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Commit e761584

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feat: update solutions to lc problems: No.2340~2347 (doocs#3838)
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‎.github/workflows/pr-add-label.yml‎

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This file was deleted.

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

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@@ -71,15 +71,15 @@ tags:
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### 方法一:维护最值下标 + 分类讨论
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我们可以用下标 $i$ 和 $j$ 分别记录数组 `nums` 第一个最小值和最后一个最大值的下标,遍历数组 `nums`,更新 $i$ 和 $j$ 的值。
74+
我们可以用下标 $i$ 和 $j$ 分别记录数组 $\textit{nums}$ 第一个最小值和最后一个最大值的下标,遍历数组 $\textit{nums}$,更新 $i$ 和 $j$ 的值。
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接下来,我们需要考虑交换的次数。
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78-
- 如果 $i = j,ドル说明数组 `nums` 已经是有效数组,不需要交换,返回 0ドル$;
79-
- 如果 $i < j,ドル说明数组 `nums` 中最小值在最大值的左边,需要交换 $i + n - 1 - j$ 次,其中 $n$ 为数组 `nums` 的长度;
80-
- 如果 $i > j,ドル说明数组 `nums` 中最小值在最大值的右边,需要交换 $i + n - 1 - j - 1$ 次。
78+
- 如果 $i = j,ドル说明数组 $\textit{nums}$ 已经是有效数组,不需要交换,返回 0ドル$;
79+
- 如果 $i < j,ドル说明数组 $\textit{nums}$ 中最小值在最大值的左边,需要交换 $i + n - 1 - j$ 次,其中 $n$ 为数组 $\textit{nums}$ 的长度;
80+
- 如果 $i > j,ドル说明数组 $\textit{nums}$ 中最小值在最大值的右边,需要交换 $i + n - 1 - j - 1$ 次。
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时间复杂度 $O(n),ドル空间复杂度 $O(1)$。其中 $n$ 为数组 `nums` 的长度。
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时间复杂度 $O(n),ドル其中 $n$ 为数组 $\textit{nums}$ 的长度。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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‎solution/2300-2399/2340.Minimum Adjacent Swaps to Make a Valid Array/README_EN.md‎

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@@ -68,7 +68,17 @@ It can be shown that 6 swaps is the minimum swaps required to make a valid array
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<!-- solution:start -->
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### Solution 1
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### Solution 1: Maintain Index of Extremes + Case Analysis
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We can use indices $i$ and $j$ to record the index of the first minimum value and the last maximum value in the array $\textit{nums},ドル respectively. Traverse the array $\textit{nums}$ to update the values of $i$ and $j$.
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Next, we need to consider the number of swaps.
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- If $i = j,ドル it means the array $\textit{nums}$ is already a valid array, and no swaps are needed. Return 0ドル$.
78+
- If $i < j,ドル it means the minimum value in the array $\textit{nums}$ is to the left of the maximum value. The number of swaps needed is $i + n - 1 - j,ドル where $n$ is the length of the array $\textit{nums}$.
79+
- If $i > j,ドル it means the minimum value in the array $\textit{nums}$ is to the right of the maximum value. The number of swaps needed is $i + n - 1 - j - 1$.
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The time complexity is $O(n),ドル where $n$ is the length of the array $\textit{nums}$. The space complexity is $O(1)$.
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<!-- tabs:start -->
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‎solution/2300-2399/2341.Maximum Number of Pairs in Array/README.md‎

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@@ -74,15 +74,15 @@ nums[0] 和 nums[1] 形成一个数对,并从 nums 中移除,nums = [2] 。
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### 方法一:计数
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我们可以统计数组 `nums` 中每个数字 $x$ 出现的次数,记录在哈希表或数组 `cnt` 中。
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我们可以统计数组 $\textit{nums}$ 中每个数字 $x$ 出现的次数,记录在哈希表或数组 $\textit{cnt}$ 中。
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然后遍历 `cnt`,对于每个数字 $x,ドル如果 $x$ 出现的次数 $v$ 大于 1ドル,ドル则可以从数组中选出两个 $x$ 形成一个数对,我们将 $v$ 除以 2ドル$ 向下取整,即可得到当前数字 $x$ 可以形成的数对数目,然后我们累加这个数目到变量 $s$ 中。
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然后遍历 $\textit{cnt}$,对于每个数字 $x,ドル如果 $x$ 出现的次数 $v$ 大于 1ドル,ドル则可以从数组中选出两个 $x$ 形成一个数对,我们将 $v$ 除以 2ドル$ 向下取整,即可得到当前数字 $x$ 可以形成的数对数目,然后我们累加这个数目到变量 $s$ 中。
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最后剩余的个数为数组 `nums` 的长度减去可以形成的数对数目乘以 2ドル,ドル即 $n - s \times 2$。
81+
最后剩余的个数为数组 $\textit{nums}$ 的长度减去可以形成的数对数目乘以 2ドル,ドル即 $n - s \times 2$。
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答案为 $[s, n - s \times 2]$。
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时间复杂度 $O(n),ドル空间复杂度 $O(C)$。其中 $n$ 为数组 `nums` 的长度;而 $C$ 为数组 `nums` 中数字的范围,本题中 $C = 101$。
85+
时间复杂度 $O(n),ドル空间复杂度 $O(C)$。其中 $n$ 为数组 $\textit{nums}$ 的长度;而 $C$ 为数组 $\textit{nums}$ 中数字的范围,本题中 $C = 101$。
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<!-- tabs:start -->
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‎solution/2300-2399/2341.Maximum Number of Pairs in Array/README_EN.md‎

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@@ -75,7 +75,17 @@ No more pairs can be formed. A total of 1 pair has been formed, and there are 0
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<!-- solution:start -->
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### Solution 1
78+
### Solution 1: Counting
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We can count the occurrences of each number $x$ in the array $\textit{nums}$ and record them in a hash table or array $\textit{cnt}$.
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Then, we traverse $\textit{cnt}$. For each number $x,ドル if the occurrence count $v$ of $x$ is greater than 1ドル,ドル we can select two $x$'s from the array to form a pair. We divide $v$ by 2ドル$ and take the floor value to get the number of pairs that can be formed by the current number $x$. We then add this number to the variable $s$.
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The remaining count is the length of the array $\textit{nums}$ minus the number of pairs formed multiplied by 2ドル,ドル i.e., $n - s \times 2$.
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The answer is $[s, n - s \times 2]$.
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The time complexity is $O(n),ドル and the space complexity is $O(C)$. Here, $n$ is the length of the array $\textit{nums},ドル and $C$ is the range of numbers in the array $\textit{nums},ドル which is 101ドル$ in this problem.
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<!-- tabs:start -->
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‎solution/2300-2399/2343.Query Kth Smallest Trimmed Number/README.md‎

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根据题意,我们可以模拟裁剪过程,然后对裁剪后的字符串进行排序,最后根据下标找到对应的数字即可。
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时间复杂度 $O(m \times \ n \times \log n \times s),ドル空间复杂度 $O(n)$。其中 $m$ 和 $n$ 分别为 `nums` 和 `queries` 的长度,而 $s$ 为 $nums[i]$ 字符串的长度。
101+
时间复杂度 $O(m \times \ n \times \log n \times s),ドル空间复杂度 $O(n)$。其中 $m$ 和 $n$ 分别为 $\textit{nums}$ 和 $\textit{queries}$ 的长度,而 $s$ 为 $\textit{nums}[i]$ 字符串的长度。
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<!-- tabs:start -->
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‎solution/2300-2399/2343.Query Kth Smallest Trimmed Number/README_EN.md‎

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Simulation
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According to the problem description, we can simulate the cropping process, then sort the cropped strings, and finally find the corresponding number based on the index.
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The time complexity is $O(m \times n \times \log n \times s),ドル and the space complexity is $O(n)$. Here, $m$ and $n$ are the lengths of $\textit{nums}$ and $\textit{queries}$ respectively, and $s$ is the length of the string $\textit{nums}[i]$.
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<!-- tabs:start -->
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‎solution/2300-2399/2345.Finding the Number of Visible Mountains/README.md‎

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@@ -100,22 +100,20 @@ class Solution {
100100
public int visibleMountains(int[][] peaks) {
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int n = peaks.length;
102102
int[][] arr = new int[n][2];
103-
Map<String, Integer> cnt = new HashMap<>();
104103
for (int i = 0; i < n; ++i) {
105104
int x = peaks[i][0], y = peaks[i][1];
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arr[i] = new int[] {x - y, x + y};
107-
cnt.merge((x - y) + "" + (x + y), 1, Integer::sum);
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}
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Arrays.sort(arr, (a, b) -> a[0] == b[0] ? b[1] - a[1] : a[0] - b[0]);
110108
int ans = 0;
111109
int cur = Integer.MIN_VALUE;
112-
for (int[] e : arr) {
113-
int l = e[0], r = e[1];
110+
for (int i =0; i < n; ++i) {
111+
int l = arr[i][0], r = arr[i][1];
114112
if (r <= cur) {
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continue;
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}
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cur = r;
118-
if (cnt.get(l +""+ r) == 1) {
116+
if (!(i < n -1&& arr[i][0] == arr[i +1][0] && arr[i][1] == arr[i +1][1])) {
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++ans;
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}
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}
@@ -182,43 +180,4 @@ func visibleMountains(peaks [][]int) (ans int) {
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<!-- solution:end -->
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<!-- solution:start -->
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### 方法二
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<!-- tabs:start -->
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#### Java
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```java
194-
class Solution {
195-
public int visibleMountains(int[][] peaks) {
196-
int n = peaks.length;
197-
int[][] arr = new int[n][2];
198-
for (int i = 0; i < n; ++i) {
199-
int x = peaks[i][0], y = peaks[i][1];
200-
arr[i] = new int[] {x - y, x + y};
201-
}
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Arrays.sort(arr, (a, b) -> a[0] == b[0] ? b[1] - a[1] : a[0] - b[0]);
203-
int ans = 0;
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int cur = Integer.MIN_VALUE;
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for (int i = 0; i < n; ++i) {
206-
int l = arr[i][0], r = arr[i][1];
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if (r <= cur) {
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continue;
209-
}
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cur = r;
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if (!(i < n - 1 && arr[i][0] == arr[i + 1][0] && arr[i][1] == arr[i + 1][1])) {
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++ans;
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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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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

‎solution/2300-2399/2345.Finding the Number of Visible Mountains/README_EN.md‎

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Interval Sorting + Traversal
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We first convert each mountain $(x, y)$ into a horizontal interval $(x - y, x + y),ドル then sort the intervals by left endpoint in ascending order and right endpoint in descending order.
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Next, we initialize the right endpoint of the current interval as $-\infty$. We traverse each mountain. If the right endpoint of the current mountain is less than or equal to the right endpoint of the current interval, we skip this mountain. Otherwise, we update the right endpoint of the current interval to the right endpoint of the current mountain. If the interval of the current mountain appears only once, we increment the answer.
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Finally, we return the answer.
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The time complexity is $O(n \times \log n),ドル and the space complexity is $O(n)$. Here, $n$ is the number of mountains.
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<!-- tabs:start -->
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public int visibleMountains(int[][] peaks) {
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int n = peaks.length;
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int[][] arr = new int[n][2];
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Map<String, Integer> cnt = new HashMap<>();
94101
for (int i = 0; i < n; ++i) {
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int x = peaks[i][0], y = peaks[i][1];
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arr[i] = new int[] {x - y, x + y};
97-
cnt.merge((x - y) + "" + (x + y), 1, Integer::sum);
98104
}
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Arrays.sort(arr, (a, b) -> a[0] == b[0] ? b[1] - a[1] : a[0] - b[0]);
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int ans = 0;
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int cur = Integer.MIN_VALUE;
102-
for (int[] e : arr) {
103-
int l = e[0], r = e[1];
108+
for (int i =0; i < n; ++i) {
109+
int l = arr[i][0], r = arr[i][1];
104110
if (r <= cur) {
105111
continue;
106112
}
107113
cur = r;
108-
if (cnt.get(l +""+ r) == 1) {
114+
if (!(i < n -1&& arr[i][0] == arr[i +1][0] && arr[i][1] == arr[i +1][1])) {
109115
++ans;
110116
}
111117
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@@ -172,43 +178,4 @@ func visibleMountains(peaks [][]int) (ans int) {
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<!-- solution:end -->
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<!-- solution:start -->
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### Solution 2
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<!-- tabs:start -->
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#### Java
182-
183-
```java
184-
class Solution {
185-
public int visibleMountains(int[][] peaks) {
186-
int n = peaks.length;
187-
int[][] arr = new int[n][2];
188-
for (int i = 0; i < n; ++i) {
189-
int x = peaks[i][0], y = peaks[i][1];
190-
arr[i] = new int[] {x - y, x + y};
191-
}
192-
Arrays.sort(arr, (a, b) -> a[0] == b[0] ? b[1] - a[1] : a[0] - b[0]);
193-
int ans = 0;
194-
int cur = Integer.MIN_VALUE;
195-
for (int i = 0; i < n; ++i) {
196-
int l = arr[i][0], r = arr[i][1];
197-
if (r <= cur) {
198-
continue;
199-
}
200-
cur = r;
201-
if (!(i < n - 1 && arr[i][0] == arr[i + 1][0] && arr[i][1] == arr[i + 1][1])) {
202-
++ans;
203-
}
204-
}
205-
return ans;
206-
}
207-
}
208-
```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

‎solution/2300-2399/2345.Finding the Number of Visible Mountains/Solution.java‎

Lines changed: 4 additions & 6 deletions
Original file line numberDiff line numberDiff line change
@@ -2,25 +2,23 @@ class Solution {
22
public int visibleMountains(int[][] peaks) {
33
int n = peaks.length;
44
int[][] arr = new int[n][2];
5-
Map<String, Integer> cnt = new HashMap<>();
65
for (int i = 0; i < n; ++i) {
76
int x = peaks[i][0], y = peaks[i][1];
87
arr[i] = new int[] {x - y, x + y};
9-
cnt.merge((x - y) + "" + (x + y), 1, Integer::sum);
108
}
119
Arrays.sort(arr, (a, b) -> a[0] == b[0] ? b[1] - a[1] : a[0] - b[0]);
1210
int ans = 0;
1311
int cur = Integer.MIN_VALUE;
14-
for (int[] e : arr) {
15-
int l = e[0], r = e[1];
12+
for (inti = 0; i < n; ++i) {
13+
int l = arr[i][0], r = arr[i][1];
1614
if (r <= cur) {
1715
continue;
1816
}
1917
cur = r;
20-
if (cnt.get(l + "" + r) == 1) {
18+
if (!(i < n - 1 && arr[i][0] == arr[i + 1][0] && arr[i][1] == arr[i + 1][1])) {
2119
++ans;
2220
}
2321
}
2422
return ans;
2523
}
26-
}
24+
}

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