feat: add solutions to lc problems: No.2643~2646

* No.2643.Row With Maximum Ones
* No.2644.Find the Maximum Divisibility Score
* No.2645.Minimum Additions to Make Valid Strin
* No.2646.Minimize the Total Price of the Trips

Author: yanglbme

solution/0000-0099/0007.Reverse Integer/Solution.py

@@ -1,7 +1,7 @@
 class Solution:
     def reverse(self, x: int) -> int:
         ans = 0
-        mi, mx = -2**31, 2**31 - 1
+        mi, mx = -(2**31), 2**31 - 1
         while x:
             if ans < mi // 10 + 1 or ans > mx // 10:
                 return 0

solution/0000-0099/0012.Integer to Roman/Solution.py

@@ -1,7 +1,6 @@
 class Solution:
     def intToRoman(self, num: int) -> str:
-        cs = ('M', 'CM', 'D', 'CD', 'C', 'XC',
-              'L', 'XL', 'X', 'IX', 'V', 'IV', 'I')
+        cs = ('M', 'CM', 'D', 'CD', 'C', 'XC', 'L', 'XL', 'X', 'IX', 'V', 'IV', 'I')
         vs = (1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1)
         ans = []
         for c, v in zip(cs, vs):

solution/1100-1199/1147.Longest Chunked Palindrome Decomposition/Solution.py

@@ -6,7 +6,7 @@ def longestDecomposition(self, text: str) -> int:
             k = 1
             ok = False
             while i + k - 1 < j - k + 1:
-                if text[i: i + k] == text[j - k + 1: j + 1]:
+                if text[i : i + k] == text[j - k + 1 : j + 1]:
                     ans += 2
                     i += k
                     j -= k

solution/1300-1399/1396.Design Underground System/README_EN.md

@@ -137,20 +137,20 @@ class UndergroundSystem {
     public UndergroundSystem() {
 
     }
-    
+
     public void checkIn(int id, String stationName, int t) {
         ts.put(id, t);
         names.put(id, stationName);
     }
-    
+
     public void checkOut(int id, String stationName, int t) {
         String key = names.get(id) + "-" + stationName;
         int[] v = d.getOrDefault(key, new int[2]);
         v[0] += t - ts.get(id);
         v[1]++;
         d.put(key, v);
     }
-    
+
     public double getAverageTime(String startStation, String endStation) {
         String key = startStation + "-" + endStation;
         int[] v = d.get(key);

solution/1300-1399/1396.Design Underground System/Solution.py

@@ -1,5 +1,4 @@
 class UndergroundSystem:
-
     def __init__(self):
         self.ts = {}
         self.d = {}

solution/1600-1699/1649.Create Sorted Array through Instructions/Solution.py

@@ -21,7 +21,7 @@ def createSortedArray(self, instructions: List[int]) -> int:
         m = max(instructions)
         tree = BinaryIndexedTree(m)
         ans = 0
-        mod = 10 ** 9 + 7
+        mod = 10**9 + 7
         for i, x in enumerate(instructions):
             cost = min(tree.query(x - 1), i - tree.query(x))
             ans += cost

solution/1700-1799/1756.Design Most Recently Used Queue/Solution.py

@@ -17,7 +17,6 @@ def query(self, x: int) -> int:
 
 
 class MRUQueue:
-
     def __init__(self, n: int):
         self.q = list(range(n + 1))
         self.tree = BinaryIndexedTree(n + 2010)
@@ -35,6 +34,7 @@ def fetch(self, k: int) -> int:
         self.tree.update(l, 1)
         return x
 
+
 # Your MRUQueue object will be instantiated and called as such:
 # obj = MRUQueue(n)
 # param_1 = obj.fetch(k)

solution/2600-2699/2641.Cousins in Binary Tree II/Solution.py

@@ -18,8 +18,9 @@ def dfs1(root, d):
         def dfs2(root, d):
             if root is None:
                 return
-            t = (root.left.val if root.left else 0) + \
-                (root.right.val if root.right else 0)
+            t = (root.left.val if root.left else 0) + (
+                root.right.val if root.right else 0
+            )
             if root.left:
                 root.left.val = s[d] - t
             if root.right:

solution/2600-2699/2642.Design Graph With Shortest Path Calculator/Solution.py

@@ -1,5 +1,4 @@
 class Graph:
-
     def __init__(self, n: int, edges: List[List[int]]):
         self.n = n
         self.g = [[inf] * n for _ in range(n)]

solution/2600-2699/2643.Row With Maximum Ones/README.md

@@ -52,6 +52,12 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：模拟**
+
+我们直接遍历矩阵，统计每一行中 $1$ 的个数，更新最大值和对应的行下标。注意，如果当前行的 $1$ 的个数与最大值相等，我们需要选择行下标较小的那一行。
+
+时间复杂度 $(m \times n)$，其中 $m$ 和 $n$ 分别为矩阵的行数和列数。空间复杂度 $O(1)$。
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -136,6 +142,22 @@ func rowAndMaximumOnes(mat [][]int) []int {
 }
 ```
 
+### **TypeScript**
+
+```ts
+function rowAndMaximumOnes(mat: number[][]): number[] {
+    const ans: number[] = [0, 0];
+    for (let i = 0; i < mat.length; ++i) {
+        const cnt = mat[i].reduce((a, b) => a + b);
+        if (ans[1] < cnt) {
+            ans[0] = i;
+            ans[1] = cnt;
+        }
+    }
+    return ans;
+}
+```
+
 ### **...**
 
 ```