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Dec 9, 2023
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feat: add solutions to lc problem: No.1536 #2074

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174 changes: 173 additions & 1 deletion solution/1500-1599/1536.Minimum Swaps to Arrange a Binary Grid/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -56,22 +56,194 @@

<!-- 这里可写通用的实现逻辑 -->

**方法一：贪心**

我们逐行处理，对于第 $i$ 行，最后一个 $1$ 所在的位置必须小于等于 $i$，我们在 $[i, n)$ 中找到第一个满足条件的行，记为 $k$。然后从第 $k$ 行开始，依次向上交换相邻的两行，直到第 $i$ 行。

时间复杂度 $O(n^2)$，空间复杂度 $O(n)$。其中 $n$ 是网格的边长。

<!-- tabs:start -->

### **Python3**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```python

class Solution:
def minSwaps(self, grid: List[List[int]]) -> int:
n = len(grid)
pos = [-1] * n
for i in range(n):
for j in range(n - 1, -1, -1):
if grid[i][j] == 1:
pos[i] = j
break
ans = 0
for i in range(n):
k = -1
for j in range(i, n):
if pos[j] <= i:
ans += j - i
k = j
break
if k == -1:
return -1
while k > i:
pos[k], pos[k - 1] = pos[k - 1], pos[k]
k -= 1
return ans
```

### **Java**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```java
class Solution {
public int minSwaps(int[][] grid) {
int n = grid.length;
int[] pos = new int[n];
Arrays.fill(pos, -1);
for (int i = 0; i < n; ++i) {
for (int j = n - 1; j >= 0; --j) {
if (grid[i][j] == 1) {
pos[i] = j;
break;
}
}
}
int ans = 0;
for (int i = 0; i < n; ++i) {
int k = -1;
for (int j = i; j < n; ++j) {
if (pos[j] <= i) {
ans += j - i;
k = j;
break;
}
}
if (k == -1) {
return -1;
}
for (; k > i; --k) {
int t = pos[k];
pos[k] = pos[k - 1];
pos[k - 1] = t;
}
}
return ans;
}
}
```

### **C++**

```cpp
class Solution {
public:
int minSwaps(vector<vector<int>>& grid) {
int n = grid.size();
vector<int> pos(n, -1);
for (int i = 0; i < n; ++i) {
for (int j = n - 1; j >= 0; --j) {
if (grid[i][j] == 1) {
pos[i] = j;
break;
}
}
}
int ans = 0;
for (int i = 0; i < n; ++i) {
int k = -1;
for (int j = i; j < n; ++j) {
if (pos[j] <= i) {
ans += j - i;
k = j;
break;
}
}
if (k == -1) {
return -1;
}
for (; k > i; --k) {
swap(pos[k], pos[k - 1]);
}
}
return ans;
}
};
```

### **Go**

```go
func minSwaps(grid [][]int) (ans int) {
n := len(grid)
pos := make([]int, n)
for i := range pos {
pos[i] = -1
}
for i := 0; i < n; i++ {
for j := n - 1; j >= 0; j-- {
if grid[i][j] == 1 {
pos[i] = j
break
}
}
}
for i := 0; i < n; i++ {
k := -1
for j := i; j < n; j++ {
if pos[j] <= i {
ans += j - i
k = j
break
}
}
if k == -1 {
return -1
}
for ; k > i; k-- {
pos[k], pos[k-1] = pos[k-1], pos[k]
}
}
return
}
```

### **TypeScript**

```ts
function minSwaps(grid: number[][]): number {
const n = grid.length;
const pos: number[] = Array(n).fill(-1);
for (let i = 0; i < n; ++i) {
for (let j = n - 1; ~j; --j) {
if (grid[i][j] === 1) {
pos[i] = j;
break;
}
}
}
let ans = 0;
for (let i = 0; i < n; ++i) {
let k = -1;
for (let j = i; j < n; ++j) {
if (pos[j] <= i) {
ans += j - i;
k = j;
break;
}
}
if (k === -1) {
return -1;
}
for (; k > i; --k) {
[pos[k], pos[k - 1]] = [pos[k - 1], pos[k]];
}
}
return ans;
}
```

### **...**
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174 changes: 173 additions & 1 deletion solution/1500-1599/1536.Minimum Swaps to Arrange a Binary Grid/README_EN.md
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Expand Up @@ -46,18 +46,190 @@

## Solutions

**Solution 1: Greedy**

We process row by row. For the $i$-th row, the position of the last '1' must be less than or equal to $i$. We find the first row that meets the condition in $[i, n)$, denoted as $k$. Then, starting from the $k$-th row, we swap the adjacent two rows upwards until the $i$-th row.

The time complexity is $O(n^2)$, and the space complexity is $O(n)$. Here, $n$ is the side length of the grid.

<!-- tabs:start -->

### **Python3**

```python

class Solution:
def minSwaps(self, grid: List[List[int]]) -> int:
n = len(grid)
pos = [-1] * n
for i in range(n):
for j in range(n - 1, -1, -1):
if grid[i][j] == 1:
pos[i] = j
break
ans = 0
for i in range(n):
k = -1
for j in range(i, n):
if pos[j] <= i:
ans += j - i
k = j
break
if k == -1:
return -1
while k > i:
pos[k], pos[k - 1] = pos[k - 1], pos[k]
k -= 1
return ans
```

### **Java**

```java
class Solution {
public int minSwaps(int[][] grid) {
int n = grid.length;
int[] pos = new int[n];
Arrays.fill(pos, -1);
for (int i = 0; i < n; ++i) {
for (int j = n - 1; j >= 0; --j) {
if (grid[i][j] == 1) {
pos[i] = j;
break;
}
}
}
int ans = 0;
for (int i = 0; i < n; ++i) {
int k = -1;
for (int j = i; j < n; ++j) {
if (pos[j] <= i) {
ans += j - i;
k = j;
break;
}
}
if (k == -1) {
return -1;
}
for (; k > i; --k) {
int t = pos[k];
pos[k] = pos[k - 1];
pos[k - 1] = t;
}
}
return ans;
}
}
```

### **C++**

```cpp
class Solution {
public:
int minSwaps(vector<vector<int>>& grid) {
int n = grid.size();
vector<int> pos(n, -1);
for (int i = 0; i < n; ++i) {
for (int j = n - 1; j >= 0; --j) {
if (grid[i][j] == 1) {
pos[i] = j;
break;
}
}
}
int ans = 0;
for (int i = 0; i < n; ++i) {
int k = -1;
for (int j = i; j < n; ++j) {
if (pos[j] <= i) {
ans += j - i;
k = j;
break;
}
}
if (k == -1) {
return -1;
}
for (; k > i; --k) {
swap(pos[k], pos[k - 1]);
}
}
return ans;
}
};
```

### **Go**

```go
func minSwaps(grid [][]int) (ans int) {
n := len(grid)
pos := make([]int, n)
for i := range pos {
pos[i] = -1
}
for i := 0; i < n; i++ {
for j := n - 1; j >= 0; j-- {
if grid[i][j] == 1 {
pos[i] = j
break
}
}
}
for i := 0; i < n; i++ {
k := -1
for j := i; j < n; j++ {
if pos[j] <= i {
ans += j - i
k = j
break
}
}
if k == -1 {
return -1
}
for ; k > i; k-- {
pos[k], pos[k-1] = pos[k-1], pos[k]
}
}
return
}
```

### **TypeScript**

```ts
function minSwaps(grid: number[][]): number {
const n = grid.length;
const pos: number[] = Array(n).fill(-1);
for (let i = 0; i < n; ++i) {
for (let j = n - 1; ~j; --j) {
if (grid[i][j] === 1) {
pos[i] = j;
break;
}
}
}
let ans = 0;
for (let i = 0; i < n; ++i) {
let k = -1;
for (let j = i; j < n; ++j) {
if (pos[j] <= i) {
ans += j - i;
k = j;
break;
}
}
if (k === -1) {
return -1;
}
for (; k > i; --k) {
[pos[k], pos[k - 1]] = [pos[k - 1], pos[k]];
}
}
return ans;
}
```

### **...**
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