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feat: add solutions to lc problem: No.2340
No.2340.Minimum Adjacent Swaps to Make a Valid Array
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‎solution/1200-1299/1271.Hexspeak/README.md

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Original file line numberDiff line numberDiff line change
@@ -45,7 +45,7 @@
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将数字转换为十六进制字符串，然后遍历字符串，将数字 0 转换为字母 O，将数字 1 转换为字母 I，最后判断转换后的字符串是否合法。
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时间复杂度 $O(\log n)$，其中 $n$ 为 `num` 的长度。
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时间复杂度 $O(\log n)$，其中 $n$ 为 `num` 所表示的十进制数字的大小。
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<!-- tabs:start -->
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‎solution/2300-2399/2340.Minimum Adjacent Swaps to Make a Valid Array/README.md

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Original file line numberDiff line numberDiff line change
@@ -55,22 +55,143 @@ It can be shown that 6 swaps is the minimum swaps required to make a valid array
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：贪心 + 分类讨论**
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我们先找出数组中第一个最小值和最后一个最大值的位置，分别记为 $i$ 和 $j$。
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如果 $i$ 和 $j$ 相等，说明数组已经是有效的，直接返回 $0$ 即可。
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否则，我们判断 $i$ 是否在 $j$ 的左边，如果是，那么交换次数为 $i + n - 1 - j$；否则，交换次数为 $i + n - 2 - j$。
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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 minimumSwaps(self, nums: List[int]) -> int:
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mi, mx = min(nums), max(nums)
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i, j = -1, -1
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for k, v in enumerate(nums):
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if v == mi and i == -1:
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i = k
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if v == mx:
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j = k
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if i == j:
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return 0
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n = len(nums)
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if i < j:
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return i + n - 1 - j
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return i + n - 2 - j
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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 minimumSwaps(int[] nums) {
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int n = nums.length;
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int mi = min(nums), mx = max(nums);
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int i = -1, j = -1;
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for (int k = 0; k < n; ++k) {
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if (nums[k] == mi && i == -1) {
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i = k;
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}
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if (nums[k] == mx) {
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j = k;
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}
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}
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if (i == j) {
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return 0;
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}
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return i < j ? i + n - 1 - j : i + n - 2 - j;
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}
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private int max(int[] nums) {
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int v = 0;
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for (int x : nums) {
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v = Math.max(v, x);
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}
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return v;
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}
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private int min(int[] nums) {
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int v = nums[0];
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for (int x : nums) {
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v = Math.min(v, x);
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}
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return v;
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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 minimumSwaps(vector<int>& nums) {
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int n = nums.size();
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int mi = *min_element(nums.begin(), nums.end());
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int mx = *max_element(nums.begin(), nums.end());
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int i = -1, j = -1;
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for (int k = 0; k < n; ++k) {
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if (nums[k] == mi && i == -1) i = k;
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if (nums[k] == mx) j = k;
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}
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if (i == j) return 0;
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return i < j ? i + n - 1 - j : i + n - 2 - j;
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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 minimumSwaps(nums []int) int {
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mi, mx := nums[0], 0
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for _, v := range nums {
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mi = min(mi, v)
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mx = max(mx, v)
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}
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i, j := -1, -1
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for k, v := range nums {
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if v == mi && i == -1 {
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i = k
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}
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if v == mx {
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j = k
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}
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}
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if i == j {
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return 0
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}
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n := len(nums)
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if i < j {
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return i + n - 1 - j
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}
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return i + n - 2 - j
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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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func min(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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### **TypeScript**

‎solution/2300-2399/2340.Minimum Adjacent Swaps to Make a Valid Array/README_EN.md

Lines changed: 112 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -56,13 +56,124 @@ It can be shown that 6 swaps is the minimum swaps required to make a valid array
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### **Python3**
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```python
59-
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class Solution:
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def minimumSwaps(self, nums: List[int]) -> int:
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mi, mx = min(nums), max(nums)
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i, j = -1, -1
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for k, v in enumerate(nums):
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if v == mi and i == -1:
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i = k
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if v == mx:
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j = k
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if i == j:
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return 0
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n = len(nums)
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if i < j:
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return i + n - 1 - j
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return i + n - 2 - j
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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 minimumSwaps(int[] nums) {
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int n = nums.length;
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int mi = min(nums), mx = max(nums);
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int i = -1, j = -1;
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for (int k = 0; k < n; ++k) {
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if (nums[k] == mi && i == -1) {
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i = k;
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}
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if (nums[k] == mx) {
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j = k;
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}
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}
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if (i == j) {
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return 0;
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}
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return i < j ? i + n - 1 - j : i + n - 2 - j;
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}
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private int max(int[] nums) {
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int v = 0;
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for (int x : nums) {
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v = Math.max(v, x);
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}
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return v;
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}
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private int min(int[] nums) {
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int v = nums[0];
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for (int x : nums) {
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v = Math.min(v, x);
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}
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return v;
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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 minimumSwaps(vector<int>& nums) {
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int n = nums.size();
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int mi = *min_element(nums.begin(), nums.end());
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int mx = *max_element(nums.begin(), nums.end());
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int i = -1, j = -1;
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for (int k = 0; k < n; ++k) {
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if (nums[k] == mi && i == -1) i = k;
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if (nums[k] == mx) j = k;
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}
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if (i == j) return 0;
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return i < j ? i + n - 1 - j : i + n - 2 - j;
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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 minimumSwaps(nums []int) int {
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mi, mx := nums[0], 0
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for _, v := range nums {
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mi = min(mi, v)
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mx = max(mx, v)
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}
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i, j := -1, -1
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for k, v := range nums {
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if v == mi && i == -1 {
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i = k
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}
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if v == mx {
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j = k
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}
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}
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if i == j {
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return 0
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}
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n := len(nums)
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if i < j {
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return i + n - 1 - j
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}
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return i + n - 2 - j
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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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func min(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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### **TypeScript**

‎solution/2300-2399/2340.Minimum Adjacent Swaps to Make a Valid Array/Solution.cpp

Lines changed: 15 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,15 @@
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class Solution {
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public:
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int minimumSwaps(vector<int>& nums) {
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int n = nums.size();
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int mi = *min_element(nums.begin(), nums.end());
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int mx = *max_element(nums.begin(), nums.end());
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int i = -1, j = -1;
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for (int k = 0; k < n; ++k) {
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if (nums[k] == mi && i == -1) i = k;
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if (nums[k] == mx) j = k;
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}
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if (i == j) return 0;
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return i < j ? i + n - 1 - j : i + n - 2 - j;
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}
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};

‎solution/2300-2399/2340.Minimum Adjacent Swaps to Make a Valid Array/Solution.go

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,38 @@
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func minimumSwaps(nums []int) int {
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mi, mx := nums[0], 0
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for _, v := range nums {
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mi = min(mi, v)
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mx = max(mx, v)
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}
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i, j := -1, -1
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for k, v := range nums {
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if v == mi && i == -1 {
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i = k
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}
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if v == mx {
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j = k
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}
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}
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if i == j {
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return 0
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}
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n := len(nums)
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if i < j {
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return i + n - 1 - j
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}
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return i + n - 2 - j
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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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func min(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/2300-2399/2340.Minimum Adjacent Swaps to Make a Valid Array/Solution.java

Lines changed: 35 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,35 @@
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class Solution {
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public int minimumSwaps(int[] nums) {
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int n = nums.length;
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int mi = min(nums), mx = max(nums);
5+
int i = -1, j = -1;
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for (int k = 0; k < n; ++k) {
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if (nums[k] == mi && i == -1) {
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i = k;
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}
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if (nums[k] == mx) {
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j = k;
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}
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}
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if (i == j) {
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return 0;
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}
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return i < j ? i + n - 1 - j : i + n - 2 - j;
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}
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private int max(int[] nums) {
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int v = 0;
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for (int x : nums) {
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v = Math.max(v, x);
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}
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return v;
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}
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private int min(int[] nums) {
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int v = nums[0];
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for (int x : nums) {
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v = Math.min(v, x);
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}
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return v;
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}
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}

‎solution/2300-2399/2340.Minimum Adjacent Swaps to Make a Valid Array/Solution.py

Lines changed: 15 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,15 @@
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class Solution:
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def minimumSwaps(self, nums: List[int]) -> int:
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mi, mx = min(nums), max(nums)
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i, j = -1, -1
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for k, v in enumerate(nums):
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if v == mi and i == -1:
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i = k
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if v == mx:
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j = k
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if i == j:
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return 0
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n = len(nums)
13+
if i < j:
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return i + n - 1 - j
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return i + n - 2 - j

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