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feat: add solutions to lc problem: No.2134 (#2562)
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No.2134.Minimum Swaps to Group All 1's Together II
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@yanglbme
yanglbme authored Apr 9, 2024
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144 changes: 88 additions & 56 deletions solution/2100-2199/2134.Minimum Swaps to Group All 1's Together II/README.md
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Expand Up @@ -57,46 +57,44 @@

## 解法

### 方法一
### 方法一：滑动窗口

我们先统计数组中 $1$ 的个数，记为 $k$，那么题目实际上就是求一个长度为 $k$ 的环形子数组，使得子数组中 $1$ 的个数最多，那么最少交换次数就是 $k$ 减去子数组中 $1$ 的个数最多的那个子数组中 $1$ 的个数。

我们可以使用滑动窗口来解决这个问题，首先统计数组中前 $k$ 个元素中 $1$ 的个数，记为 $cnt$，然后我们维护一个长度为 $k$ 的滑动窗口，每次向右移动一个位置，更新 $cnt$，同时更新最大的 $cnt$ 值，即 $mx = \max(mx, cnt)$，最后答案就是 $k - mx$。

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

<!-- tabs:start -->

```python
class Solution:
def minSwaps(self, nums: List[int]) -> int:
cnt = nums.count(1)
k = nums.count(1)
mx = cnt = sum(nums[:k])
n = len(nums)
s = [0] * ((n << 1) + 1)
for i in range(n << 1):
s[i + 1] = s[i] + nums[i % n]
mx = 0
for i in range(n << 1):
j = i + cnt - 1
if j < (n << 1):
mx = max(mx, s[j + 1] - s[i])
return cnt - mx
for i in range(k, n + k):
cnt += nums[i % n]
cnt -= nums[(i - k + n) % n]
mx = max(mx, cnt)
return k - mx
```

```java
class Solution {
public int minSwaps(int[] nums) {
int cnt = 0;
for (int v : nums) {
cnt += v;
}
int k = Arrays.stream(nums).sum();
int n = nums.length;
int[] s = new int[(n << 1) + 1];
for (int i = 0; i < (n << 1); ++i) {
s[i + 1] = s[i] + nums[i % n];
int cnt = 0;
for (int i = 0; i < k; ++i) {
cnt += nums[i];
}
int mx = 0;
for (int i = 0; i < (n << 1); ++i) {
int j = i + cnt - 1;
if (j < (n << 1)) {
mx = Math.max(mx, s[j + 1] - s[i]);
}
int mx = cnt;
for (int i = k; i < n + k; ++i) {
cnt += nums[i % n] - nums[(i - k + n) % n];
mx = Math.max(mx, cnt);
}
return cnt - mx;
return k - mx;
}
}
```
Expand All @@ -105,56 +103,90 @@ class Solution {
class Solution {
public:
int minSwaps(vector<int>& nums) {
int cnt = 0;
for (int& v : nums) cnt += v;
int k = accumulate(nums.begin(), nums.end(), 0);
int n = nums.size();
vector<int> s((n << 1) + 1);
for (int i = 0; i < (n << 1); ++i) s[i + 1] = s[i] + nums[i % n];
int mx = 0;
for (int i = 0; i < (n << 1); ++i) {
int j = i + cnt - 1;
if (j < (n << 1)) mx = max(mx, s[j + 1] - s[i]);
int cnt = accumulate(nums.begin(), nums.begin() + k, 0);
int mx = cnt;
for (int i = k; i < n + k; ++i) {
cnt += nums[i % n] - nums[(i - k + n) % n];
mx = max(mx, cnt);
}
return cnt - mx;
return k - mx;
}
};
```
```go
func minSwaps(nums []int) int {
k := 0
for _, x := range nums {
k += x
}
cnt := 0
for _, v := range nums {
cnt += v
for i := 0; i < k; i++ {
cnt += nums[i]
}
mx := cnt
n := len(nums)
s := make([]int, (n<<1)+1)
for i := 0; i < (n << 1); i++ {
s[i+1] = s[i] + nums[i%n]
}
mx := 0
for i := 0; i < (n << 1); i++ {
j := i + cnt - 1
if j < (n << 1) {
mx = max(mx, s[j+1]-s[i])
}
for i := k; i < n+k; i++ {
cnt += nums[i%n] - nums[(i-k+n)%n]
mx = max(mx, cnt)
}
return cnt - mx
return k - mx
}
```

```ts
function minSwaps(nums: number[]): number {
const k = nums.reduce((a, b) => a + b, 0);
let cnt = nums.slice(0, k).reduce((a, b) => a + b, 0);
let mx = cnt;
const n = nums.length;
const m = nums.reduce((a, c) => a + c, 0);
let cnt = nums.reduce((a, c, i) => a + (i < m ? c : 0), 0);
let ans = cnt;
for (let i = m; i < m + n; i++) {
let prev = nums[i - m];
let post = nums[i % n];
cnt += post - prev;
ans = Math.max(cnt, ans);
for (let i = k; i < n + k; ++i) {
cnt += nums[i % n] - nums[(i - k + n) % n];
mx = Math.max(mx, cnt);
}
return k - mx;
}
```

```rust
impl Solution {
pub fn min_swaps(nums: Vec<i32>) -> i32 {
let k: i32 = nums.iter().sum();
let n: usize = nums.len();
let mut cnt: i32 = 0;
for i in 0..k {
cnt += nums[i as usize];
}
let mut mx: i32 = cnt;
for i in k..(n as i32) + k {
cnt +=
nums[(i % (n as i32)) as usize] -
nums[((i - k + (n as i32)) % (n as i32)) as usize];
mx = mx.max(cnt);
}
return k - mx;
}
}
```

```cs
public class Solution {
public int MinSwaps(int[] nums) {
int k = nums.Sum();
int n = nums.Length;
int cnt = 0;
for (int i = 0; i < k; ++i) {
cnt += nums[i];
}
int mx = cnt;
for (int i = k; i < n + k; ++i) {
cnt += nums[i % n] - nums[(i - k + n) % n];
mx = Math.Max(mx, cnt);
}
return k - mx;
}
return m - ans;
}
```

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144 changes: 88 additions & 56 deletions solution/2100-2199/2134.Minimum Swaps to Group All 1's Together II/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -57,46 +57,44 @@ Thus, the minimum number of swaps required is 0.

## Solutions

### Solution 1
### Solution 1: Sliding Window

First, we count the number of $1$s in the array, denoted as $k$. The problem is actually asking for a circular subarray of length $k$ that contains the maximum number of $1$s. Therefore, the minimum number of swaps is $k$ minus the maximum number of $1$s in that subarray.

We can solve this problem using a sliding window. First, we count the number of $1$s in the first $k$ elements of the array, denoted as $cnt$. Then, we maintain a sliding window of length $k$. Each time we move the window one position to the right, we update $cnt$ and simultaneously update the maximum $cnt$ value, i.e., $mx = \max(mx, cnt)$. Finally, the answer is $k - mx$.

The time complexity is $O(n)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$.

<!-- tabs:start -->

```python
class Solution:
def minSwaps(self, nums: List[int]) -> int:
cnt = nums.count(1)
k = nums.count(1)
mx = cnt = sum(nums[:k])
n = len(nums)
s = [0] * ((n << 1) + 1)
for i in range(n << 1):
s[i + 1] = s[i] + nums[i % n]
mx = 0
for i in range(n << 1):
j = i + cnt - 1
if j < (n << 1):
mx = max(mx, s[j + 1] - s[i])
return cnt - mx
for i in range(k, n + k):
cnt += nums[i % n]
cnt -= nums[(i - k + n) % n]
mx = max(mx, cnt)
return k - mx
```

```java
class Solution {
public int minSwaps(int[] nums) {
int cnt = 0;
for (int v : nums) {
cnt += v;
}
int k = Arrays.stream(nums).sum();
int n = nums.length;
int[] s = new int[(n << 1) + 1];
for (int i = 0; i < (n << 1); ++i) {
s[i + 1] = s[i] + nums[i % n];
int cnt = 0;
for (int i = 0; i < k; ++i) {
cnt += nums[i];
}
int mx = 0;
for (int i = 0; i < (n << 1); ++i) {
int j = i + cnt - 1;
if (j < (n << 1)) {
mx = Math.max(mx, s[j + 1] - s[i]);
}
int mx = cnt;
for (int i = k; i < n + k; ++i) {
cnt += nums[i % n] - nums[(i - k + n) % n];
mx = Math.max(mx, cnt);
}
return cnt - mx;
return k - mx;
}
}
```
Expand All @@ -105,56 +103,90 @@ class Solution {
class Solution {
public:
int minSwaps(vector<int>& nums) {
int cnt = 0;
for (int& v : nums) cnt += v;
int k = accumulate(nums.begin(), nums.end(), 0);
int n = nums.size();
vector<int> s((n << 1) + 1);
for (int i = 0; i < (n << 1); ++i) s[i + 1] = s[i] + nums[i % n];
int mx = 0;
for (int i = 0; i < (n << 1); ++i) {
int j = i + cnt - 1;
if (j < (n << 1)) mx = max(mx, s[j + 1] - s[i]);
int cnt = accumulate(nums.begin(), nums.begin() + k, 0);
int mx = cnt;
for (int i = k; i < n + k; ++i) {
cnt += nums[i % n] - nums[(i - k + n) % n];
mx = max(mx, cnt);
}
return cnt - mx;
return k - mx;
}
};
```
```go
func minSwaps(nums []int) int {
k := 0
for _, x := range nums {
k += x
}
cnt := 0
for _, v := range nums {
cnt += v
for i := 0; i < k; i++ {
cnt += nums[i]
}
mx := cnt
n := len(nums)
s := make([]int, (n<<1)+1)
for i := 0; i < (n << 1); i++ {
s[i+1] = s[i] + nums[i%n]
}
mx := 0
for i := 0; i < (n << 1); i++ {
j := i + cnt - 1
if j < (n << 1) {
mx = max(mx, s[j+1]-s[i])
}
for i := k; i < n+k; i++ {
cnt += nums[i%n] - nums[(i-k+n)%n]
mx = max(mx, cnt)
}
return cnt - mx
return k - mx
}
```

```ts
function minSwaps(nums: number[]): number {
const k = nums.reduce((a, b) => a + b, 0);
let cnt = nums.slice(0, k).reduce((a, b) => a + b, 0);
let mx = cnt;
const n = nums.length;
const m = nums.reduce((a, c) => a + c, 0);
let cnt = nums.reduce((a, c, i) => a + (i < m ? c : 0), 0);
let ans = cnt;
for (let i = m; i < m + n; i++) {
let prev = nums[i - m];
let post = nums[i % n];
cnt += post - prev;
ans = Math.max(cnt, ans);
for (let i = k; i < n + k; ++i) {
cnt += nums[i % n] - nums[(i - k + n) % n];
mx = Math.max(mx, cnt);
}
return k - mx;
}
```

```rust
impl Solution {
pub fn min_swaps(nums: Vec<i32>) -> i32 {
let k: i32 = nums.iter().sum();
let n: usize = nums.len();
let mut cnt: i32 = 0;
for i in 0..k {
cnt += nums[i as usize];
}
let mut mx: i32 = cnt;
for i in k..(n as i32) + k {
cnt +=
nums[(i % (n as i32)) as usize] -
nums[((i - k + (n as i32)) % (n as i32)) as usize];
mx = mx.max(cnt);
}
return k - mx;
}
}
```

```cs
public class Solution {
public int MinSwaps(int[] nums) {
int k = nums.Sum();
int n = nums.Length;
int cnt = 0;
for (int i = 0; i < k; ++i) {
cnt += nums[i];
}
int mx = cnt;
for (int i = k; i < n + k; ++i) {
cnt += nums[i % n] - nums[(i - k + n) % n];
mx = Math.Max(mx, cnt);
}
return k - mx;
}
return m - ans;
}
```

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