '0' + m : m}`);
const ans: string[] = [];

for (let i = 0; i < 12; ++i) {
for (let j = 0; j < 60; ++j) {
if (bitCount(i) + bitCount(j) === turnedOn) {
ans.push(`${i}:${j.toString().padStart(2, '0')}`);
}
}
helper(i + 1, count - 1);
bitArr[i] = false;
helper(i + 1, count);
};
helper(0, turnedOn);
return res;
}

return ans;
}

function bitCount(i: number): number {
i = i - ((i >>> 1) & 0x55555555);
i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);
i = (i + (i >>> 4)) & 0x0f0f0f0f;
i = i + (i >>> 8);
i = i + (i >>> 16);
return i & 0x3f;
}
```

#### Rust

```rust
impl Solution {
fn create_time(bit_arr: &[bool; 10]) -> (i32, i32) {
let mut h = 0;
let mut m = 0;
for i in 0..4 {
h <<= 1;
h |= if bit_arr[i] { 1 } else { 0 };
}
for i in 4..10 {
m <<= 1;
m |= if bit_arr[i] { 1 } else { 0 };
}

(h, m)
}
pub fn read_binary_watch(turned_on: i32) -> Vec<String> {
let mut ans: Vec<String> = Vec::new();

fn helper(res: &mut Vec<String>, bit_arr: &mut [bool; 10], i: usize, count: usize) {
if i + count > 10 || count == 0 {
return;
}
bit_arr[i] = true;
if count == 1 {
let (h, m) = Self::create_time(bit_arr);
if h < 12 && m < 60 {
if m < 10 {
res.push(format!("{}:0{}", h, m));
} else {
res.push(format!("{}:{}", h, m));
for i in 0u32..12 {
for j in 0u32..60 {
if (i.count_ones() + j.count_ones()) as i32 == turned_on {
ans.push(format!("{}:{:02}", i, j));
}
}
}
Self::helper(res, bit_arr, i + 1, count - 1);
bit_arr[i] = false;
Self::helper(res, bit_arr, i + 1, count);
}

pub fn read_binary_watch(turned_on: i32) -> Vec<String> {
if turned_on == 0 {
return vec![String::from("0:00")];
}
let mut res = vec![];
let mut bit_arr = [false; 10];
Self::helper(&mut res, &mut bit_arr, 0, turned_on as usize);
res
ans
}
}
```
Expand All @@ -238,7 +200,9 @@ impl Solution {

### 方法二：二进制枚举

利用 10 个二进制位表示手表，其中前 4 位代表小时，后 6 位代表分钟。枚举 `[0, 1 << 10)` 的所有数，找出合法的数。
我们可以利用 $10$ 个二进制位表示手表，其中前 $4$ 位代表小时，后 $6$ 位代表分钟。枚举 $[0, 2^{10})$ 中的每个数，判断其二进制表示中 1 的个数是否等于 $\textit{turnedOn}$，如果是，则将其转换为时间格式加入答案中。

时间复杂度 $O(1)$，空间复杂度 $O(1)$。

<!-- tabs:start -->

Expand Down Expand Up @@ -305,6 +269,55 @@ func readBinaryWatch(turnedOn int) []string {
}
```

#### TypeScript

```ts
function readBinaryWatch(turnedOn: number): string[] {
const ans: string[] = [];

for (let i = 0; i < (1 << 10); ++i) {
const h = i >> 6;
const m = i & 0b111111;

if (h < 12 && m < 60 && bitCount(i) === turnedOn) {
ans.push(`${h}:${m < 10 ? "0" : ""}${m}`);
}
}

return ans;
}

function bitCount(i: number): number {
i = i - ((i >>> 1) & 0x55555555);
i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);
i = (i + (i >>> 4)) & 0x0f0f0f0f;
i = i + (i >>> 8);
i = i + (i >>> 16);
return i & 0x3f;
}
```

#### Rust

```rust
impl Solution {
pub fn read_binary_watch(turned_on: i32) -> Vec<String> {
let mut ans: Vec<String> = Vec::new();

for i in 0u32..(1 << 10) {
let h = i >> 6;
let m = i & 0b111111;

if h < 12 && m < 60 && i.count_ones() as i32 == turned_on {
ans.push(format!("{}:{:02}", h, m));
}
}

ans
}
}
```

<!-- tabs:end -->

<!-- solution:end -->
Expand Down
153 changes: 86 additions & 67 deletions solution/0400-0499/0401.Binary Watch/README_EN.md
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -60,7 +60,13 @@ tags:

<!-- solution:start -->

### Solution 1
### Solution 1: Enumerate Combinations

The problem can be converted to finding all possible combinations of $i \in [0, 12)$ and $j \in [0, 60)$.

A valid combination must satisfy the condition that the number of 1s in the binary representation of $i$ plus the number of 1s in the binary representation of $j$ equals $\textit{turnedOn}$.

The time complexity is $O(1)$, and the space complexity is $O(1)$.

<!-- tabs:start -->

Expand Down Expand Up @@ -134,85 +140,45 @@ func readBinaryWatch(turnedOn int) []string {

```ts
function readBinaryWatch(turnedOn: number): string[] {
if (turnedOn === 0) {
return ['0:00'];
}
const n = 10;
const res = [];
const bitArr = new Array(10).fill(false);
const createTime = () => {
return [
bitArr.slice(0, 4).reduce((p, v) => (p << 1) | Number(v), 0),
bitArr.slice(4).reduce((p, v) => (p << 1) | Number(v), 0),
];
};
const helper = (i: number, count: number) => {
if (i + count > n || count === 0) {
return;
}
bitArr[i] = true;
if (count === 1) {
const [h, m] = createTime();
if (h < 12 && m < 60) {
res.push(`${h}:${m < 10 ? '0' + m : m}`);
const ans: string[] = [];

for (let i = 0; i < 12; ++i) {
for (let j = 0; j < 60; ++j) {
if (bitCount(i) + bitCount(j) === turnedOn) {
ans.push(`${i}:${j.toString().padStart(2, '0')}`);
}
}
helper(i + 1, count - 1);
bitArr[i] = false;
helper(i + 1, count);
};
helper(0, turnedOn);
return res;
}

return ans;
}

function bitCount(i: number): number {
i = i - ((i >>> 1) & 0x55555555);
i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);
i = (i + (i >>> 4)) & 0x0f0f0f0f;
i = i + (i >>> 8);
i = i + (i >>> 16);
return i & 0x3f;
}
```

#### Rust

```rust
impl Solution {
fn create_time(bit_arr: &[bool; 10]) -> (i32, i32) {
let mut h = 0;
let mut m = 0;
for i in 0..4 {
h <<= 1;
h |= if bit_arr[i] { 1 } else { 0 };
}
for i in 4..10 {
m <<= 1;
m |= if bit_arr[i] { 1 } else { 0 };
}

(h, m)
}
pub fn read_binary_watch(turned_on: i32) -> Vec<String> {
let mut ans: Vec<String> = Vec::new();

fn helper(res: &mut Vec<String>, bit_arr: &mut [bool; 10], i: usize, count: usize) {
if i + count > 10 || count == 0 {
return;
}
bit_arr[i] = true;
if count == 1 {
let (h, m) = Self::create_time(bit_arr);
if h < 12 && m < 60 {
if m < 10 {
res.push(format!("{}:0{}", h, m));
} else {
res.push(format!("{}:{}", h, m));
for i in 0u32..12 {
for j in 0u32..60 {
if (i.count_ones() + j.count_ones()) as i32 == turned_on {
ans.push(format!("{}:{:02}", i, j));
}
}
}
Self::helper(res, bit_arr, i + 1, count - 1);
bit_arr[i] = false;
Self::helper(res, bit_arr, i + 1, count);
}

pub fn read_binary_watch(turned_on: i32) -> Vec<String> {
if turned_on == 0 {
return vec![String::from("0:00")];
}
let mut res = vec![];
let mut bit_arr = [false; 10];
Self::helper(&mut res, &mut bit_arr, 0, turned_on as usize);
res
ans
}
}
```
Expand All @@ -223,7 +189,11 @@ impl Solution {

<!-- solution:start -->

### Solution 2
### Solution 2: Binary Enumeration

We can use $10$ binary bits to represent the watch, where the first $4$ bits represent hours and the last $6$ bits represent minutes. Enumerate each number in $[0, 2^{10})$, check if the number of 1s in its binary representation equals $\textit{turnedOn}$, and if so, convert it to time format and add it to the answer.

The time complexity is $O(1)$, and the space complexity is $O(1)$.

<!-- tabs:start -->

Expand Down Expand Up @@ -290,6 +260,55 @@ func readBinaryWatch(turnedOn int) []string {
}
```

#### TypeScript

```ts
function readBinaryWatch(turnedOn: number): string[] {
const ans: string[] = [];

for (let i = 0; i < (1 << 10); ++i) {
const h = i >> 6;
const m = i & 0b111111;

if (h < 12 && m < 60 && bitCount(i) === turnedOn) {
ans.push(`${h}:${m < 10 ? "0" : ""}${m}`);
}
}

return ans;
}

function bitCount(i: number): number {
i = i - ((i >>> 1) & 0x55555555);
i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);
i = (i + (i >>> 4)) & 0x0f0f0f0f;
i = i + (i >>> 8);
i = i + (i >>> 16);
return i & 0x3f;
}
```

#### Rust

```rust
impl Solution {
pub fn read_binary_watch(turned_on: i32) -> Vec<String> {
let mut ans: Vec<String> = Vec::new();

for i in 0u32..(1 << 10) {
let h = i >> 6;
let m = i & 0b111111;

if h < 12 && m < 60 && i.count_ones() as i32 == turned_on {
ans.push(format!("{}:{:02}", h, m));
}
}

ans
}
}
```

<!-- tabs:end -->

<!-- solution:end -->
Expand Down
Loading