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feat: add solutions to lc problem: No.1151
No.1151.Minimum Swaps to Group All 1's Together
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‎solution/1100-1199/1151.Minimum Swaps to Group All 1's Together/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：滑动窗口**
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我们先统计数组中 $1$ 的个数，记为 $k$。然后我们使用滑动窗口，窗口大小为 $k$，窗口右边界从左向右移动，统计窗口内 $1$ 的个数，记为 $t$。每次移动窗口时，都更新 $t$ 的值，最后窗口右边界移动到数组末尾时，窗口内 $1$ 的个数最多，记为 $mx$。最后答案为 $k - mx$。
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时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为数组长度。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def minSwaps(self, data: List[int]) -> int:
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k = data.count(1)
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t = sum(data[:k])
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mx = t
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for i in range(k, len(data)):
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t += data[i]
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t -= data[i - k]
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mx = max(mx, t)
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return k - mx
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int minSwaps(int[] data) {
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int k = 0;
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for (int v : data) {
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k += v;
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}
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int t = 0;
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for (int i = 0; i < k; ++i) {
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t += data[i];
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}
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int mx = t;
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for (int i = k; i < data.length; ++i) {
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t += data[i];
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t -= data[i - k];
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mx = Math.max(mx, t);
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}
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return k - mx;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int minSwaps(vector<int>& data) {
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int k = 0;
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for (int& v : data) {
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k += v;
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}
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int t = 0;
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for (int i = 0; i < k; ++i) {
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t += data[i];
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}
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int mx = t;
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for (int i = k; i < data.size(); ++i) {
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t += data[i];
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t -= data[i - k];
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mx = max(mx, t);
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}
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return k - mx;
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}
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};
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```
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### **Go**
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```go
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func minSwaps(data []int) int {
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k := 0
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for _, v := range data {
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k += v
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}
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t := 0
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for _, v := range data[:k] {
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t += v
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}
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mx := t
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for i := k; i < len(data); i++ {
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t += data[i]
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t -= data[i-k]
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mx = max(mx, t)
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}
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return k - mx
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **...**

‎solution/1100-1199/1151.Minimum Swaps to Group All 1's Together/README_EN.md

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### **Python3**
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```python
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class Solution:
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def minSwaps(self, data: List[int]) -> int:
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k = data.count(1)
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t = sum(data[:k])
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mx = t
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for i in range(k, len(data)):
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t += data[i]
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t -= data[i - k]
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mx = max(mx, t)
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return k - mx
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```
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### **Java**
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```java
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class Solution {
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public int minSwaps(int[] data) {
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int k = 0;
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for (int v : data) {
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k += v;
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}
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int t = 0;
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for (int i = 0; i < k; ++i) {
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t += data[i];
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}
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int mx = t;
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for (int i = k; i < data.length; ++i) {
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t += data[i];
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t -= data[i - k];
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mx = Math.max(mx, t);
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}
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return k - mx;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int minSwaps(vector<int>& data) {
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int k = 0;
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for (int& v : data) {
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k += v;
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}
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int t = 0;
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for (int i = 0; i < k; ++i) {
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t += data[i];
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}
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int mx = t;
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for (int i = k; i < data.size(); ++i) {
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t += data[i];
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t -= data[i - k];
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mx = max(mx, t);
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}
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return k - mx;
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}
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};
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```
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### **Go**
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```go
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func minSwaps(data []int) int {
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k := 0
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for _, v := range data {
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k += v
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}
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t := 0
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for _, v := range data[:k] {
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t += v
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}
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mx := t
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for i := k; i < len(data); i++ {
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t += data[i]
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t -= data[i-k]
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mx = max(mx, t)
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}
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return k - mx
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **...**

‎solution/1100-1199/1151.Minimum Swaps to Group All 1's Together/Solution.cpp

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class Solution {
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public:
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int minSwaps(vector<int>& data) {
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int k = 0;
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for (int& v : data) {
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k += v;
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}
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int t = 0;
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for (int i = 0; i < k; ++i) {
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t += data[i];
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}
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int mx = t;
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for (int i = k; i < data.size(); ++i) {
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t += data[i];
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t -= data[i - k];
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mx = max(mx, t);
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}
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return k - mx;
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}
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};

‎solution/1100-1199/1151.Minimum Swaps to Group All 1's Together/Solution.go

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func minSwaps(data []int) int {
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k := 0
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for _, v := range data {
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k += v
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}
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t := 0
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for _, v := range data[:k] {
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t += v
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}
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mx := t
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for i := k; i < len(data); i++ {
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t += data[i]
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t -= data[i-k]
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mx = max(mx, t)
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}
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return k - mx
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}

‎solution/1100-1199/1151.Minimum Swaps to Group All 1's Together/Solution.java

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class Solution {
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public int minSwaps(int[] data) {
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int k = 0;
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for (int v : data) {
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k += v;
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}
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int t = 0;
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for (int i = 0; i < k; ++i) {
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t += data[i];
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}
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int mx = t;
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for (int i = k; i < data.length; ++i) {
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t += data[i];
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t -= data[i - k];
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mx = Math.max(mx, t);
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}
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return k - mx;
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}
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}

‎solution/1100-1199/1151.Minimum Swaps to Group All 1's Together/Solution.py

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class Solution:
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def minSwaps(self, data: List[int]) -> int:
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k = data.count(1)
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t = sum(data[:k])
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mx = t
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for i in range(k, len(data)):
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t += data[i]
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t -= data[i - k]
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mx = max(mx, t)
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return k - mx

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