feat: update solutions to lc problmes

solution/2700-2799/2733.Neither Minimum nor Maximum/README.md

@@ -47,14 +47,19 @@
 
 ## 解法
 
-### 方法一
+### 方法一：模拟
+
+我们可以先找到数组中的最小值和最大值，分别记为 $mi$ 和 $mx$。然后遍历数组，找到第一个不等于 $mi$ 且不等于 $mx$ 的数字，返回即可。
+
+时间复杂度 $O(n)$，其中 $n$ 是数组的长度。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
 ```python
 class Solution:
     def findNonMinOrMax(self, nums: List[int]) -> int:
-        return -1 if len(nums) < 3 else sorted(nums)[1]
+        mi, mx = min(nums), max(nums)
+        return next((x for x in nums if x != mi and x != mx), -1)
 ```
 
 ```java
@@ -79,10 +84,9 @@ class Solution {
 class Solution {
 public:
     int findNonMinOrMax(vector<int>& nums) {
-        int mi = *min_element(nums.begin(), nums.end());
-        int mx = *max_element(nums.begin(), nums.end());
+        auto [mi, mx] = minmax_element(nums.begin(), nums.end());
         for (int x : nums) {
-            if (x != mi && x != mx) {
+            if (x != *mi && x != *mx) {
                 return x;
             }
         }

solution/2700-2799/2733.Neither Minimum nor Maximum/README_EN.md

@@ -46,14 +46,19 @@
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Simulation
+
+First, we find the minimum and maximum values in the array, denoted as $mi$ and $mx$ respectively. Then, we traverse the array and find the first number that is not equal to $mi$ and not equal to $mx$, and return it.
+
+The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.
 
 <!-- tabs:start -->
 
 ```python
 class Solution:
     def findNonMinOrMax(self, nums: List[int]) -> int:
-        return -1 if len(nums) < 3 else sorted(nums)[1]
+        mi, mx = min(nums), max(nums)
+        return next((x for x in nums if x != mi and x != mx), -1)
 ```
 
 ```java
@@ -78,10 +83,9 @@ class Solution {
 class Solution {
 public:
     int findNonMinOrMax(vector<int>& nums) {
-        int mi = *min_element(nums.begin(), nums.end());
-        int mx = *max_element(nums.begin(), nums.end());
+        auto [mi, mx] = minmax_element(nums.begin(), nums.end());
         for (int x : nums) {
-            if (x != mi && x != mx) {
+            if (x != *mi && x != *mx) {
                 return x;
             }
         }

solution/2700-2799/2733.Neither Minimum nor Maximum/Solution.cpp

@@ -1,10 +1,9 @@
 class Solution {
 public:
     int findNonMinOrMax(vector<int>& nums) {
-        int mi = *min_element(nums.begin(), nums.end());
-        int mx = *max_element(nums.begin(), nums.end());
+        auto [mi, mx] = minmax_element(nums.begin(), nums.end());
         for (int x : nums) {
-            if (x != mi && x != mx) {
+            if (x != *mi && x != *mx) {
                 return x;
             }
         }

solution/2700-2799/2733.Neither Minimum nor Maximum/Solution.py

@@ -1,3 +1,4 @@
 class Solution:
     def findNonMinOrMax(self, nums: List[int]) -> int:
-        return -1 if len(nums) < 3 else sorted(nums)[1]
+        mi, mx = min(nums), max(nums)
+        return next((x for x in nums if x != mi and x != mx), -1)

solution/2700-2799/2736.Maximum Sum Queries/README.md

@@ -186,6 +186,51 @@ class Solution {
 }
 ```
 
+```java
+class Solution {
+    public int[] maximumSumQueries(int[] nums1, int[] nums2, int[][] q) {
+        int n = nums1.length, m = q.length;
+        int[][] a = new int[n][2];
+        for (int i = 0; i < n; i++) {
+            a[i][0] = nums1[i];
+            a[i][1] = nums2[i];
+        }
+        int[][] b = new int[m][3];
+        for (int i = 0; i < m; i++) {
+            b[i][0] = q[i][0];
+            b[i][1] = q[i][1];
+            b[i][2] = i;
+        }
+        Arrays.sort(a, (o1, o2) -> o1[0] - o2[0]);
+        Arrays.sort(b, (o1, o2) -> o1[0] - o2[0]);
+        TreeMap<Integer, Integer> map = new TreeMap<>();
+        int[] res = new int[m];
+        int max = -1;
+        for (int i = m - 1, j = n - 1; i >= 0; i--) {
+            int x = b[i][0], y = b[i][1], idx = b[i][2];
+            while (j >= 0 && a[j][0] >= x) {
+                if (max < a[j][1]) {
+                    max = a[j][1];
+                    Integer key = map.floorKey(a[j][1]);
+                    while (key != null && map.get(key) <= a[j][0] + a[j][1]) {
+                        map.remove(key);
+                        key = map.floorKey(key);
+                    }
+                    map.put(max, a[j][0] + a[j][1]);
+                }
+                j--;
+            }
+            Integer key = map.ceilingKey(y);
+            if (key == null)
+                res[idx] = -1;
+            else
+                res[idx] = map.get(key);
+        }
+        return res;
+    }
+}
+```
+
 ```cpp
 class BinaryIndexedTree {
 private:
@@ -374,55 +419,4 @@ function maximumSumQueries(nums1: number[], nums2: number[], queries: number[][]
 
 <!-- tabs:end -->
 
-### 方法二
-
-<!-- tabs:start -->
-
-```java
-class Solution {
-    public int[] maximumSumQueries(int[] nums1, int[] nums2, int[][] q) {
-        int n = nums1.length, m = q.length;
-        int[][] a = new int[n][2];
-        for (int i = 0; i < n; i++) {
-            a[i][0] = nums1[i];
-            a[i][1] = nums2[i];
-        }
-        int[][] b = new int[m][3];
-        for (int i = 0; i < m; i++) {
-            b[i][0] = q[i][0];
-            b[i][1] = q[i][1];
-            b[i][2] = i;
-        }
-        Arrays.sort(a, (o1, o2) -> o1[0] - o2[0]);
-        Arrays.sort(b, (o1, o2) -> o1[0] - o2[0]);
-        TreeMap<Integer, Integer> map = new TreeMap<>();
-        int[] res = new int[m];
-        int max = -1;
-        for (int i = m - 1, j = n - 1; i >= 0; i--) {
-            int x = b[i][0], y = b[i][1], idx = b[i][2];
-            while (j >= 0 && a[j][0] >= x) {
-                if (max < a[j][1]) {
-                    max = a[j][1];
-                    Integer key = map.floorKey(a[j][1]);
-                    while (key != null && map.get(key) <= a[j][0] + a[j][1]) {
-                        map.remove(key);
-                        key = map.floorKey(key);
-                    }
-                    map.put(max, a[j][0] + a[j][1]);
-                }
-                j--;
-            }
-            Integer key = map.ceilingKey(y);
-            if (key == null)
-                res[idx] = -1;
-            else
-                res[idx] = map.get(key);
-        }
-        return res;
-    }
-}
-```
-
-<!-- tabs:end -->
-
 <!-- end -->

solution/2700-2799/2736.Maximum Sum Queries/README_EN.md

@@ -188,6 +188,51 @@ class Solution {
 }
 ```
 
+```java
+class Solution {
+    public int[] maximumSumQueries(int[] nums1, int[] nums2, int[][] q) {
+        int n = nums1.length, m = q.length;
+        int[][] a = new int[n][2];
+        for (int i = 0; i < n; i++) {
+            a[i][0] = nums1[i];
+            a[i][1] = nums2[i];
+        }
+        int[][] b = new int[m][3];
+        for (int i = 0; i < m; i++) {
+            b[i][0] = q[i][0];
+            b[i][1] = q[i][1];
+            b[i][2] = i;
+        }
+        Arrays.sort(a, (o1, o2) -> o1[0] - o2[0]);
+        Arrays.sort(b, (o1, o2) -> o1[0] - o2[0]);
+        TreeMap<Integer, Integer> map = new TreeMap<>();
+        int[] res = new int[m];
+        int max = -1;
+        for (int i = m - 1, j = n - 1; i >= 0; i--) {
+            int x = b[i][0], y = b[i][1], idx = b[i][2];
+            while (j >= 0 && a[j][0] >= x) {
+                if (max < a[j][1]) {
+                    max = a[j][1];
+                    Integer key = map.floorKey(a[j][1]);
+                    while (key != null && map.get(key) <= a[j][0] + a[j][1]) {
+                        map.remove(key);
+                        key = map.floorKey(key);
+                    }
+                    map.put(max, a[j][0] + a[j][1]);
+                }
+                j--;
+            }
+            Integer key = map.ceilingKey(y);
+            if (key == null)
+                res[idx] = -1;
+            else
+                res[idx] = map.get(key);
+        }
+        return res;
+    }
+}
+```
+
 ```cpp
 class BinaryIndexedTree {
 private:
@@ -376,55 +421,4 @@ function maximumSumQueries(nums1: number[], nums2: number[], queries: number[][]
 
 <!-- tabs:end -->
 
-### Solution 2
-
-<!-- tabs:start -->
-
-```java
-class Solution {
-    public int[] maximumSumQueries(int[] nums1, int[] nums2, int[][] q) {
-        int n = nums1.length, m = q.length;
-        int[][] a = new int[n][2];
-        for (int i = 0; i < n; i++) {
-            a[i][0] = nums1[i];
-            a[i][1] = nums2[i];
-        }
-        int[][] b = new int[m][3];
-        for (int i = 0; i < m; i++) {
-            b[i][0] = q[i][0];
-            b[i][1] = q[i][1];
-            b[i][2] = i;
-        }
-        Arrays.sort(a, (o1, o2) -> o1[0] - o2[0]);
-        Arrays.sort(b, (o1, o2) -> o1[0] - o2[0]);
-        TreeMap<Integer, Integer> map = new TreeMap<>();
-        int[] res = new int[m];
-        int max = -1;
-        for (int i = m - 1, j = n - 1; i >= 0; i--) {
-            int x = b[i][0], y = b[i][1], idx = b[i][2];
-            while (j >= 0 && a[j][0] >= x) {
-                if (max < a[j][1]) {
-                    max = a[j][1];
-                    Integer key = map.floorKey(a[j][1]);
-                    while (key != null && map.get(key) <= a[j][0] + a[j][1]) {
-                        map.remove(key);
-                        key = map.floorKey(key);
-                    }
-                    map.put(max, a[j][0] + a[j][1]);
-                }
-                j--;
-            }
-            Integer key = map.ceilingKey(y);
-            if (key == null)
-                res[idx] = -1;
-            else
-                res[idx] = map.get(key);
-        }
-        return res;
-    }
-}
-```
-
-<!-- tabs:end -->
-
 <!-- end -->

solution/2700-2799/2737.Find the Closest Marked Node/README_EN.md

@@ -66,7 +66,15 @@ So the answer is 3.
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Dijkstra's Algorithm
+
+First, we construct an adjacency matrix $g$ based on the edge information provided in the problem, where $g[i][j]$ represents the distance from node $i$ to node $j$. If such an edge does not exist, then $g[i][j]$ is positive infinity.
+
+Then, we can use Dijkstra's algorithm to find the shortest distance from the starting point $s$ to all nodes, denoted as $dist$.
+
+Finally, we traverse all the marked nodes and find the marked node with the smallest distance. If the distance is positive infinity, we return $-1$.
+
+The time complexity is $O(n^2)$, and the space complexity is $O(n^2)$. Here, $n$ is the number of nodes.
 
 <!-- tabs:start -->
 

solution/2700-2799/2739.Total Distance Traveled/README_EN.md

@@ -46,7 +46,11 @@ Total distance traveled is 10km.
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Simulation
+
+We can simulate the process of the truck's movement. Each time, it consumes 1 liter of fuel from the main fuel tank and travels 10 kilometers. Whenever the fuel in the main fuel tank is consumed by 5 liters, if there is fuel in the auxiliary fuel tank, 1 liter of fuel is transferred from the auxiliary fuel tank to the main fuel tank. The simulation continues until the fuel in the main fuel tank is exhausted.
+
+The time complexity is $O(n + m)$, where $n$ and $m$ are the amounts of fuel in the main and auxiliary fuel tanks, respectively. The space complexity is $O(1)$.
 
 <!-- tabs:start -->
 

solution/2700-2799/2740.Find the Value of the Partition/README_EN.md

@@ -57,7 +57,11 @@ It can be proven that 9 is the minimum value out of all partitions.
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Sorting
+
+The problem requires us to minimize the partition value. Therefore, we can sort the array and then take the minimum difference between two adjacent numbers.
+
+The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the length of the array.
 
 <!-- tabs:start -->
 

solution/2700-2799/2741.Special Permutations/README_EN.md

@@ -41,7 +41,25 @@
 
 ## Solutions
 
-### Solution 1
+### Solution 1: State Compression Dynamic Programming
+
+We notice that the maximum length of the array in the problem does not exceed $14$. Therefore, we can use an integer to represent the current state, where the $i$-th bit is $1$ if the $i$-th number in the array has been selected, and $0$ if it has not been selected.
+
+We define $f[i][j]$ as the number of schemes where the current selected integer state is $i$, and the index of the last selected integer is $j$. Initially, $f[0][0]=0$, and the answer is $\sum_{j=0}^{n-1}f[2^n-1][j]$.
+
+Considering $f[i][j]$, if only one number is currently selected, then $f[i][j]=1$. Otherwise, we can enumerate the index $k$ of the last selected number. If the numbers corresponding to $k$ and $j$ meet the requirements of the problem, then $f[i][j]$ can be transferred from $f[i \oplus 2^j][k]$. That is:
+
+$$
+f[i][j]=
+\begin{cases}
+1, & i=2^j\\
+\sum_{k=0}^{n-1}f[i \oplus 2^j][k], & i \neq 2^j \text{ and nums}[j] \text{ and nums}[k] \text{ meet the requirements of the problem}\\
+\end{cases}
+$$
+
+The final answer is $\sum_{j=0}^{n-1}f[2^n-1][j]$. Note that the answer may be very large, so we need to take the modulus of $10^9+7$.
+
+The time complexity is $O(n^2 \times 2^n)$, and the space complexity is $O(n \times 2^n)$. Here, $n$ is the length of the array.
 
 <!-- tabs:start -->
 

solution/2700-2799/2743.Count Substrings Without Repeating Character/README_EN.md

@@ -49,7 +49,11 @@ And it can be shown that there are no special substrings with a length of at lea
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Counting + Two Pointers
+
+We use two pointers $j$ and $i$ to represent the left and right boundaries of the current substring, and an array $cnt$ of length $26$ to count the occurrence of each character in the current substring. We traverse the string from left to right. Each time we traverse to position $i$, we increase the occurrence of $s[i]$, and then check whether $s[i]$ appears at least twice. If so, we need to decrease the occurrence of $s[j]$ and move $j$ one step to the right, until the occurrence of $s[i]$ does not exceed once. In this way, we get the length of the longest special substring ending with $s[i]$, which is $i - j + 1$, so the number of special substrings ending with $s[i]$ is $i - j + 1$. Finally, we add up the number of special substrings ending at each position to get the answer.
+
+The time complexity is $O(n)$, and the space complexity is $O(C)$. Here, $n$ is the length of the string $s$, and $C$ is the size of the character set. In this problem, the character set consists of lowercase English letters, so $C = 26$.
 
 <!-- tabs:start -->