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feat: add 2nd ts solution to lc problem: No.0995 #3161

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Merged
yanglbme merged 14 commits into doocs:main from rain84:feature/lc-995
Jun 25, 2024
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Original file line number Diff line number Diff line change
Expand Up @@ -96,7 +96,7 @@ class Solution:
ans = s = 0
for i, x in enumerate(nums):
s += d[i]
if x % 2 == s % 2:
if s % 2 == x:
if i + k > n:
return -1
d[i] += 1
Expand All @@ -116,7 +116,7 @@ class Solution {
int ans = 0, s = 0;
for (int i = 0; i < n; ++i) {
s += d[i];
if (nums[i] % 2 == s % 2) {
if (s % 2 == nums[i]) {
if (i + k > n) {
return -1;
}
Expand All @@ -143,7 +143,7 @@ public:
int ans = 0, s = 0;
for (int i = 0; i < n; ++i) {
s += d[i];
if (s % 2 == nums[i] % 2) {
if (s % 2 == nums[i]) {
if (i + k > n) {
return -1;
}
Expand All @@ -161,13 +161,13 @@ public:
#### Go

```go
func minKBitFlips(nums []int, k int) int {
func minKBitFlips(nums []int, k int) (ans int) {
n := len(nums)
d := make([]int, n+1)
ans, s := 0, 0
s := 0
for i, x := range nums {
s += d[i]
if s%2 == x%2 {
if s%2 == x {
if i+k > n {
return -1
}
Expand All @@ -177,7 +177,7 @@ func minKBitFlips(nums []int, k int) int {
ans++
}
}
return ans
return
}
```

Expand All @@ -190,7 +190,7 @@ function minKBitFlips(nums: number[], k: number): number {
let [ans, s] = [0, 0];
for (let i = 0; i < n; ++i) {
s += d[i];
if (s % 2 === nums[i] % 2) {
if (s % 2 === nums[i]) {
if (i + k > n) {
return -1;
}
Expand All @@ -215,7 +215,7 @@ impl Solution {
let mut s = 0;
for i in 0..n {
s += d[i];
if nums[i] % 2 == s % 2 {
if s % 2 == nums[i] {
if i + (k as usize) > n {
return -1;
}
Expand All @@ -234,4 +234,164 @@ impl Solution {

<!-- solution:end -->

<!-- solution:start -->

### 方法二:滑动窗口

我们可以用一个变量 $\text{flipped}$ 来表示当前位置是否翻转,如果 $\text{flipped}$ 为 1ドル,ドル表示当前位置已经翻转,否则表示当前位置未翻转。对于翻转过的位置,我们可以将其值设置为 $-1,ドル这样我们就可以区分出哪些位置已经翻转过了。

接下来我们从左到右遍历数组,对于每个位置 $i,ドル如果 $i \geq k$ 且 $i-k$ 位置的元素为 $-1,ドル那么当前位置的翻转状态应该与前一个位置的翻转状态相反。即 $\text{flipped} = \text{flipped} \oplus 1$。如果当前位置的元素与当前位置的翻转状态相同,那么我们需要翻转当前位置,此时我们判断一下 $i+k$ 是否超出了数组的长度,如果超出了数组的长度,那么就无法完成目标,返回 $-1$。否则我们将当前位置的翻转状态取反,同时将答案增加 1ドル,ドル并且将当前位置的元素设置为 $-1$。

这样当我们处理完数组中的所有元素时,返回答案即可。

时间复杂度 $O(n),ドル其中 $n$ 是数组 $nums$ 的长度。空间复杂度 $O(1)$。

<!-- tabs:start -->

#### Python3

```python
class Solution:
def minKBitFlips(self, nums: List[int], k: int) -> int:
ans = flipped = 0
for i, x in enumerate(nums):
if i >= k and nums[i - k] == -1:
flipped ^= 1
if x == flipped:
if i + k > len(nums):
return -1
flipped ^= 1
ans += 1
nums[i] = -1
return ans
```

#### Java

```java
class Solution {
public int minKBitFlips(int[] nums, int k) {
int n = nums.length;
int ans = 0, flipped = 0;
for (int i = 0; i < n; ++i) {
if (i >= k && nums[i - k] == -1) {
flipped ^= 1;
}
if (flipped == nums[i]) {
if (i + k > n) {
return -1;
}
flipped ^= 1;
++ans;
nums[i] = -1;
}
}
return ans;
}
}
```

#### C++

```cpp
class Solution {
public:
int minKBitFlips(vector<int>& nums, int k) {
int n = nums.size();
int ans = 0, flipped = 0;
for (int i = 0; i < n; ++i) {
if (i >= k && nums[i - k] == -1) {
flipped ^= 1;
}
if (flipped == nums[i]) {
if (i + k > n) {
return -1;
}
flipped ^= 1;
++ans;
nums[i] = -1;
}
}
return ans;
}
};
```

#### Go

```go
func minKBitFlips(nums []int, k int) (ans int) {
flipped := 0
for i, x := range nums {
if i >= k && nums[i-k] == -1 {
flipped ^= 1
}
if flipped == x {
if i+k > len(nums) {
return -1
}
flipped ^= 1
ans++
nums[i] = -1
}
}
return
}
```

#### TypeScript

```ts
function minKBitFlips(nums: number[], k: number): number {
const n = nums.length;
let [ans, flipped] = [0, 0];
for (let i = 0; i < n; i++) {
if (nums[i - k] === -1) {
flipped ^= 1;
}
if (nums[i] === flipped) {
if (i + k > n) {
return -1;
}
flipped ^= 1;
++ans;
nums[i] = -1;
}
}
return ans;
}
```

#### Rust

```rust
impl Solution {
pub fn min_k_bit_flips(mut nums: Vec<i32>, k: i32) -> i32 {
let mut ans = 0;
let mut flipped = 0;
let k = k as usize;

for i in 0..nums.len() {
if i >= k && nums[i - k] == -1 {
flipped ^= 1;
}
if flipped == nums[i] {
if i + k > nums.len() {
return -1;
}
flipped ^= 1;
ans += 1;
nums[i] = -1;
}
}

ans
}
}
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- problem:end -->
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