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Commit 6aef81e

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feat: add solutions to lc problem: No.1536 (#2074)
No.1536.Minimum Swaps to Arrange a Binary Grid
1 parent 60c30be commit 6aef81e

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7 files changed

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7 files changed

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‎solution/1500-1599/1536.Minimum Swaps to Arrange a Binary Grid/README.md‎

Lines changed: 173 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -56,22 +56,194 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:贪心**
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我们逐行处理,对于第 $i$ 行,最后一个 1ドル$ 所在的位置必须小于等于 $i,ドル我们在 $[i, n)$ 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。
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时间复杂度 $O(n^2),ドル空间复杂度 $O(n)$。其中 $n$ 是网格的边长。
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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, grid: List[List[int]]) -> int:
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n = len(grid)
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pos = [-1] * n
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for i in range(n):
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for j in range(n - 1, -1, -1):
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if grid[i][j] == 1:
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pos[i] = j
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break
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ans = 0
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for i in range(n):
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k = -1
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for j in range(i, n):
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if pos[j] <= i:
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ans += j - i
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k = j
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break
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if k == -1:
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return -1
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while k > i:
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pos[k], pos[k - 1] = pos[k - 1], pos[k]
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k -= 1
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return ans
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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[][] grid) {
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int n = grid.length;
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int[] pos = new int[n];
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Arrays.fill(pos, -1);
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for (int i = 0; i < n; ++i) {
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for (int j = n - 1; j >= 0; --j) {
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if (grid[i][j] == 1) {
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pos[i] = j;
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break;
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}
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}
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}
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int ans = 0;
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for (int i = 0; i < n; ++i) {
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int k = -1;
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for (int j = i; j < n; ++j) {
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if (pos[j] <= i) {
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ans += j - i;
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k = j;
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break;
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}
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}
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if (k == -1) {
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return -1;
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}
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for (; k > i; --k) {
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int t = pos[k];
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pos[k] = pos[k - 1];
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pos[k - 1] = t;
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}
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}
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return ans;
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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<vector<int>>& grid) {
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int n = grid.size();
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vector<int> pos(n, -1);
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for (int i = 0; i < n; ++i) {
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for (int j = n - 1; j >= 0; --j) {
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if (grid[i][j] == 1) {
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pos[i] = j;
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break;
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}
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}
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}
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int ans = 0;
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for (int i = 0; i < n; ++i) {
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int k = -1;
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for (int j = i; j < n; ++j) {
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if (pos[j] <= i) {
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ans += j - i;
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k = j;
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break;
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}
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}
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if (k == -1) {
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return -1;
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}
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for (; k > i; --k) {
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swap(pos[k], pos[k - 1]);
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}
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}
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return ans;
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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(grid [][]int) (ans int) {
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n := len(grid)
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pos := make([]int, n)
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for i := range pos {
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pos[i] = -1
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}
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for i := 0; i < n; i++ {
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for j := n - 1; j >= 0; j-- {
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if grid[i][j] == 1 {
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pos[i] = j
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break
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}
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}
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}
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for i := 0; i < n; i++ {
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k := -1
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for j := i; j < n; j++ {
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if pos[j] <= i {
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ans += j - i
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k = j
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break
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}
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}
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if k == -1 {
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return -1
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}
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for ; k > i; k-- {
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pos[k], pos[k-1] = pos[k-1], pos[k]
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}
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}
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return
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}
212+
```
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### **TypeScript**
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```ts
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function minSwaps(grid: number[][]): number {
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const n = grid.length;
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const pos: number[] = Array(n).fill(-1);
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for (let i = 0; i < n; ++i) {
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for (let j = n - 1; ~j; --j) {
222+
if (grid[i][j] === 1) {
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pos[i] = j;
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break;
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}
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}
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}
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let ans = 0;
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for (let i = 0; i < n; ++i) {
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let k = -1;
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for (let j = i; j < n; ++j) {
232+
if (pos[j] <= i) {
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ans += j - i;
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k = j;
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break;
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}
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}
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if (k === -1) {
239+
return -1;
240+
}
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for (; k > i; --k) {
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[pos[k], pos[k - 1]] = [pos[k - 1], pos[k]];
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}
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}
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return ans;
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}
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```
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### **...**

‎solution/1500-1599/1536.Minimum Swaps to Arrange a Binary Grid/README_EN.md‎

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Original file line numberDiff line numberDiff line change
@@ -46,18 +46,190 @@
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## Solutions
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**Solution 1: Greedy**
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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),ドル denoted as $k$. Then, starting from the $k$-th row, we swap the adjacent two rows upwards until the $i$-th row.
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The time complexity is $O(n^2),ドル and the space complexity is $O(n)$. Here, $n$ is the side length of the grid.
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<!-- tabs:start -->
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### **Python3**
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```python
54-
60+
class Solution:
61+
def minSwaps(self, grid: List[List[int]]) -> int:
62+
n = len(grid)
63+
pos = [-1] * n
64+
for i in range(n):
65+
for j in range(n - 1, -1, -1):
66+
if grid[i][j] == 1:
67+
pos[i] = j
68+
break
69+
ans = 0
70+
for i in range(n):
71+
k = -1
72+
for j in range(i, n):
73+
if pos[j] <= i:
74+
ans += j - i
75+
k = j
76+
break
77+
if k == -1:
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return -1
79+
while k > i:
80+
pos[k], pos[k - 1] = pos[k - 1], pos[k]
81+
k -= 1
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return ans
5583
```
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### **Java**
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```java
88+
class Solution {
89+
public int minSwaps(int[][] grid) {
90+
int n = grid.length;
91+
int[] pos = new int[n];
92+
Arrays.fill(pos, -1);
93+
for (int i = 0; i < n; ++i) {
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for (int j = n - 1; j >= 0; --j) {
95+
if (grid[i][j] == 1) {
96+
pos[i] = j;
97+
break;
98+
}
99+
}
100+
}
101+
int ans = 0;
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for (int i = 0; i < n; ++i) {
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int k = -1;
104+
for (int j = i; j < n; ++j) {
105+
if (pos[j] <= i) {
106+
ans += j - i;
107+
k = j;
108+
break;
109+
}
110+
}
111+
if (k == -1) {
112+
return -1;
113+
}
114+
for (; k > i; --k) {
115+
int t = pos[k];
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pos[k] = pos[k - 1];
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pos[k - 1] = t;
118+
}
119+
}
120+
return ans;
121+
}
122+
}
123+
```
124+
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### **C++**
126+
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```cpp
128+
class Solution {
129+
public:
130+
int minSwaps(vector<vector<int>>& grid) {
131+
int n = grid.size();
132+
vector<int> pos(n, -1);
133+
for (int i = 0; i < n; ++i) {
134+
for (int j = n - 1; j >= 0; --j) {
135+
if (grid[i][j] == 1) {
136+
pos[i] = j;
137+
break;
138+
}
139+
}
140+
}
141+
int ans = 0;
142+
for (int i = 0; i < n; ++i) {
143+
int k = -1;
144+
for (int j = i; j < n; ++j) {
145+
if (pos[j] <= i) {
146+
ans += j - i;
147+
k = j;
148+
break;
149+
}
150+
}
151+
if (k == -1) {
152+
return -1;
153+
}
154+
for (; k > i; --k) {
155+
swap(pos[k], pos[k - 1]);
156+
}
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}
158+
return ans;
159+
}
160+
};
161+
```
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### **Go**
164+
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```go
166+
func minSwaps(grid [][]int) (ans int) {
167+
n := len(grid)
168+
pos := make([]int, n)
169+
for i := range pos {
170+
pos[i] = -1
171+
}
172+
for i := 0; i < n; i++ {
173+
for j := n - 1; j >= 0; j-- {
174+
if grid[i][j] == 1 {
175+
pos[i] = j
176+
break
177+
}
178+
}
179+
}
180+
for i := 0; i < n; i++ {
181+
k := -1
182+
for j := i; j < n; j++ {
183+
if pos[j] <= i {
184+
ans += j - i
185+
k = j
186+
break
187+
}
188+
}
189+
if k == -1 {
190+
return -1
191+
}
192+
for ; k > i; k-- {
193+
pos[k], pos[k-1] = pos[k-1], pos[k]
194+
}
195+
}
196+
return
197+
}
198+
```
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200+
### **TypeScript**
201+
202+
```ts
203+
function minSwaps(grid: number[][]): number {
204+
const n = grid.length;
205+
const pos: number[] = Array(n).fill(-1);
206+
for (let i = 0; i < n; ++i) {
207+
for (let j = n - 1; ~j; --j) {
208+
if (grid[i][j] === 1) {
209+
pos[i] = j;
210+
break;
211+
}
212+
}
213+
}
214+
let ans = 0;
215+
for (let i = 0; i < n; ++i) {
216+
let k = -1;
217+
for (let j = i; j < n; ++j) {
218+
if (pos[j] <= i) {
219+
ans += j - i;
220+
k = j;
221+
break;
222+
}
223+
}
224+
if (k === -1) {
225+
return -1;
226+
}
227+
for (; k > i; --k) {
228+
[pos[k], pos[k - 1]] = [pos[k - 1], pos[k]];
229+
}
230+
}
231+
return ans;
232+
}
61233
```
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### **...**

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