feat: update solutions to lc problems: No.1614,1615

Author: yanglbme

solution/1600-1699/1614.Maximum Nesting Depth of the Parentheses/README.md

@@ -69,11 +69,13 @@ tags:
 
 ### 方法一：遍历
 
-我们可以遍历字符串，维护当前的嵌套深度，遇到左括号时深度加一，并且更新组最大深大；遇到右括号时深度减一。
+我们用一个变量 $d$ 记录当前的深度，初始时 $d = 0$。
 
-遍历结束后，返回最大深度即可。
+遍历字符串 $s$，当遇到左括号时，深度 $d$ 加一，同时更新答案为当前深度 $d$ 和答案的最大值。当遇到右括号时，深度 $d$ 减一。
 
-时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为字符串长度。
+最后返回答案即可。
+
+时间复杂度 $O(n)$，其中 $n$ 是字符串 $s$ 的长度。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 

solution/1600-1699/1614.Maximum Nesting Depth of the Parentheses/README_EN.md

@@ -69,7 +69,15 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Traversal
+
+We use a variable $d$ to record the current depth, initially $d = 0$.
+
+Traverse the string $s$. When encountering a left parenthesis, increment the depth $d$ by one and update the answer to be the maximum of the current depth $d$ and the answer. When encountering a right parenthesis, decrement the depth $d$ by one.
+
+Finally, return the answer.
+
+The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(1)$.
 
 <!-- tabs:start -->
 

solution/1600-1699/1614.Maximum Nesting Depth of the Parentheses/Solution.cpp

@@ -11,4 +11,4 @@ class Solution {
         }
         return ans;
     }
-};
+};

solution/1600-1699/1615.Maximal Network Rank/README.md

@@ -77,9 +77,9 @@ tags:
 
 ### 方法一：计数
 
-我们可以用一维数组 $cnt$ 记录每个城市的度，用二维数组 $g$ 记录每对城市之间是否有道路相连，如果城市 $a$ 和城市 $b$ 之间有道路相连，则 $g[a][b] = g[b][a] = 1$，否则 $g[a][b] = g[b][a] = 0$。
+我们可以用一维数组 $\textit{cnt}$ 记录每个城市的度，用二维数组 $\textit{g}$ 记录每对城市之间是否有道路相连，如果城市 $a$ 和城市 $b$ 之间有道路相连，则 $\textit{g}[a][b] = \textit{g}[b][a] = 1$，否则 $\textit{g}[a][b] = \textit{g}[b][a] = 0$。
 
-接下来，我们枚举每对城市 $(a, b)$，其中 $a \lt b$，计算它们的网络秩，即 $cnt[a] + cnt[b] - g[a][b]$，取其中的最大值即为答案。
+接下来，我们枚举每对城市 $(a, b)$，其中 $a \lt b$，计算它们的网络秩，即 $\textit{cnt}[a] + \textit{cnt}[b] - \textit{g}[a][b]$，取其中的最大值即为答案。
 
 时间复杂度 $O(n^2)$，空间复杂度 $O(n^2)$。其中 $n$ 是城市的数量。
 
@@ -90,16 +90,13 @@ tags:
 ```python
 class Solution:
     def maximalNetworkRank(self, n: int, roads: List[List[int]]) -> int:
-        g = defaultdict(set)
+        g = [[0] * n for _ in range(n)]
+        cnt = [0] * n
         for a, b in roads:
-            g[a].add(b)
-            g[b].add(a)
-        ans = 0
-        for a in range(n):
-            for b in range(a + 1, n):
-                if (t := len(g[a]) + len(g[b]) - (a in g[b])) > ans:
-                    ans = t
-        return ans
+            g[a][b] = g[b][a] = 1
+            cnt[a] += 1
+            cnt[b] += 1
+        return max(cnt[a] + cnt[b] - g[a][b] for a in range(n) for b in range(a + 1, n))
 ```
 
 #### Java
@@ -182,8 +179,8 @@ func maximalNetworkRank(n int, roads [][]int) (ans int) {
 
 ```ts
 function maximalNetworkRank(n: number, roads: number[][]): number {
-    const g: number[][] = Array.from(new Array(n), () => new Array(n).fill(0));
-    const cnt: number[] = new Array(n).fill(0);
+    const g: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
+    const cnt: number[] = Array(n).fill(0);
     for (const [a, b] of roads) {
         g[a][b] = 1;
         g[b][a] = 1;
@@ -204,28 +201,4 @@ function maximalNetworkRank(n: number, roads: number[][]): number {
 
 <!-- solution:end -->
 
-<!-- solution:start -->
-
-### 方法二
-
-<!-- tabs:start -->
-
-#### Python3
-
-```python
-class Solution:
-    def maximalNetworkRank(self, n: int, roads: List[List[int]]) -> int:
-        g = [[0] * n for _ in range(n)]
-        cnt = [0] * n
-        for a, b in roads:
-            g[a][b] = g[b][a] = 1
-            cnt[a] += 1
-            cnt[b] += 1
-        return max(cnt[a] + cnt[b] - g[a][b] for a in range(n) for b in range(a + 1, n))
-```
-
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
 <!-- problem:end -->

solution/1600-1699/1615.Maximal Network Rank/README_EN.md

@@ -73,7 +73,13 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Counting
+
+We can use a one-dimensional array $\textit{cnt}$ to record the degree of each city and a two-dimensional array $\textit{g}$ to record whether there is a road between each pair of cities. If there is a road between city $a$ and city $b$, then $\textit{g}[a][b] = \textit{g}[b][a] = 1$; otherwise, $\textit{g}[a][b] = \textit{g}[b][a] = 0$.
+
+Next, we enumerate each pair of cities $(a, b)$, where $a \lt b$, and calculate their network rank, which is $\textit{cnt}[a] + \textit{cnt}[b] - \textit{g}[a][b]$. The maximum value among these is the answer.
+
+The time complexity is $O(n^2)$, and the space complexity is $O(n^2)$. Here, $n$ is the number of cities.
 
 <!-- tabs:start -->
 
@@ -82,16 +88,13 @@ tags:
 ```python
 class Solution:
     def maximalNetworkRank(self, n: int, roads: List[List[int]]) -> int:
-        g = defaultdict(set)
+        g = [[0] * n for _ in range(n)]
+        cnt = [0] * n
         for a, b in roads:
-            g[a].add(b)
-            g[b].add(a)
-        ans = 0
-        for a in range(n):
-            for b in range(a + 1, n):
-                if (t := len(g[a]) + len(g[b]) - (a in g[b])) > ans:
-                    ans = t
-        return ans
+            g[a][b] = g[b][a] = 1
+            cnt[a] += 1
+            cnt[b] += 1
+        return max(cnt[a] + cnt[b] - g[a][b] for a in range(n) for b in range(a + 1, n))
 ```
 
 #### Java
@@ -174,8 +177,8 @@ func maximalNetworkRank(n int, roads [][]int) (ans int) {
 
 ```ts
 function maximalNetworkRank(n: number, roads: number[][]): number {
-    const g: number[][] = Array.from(new Array(n), () => new Array(n).fill(0));
-    const cnt: number[] = new Array(n).fill(0);
+    const g: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
+    const cnt: number[] = Array(n).fill(0);
     for (const [a, b] of roads) {
         g[a][b] = 1;
         g[b][a] = 1;
@@ -196,28 +199,4 @@ function maximalNetworkRank(n: number, roads: number[][]): number {
 
 <!-- solution:end -->
 
-<!-- solution:start -->
-
-### Solution 2
-
-<!-- tabs:start -->
-
-#### Python3
-
-```python
-class Solution:
-    def maximalNetworkRank(self, n: int, roads: List[List[int]]) -> int:
-        g = [[0] * n for _ in range(n)]
-        cnt = [0] * n
-        for a, b in roads:
-            g[a][b] = g[b][a] = 1
-            cnt[a] += 1
-            cnt[b] += 1
-        return max(cnt[a] + cnt[b] - g[a][b] for a in range(n) for b in range(a + 1, n))
-```
-
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
 <!-- problem:end -->

solution/1600-1699/1615.Maximal Network Rank/Solution.ts

@@ -1,6 +1,6 @@
 function maximalNetworkRank(n: number, roads: number[][]): number {
-    const g: number[][] = Array.from(new Array(n), () => new Array(n).fill(0));
-    const cnt: number[] = new Array(n).fill(0);
+    const g: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
+    const cnt: number[] = Array(n).fill(0);
     for (const [a, b] of roads) {
         g[a][b] = 1;
         g[b][a] = 1;