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Jan 21, 2025
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feat: add solutions to lc problem: No.2220 #3977

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118 changes: 46 additions & 72 deletions solution/2200-2299/2220.Minimum Bit Flips to Convert Number/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -69,7 +69,11 @@ tags:

<!-- solution:start -->

### 方法一
### 方法一：位运算

根据题目描述，我们只需要计算 $\textit{start} \oplus \textit{goal}$ 的二进制表示中有多少个 1 即可。

时间复杂度 $O(\log n)$，其中 $n$ 是题目中整数的大小。空间复杂度 $O(1)$。

<!-- tabs:start -->

Expand All @@ -78,26 +82,15 @@ tags:
```python
class Solution:
def minBitFlips(self, start: int, goal: int) -> int:
t = start ^ goal
ans = 0
while t:
ans += t & 1
t >>= 1
return ans
return (start ^ goal).bit_count()
```

#### Java

```java
class Solution {
public int minBitFlips(int start, int goal) {
int t = start ^ goal;
int ans = 0;
while (t != 0) {
ans += t & 1;
t >>= 1;
}
return ans;
return Integer.bitCount(start ^ goal);
}
}
```
Expand All @@ -108,13 +101,7 @@ class Solution {
class Solution {
public:
int minBitFlips(int start, int goal) {
int t = start ^ goal;
int ans = 0;
while (t) {
ans += t & 1;
t >>= 1;
}
return ans;
return __builtin_popcount(start ^ goal);
}
};
```
Expand All @@ -123,27 +110,24 @@ public:

```go
func minBitFlips(start int, goal int) int {
t := start ^ goal
ans := 0
for t != 0 {
ans += t & 1
t >>= 1
}
return ans
return bits.OnesCount(uint(start ^ goal))
}
```

#### TypeScript

```ts
function minBitFlips(start: number, goal: number): number {
let tmp = start ^ goal;
let ans = 0;
while (tmp !== 0) {
ans += tmp & 1;
tmp >>= 1;
}
return ans;
return bitCount(start ^ goal);
}

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;
}
```

Expand All @@ -152,26 +136,42 @@ function minBitFlips(start: number, goal: number): number {
```rust
impl Solution {
pub fn min_bit_flips(start: i32, goal: i32) -> i32 {
let mut tmp = start ^ goal;
let mut ans = 0;
while tmp != 0 {
ans += tmp & 1;
tmp >>= 1;
}
ans
(start ^ goal).count_ones() as i32
}
}
```

#### JavaScript

```js
/**
* @param {number} start
* @param {number} goal
* @return {number}
*/
var minBitFlips = function (start, goal) {
return bitCount(start ^ goal);
};

function bitCount(i) {
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;
}
```

#### C

```c
int minBitFlips(int start, int goal) {
int tmp = start ^ goal;
int x = start ^ goal;
int ans = 0;
while (tmp) {
ans += tmp & 1;
tmp >>= 1;
while (x) {
ans += (x & 1);
x >>= 1;
}
return ans;
}
Expand All @@ -181,30 +181,4 @@ int minBitFlips(int start, int goal) {

<!-- solution:end -->

<!-- solution:start -->

### Solution 2

<!-- tabs:start -->

#### TypeScript

```ts
function minBitFlips(start: number, goal: number): number {
return (start ^ goal).toString(2).replace(/0/g, '').length;
}
```

#### JavaScript

```js
function minBitFlips(start, goal) {
return (start ^ goal).toString(2).replace(/0/g, '').length;
}
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- problem:end -->
118 changes: 46 additions & 72 deletions solution/2200-2299/2220.Minimum Bit Flips to Convert Number/README_EN.md
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Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -66,7 +66,11 @@ It can be shown we cannot convert 3 to 4 in less than 3 steps. Hence, we return

<!-- solution:start -->

### Solution 1
### Solution 1: Bit Manipulation

According to the problem description, we only need to count the number of 1s in the binary representation of $\textit{start} \oplus \textit{goal}$.

The time complexity is $O(\log n)$, where $n$ is the size of the integers in the problem. The space complexity is $O(1)$.

<!-- tabs:start -->

Expand All @@ -75,26 +79,15 @@ It can be shown we cannot convert 3 to 4 in less than 3 steps. Hence, we return
```python
class Solution:
def minBitFlips(self, start: int, goal: int) -> int:
t = start ^ goal
ans = 0
while t:
ans += t & 1
t >>= 1
return ans
return (start ^ goal).bit_count()
```

#### Java

```java
class Solution {
public int minBitFlips(int start, int goal) {
int t = start ^ goal;
int ans = 0;
while (t != 0) {
ans += t & 1;
t >>= 1;
}
return ans;
return Integer.bitCount(start ^ goal);
}
}
```
Expand All @@ -105,13 +98,7 @@ class Solution {
class Solution {
public:
int minBitFlips(int start, int goal) {
int t = start ^ goal;
int ans = 0;
while (t) {
ans += t & 1;
t >>= 1;
}
return ans;
return __builtin_popcount(start ^ goal);
}
};
```
Expand All @@ -120,27 +107,24 @@ public:

```go
func minBitFlips(start int, goal int) int {
t := start ^ goal
ans := 0
for t != 0 {
ans += t & 1
t >>= 1
}
return ans
return bits.OnesCount(uint(start ^ goal))
}
```

#### TypeScript

```ts
function minBitFlips(start: number, goal: number): number {
let tmp = start ^ goal;
let ans = 0;
while (tmp !== 0) {
ans += tmp & 1;
tmp >>= 1;
}
return ans;
return bitCount(start ^ goal);
}

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;
}
```

Expand All @@ -149,26 +133,42 @@ function minBitFlips(start: number, goal: number): number {
```rust
impl Solution {
pub fn min_bit_flips(start: i32, goal: i32) -> i32 {
let mut tmp = start ^ goal;
let mut ans = 0;
while tmp != 0 {
ans += tmp & 1;
tmp >>= 1;
}
ans
(start ^ goal).count_ones() as i32
}
}
```

#### JavaScript

```js
/**
* @param {number} start
* @param {number} goal
* @return {number}
*/
var minBitFlips = function (start, goal) {
return bitCount(start ^ goal);
};

function bitCount(i) {
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;
}
```

#### C

```c
int minBitFlips(int start, int goal) {
int tmp = start ^ goal;
int x = start ^ goal;
int ans = 0;
while (tmp) {
ans += tmp & 1;
tmp >>= 1;
while (x) {
ans += (x & 1);
x >>= 1;
}
return ans;
}
Expand All @@ -178,30 +178,4 @@ int minBitFlips(int start, int goal) {

<!-- solution:end -->

<!-- solution:start -->

### Solution 2

<!-- tabs:start -->

#### TypeScript

```ts
function minBitFlips(start: number, goal: number): number {
return (start ^ goal).toString(2).replace(/0/g, '').length;
}
```

#### JavaScript

```js
function minBitFlips(start, goal) {
return (start ^ goal).toString(2).replace(/0/g, '').length;
}
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- problem:end -->
10 changes: 5 additions & 5 deletions solution/2200-2299/2220.Minimum Bit Flips to Convert Number/Solution.c
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Open in desktop
Original file line number Diff line number Diff line change
@@ -1,9 +1,9 @@
int minBitFlips(int start, int goal) {
int tmp = start ^ goal;
int x = start ^ goal;
int ans = 0;
while (tmp) {
ans += tmp & 1;
tmp >>= 1;
while (x) {
ans += (x & 1);
x >>= 1;
}
return ans;
}
}
10 changes: 2 additions & 8 deletions solution/2200-2299/2220.Minimum Bit Flips to Convert Number/Solution.cpp
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,12 +1,6 @@
class Solution {
public:
int minBitFlips(int start, int goal) {
int t = start ^ goal;
int ans = 0;
while (t) {
ans += t & 1;
t >>= 1;
}
return ans;
return __builtin_popcount(start ^ goal);
}
};
};
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