feat: update solutions to lc problems: No.2399,2400

solution/2300-2399/2399.Check Distances Between Same Letters/README.md

@@ -73,7 +73,7 @@ tags:
 
 我们可以用哈希表 $d$ 记录每个字母出现的下标，然后遍历哈希表，判断每个字母的下标之差是否等于 `distance` 中对应的值。如果出现不等的情况，直接返回 `false`。如果遍历结束后，没有出现不等的情况，返回 `true`。
 
-时间复杂度 $O(n)$，空间复杂度 $O(C)$。其中 $n$ 为字符串 $s$ 的长度，而 $C$ 为字符集大小，本题中 $C = 26$。
+时间复杂度 $O(n)$，其中 $n$ 为字符串 $s$ 的长度。空间复杂度 $O(|\Sigma|)$，其中 $\Sigma$ 为字符集，这里为小写字母集合。
 
 <!-- tabs:start -->
 
@@ -83,10 +83,11 @@ tags:
 class Solution:
     def checkDistances(self, s: str, distance: List[int]) -> bool:
         d = defaultdict(int)
-        for i, c in enumerate(s, 1):
-            if d[c] and i - d[c] - 1 != distance[ord(c) - ord('a')]:
+        for i, c in enumerate(map(ord, s), 1):
+            j = c - ord("a")
+            if d[j] and i - d[j] - 1 != distance[j]:
                 return False
-            d[c] = i
+            d[j] = i
         return True
 ```
 
@@ -96,7 +97,7 @@ class Solution:
 class Solution {
     public boolean checkDistances(String s, int[] distance) {
         int[] d = new int[26];
-        for (int i = 1, n = s.length(); i <= n; ++i) {
+        for (int i = 1; i <= s.length(); ++i) {
             int j = s.charAt(i - 1) - 'a';
             if (d[j] > 0 && i - d[j] - 1 != distance[j]) {
                 return false;
@@ -147,8 +148,8 @@ func checkDistances(s string, distance []int) bool {
 
 ```ts
 function checkDistances(s: string, distance: number[]): boolean {
+    const d: number[] = Array(26).fill(0);
     const n = s.length;
-    const d: number[] = new Array(26).fill(0);
     for (let i = 1; i <= n; ++i) {
         const j = s.charCodeAt(i - 1) - 97;
         if (d[j] && i - d[j] - 1 !== distance[j]) {

solution/2300-2399/2399.Check Distances Between Same Letters/README_EN.md

@@ -69,7 +69,11 @@ Because distance[0] = 1, s is not a well-spaced string.
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Array or Hash Table
+
+We can use a hash table $d$ to record the indices of each letter's occurrences. Then, traverse the hash table and check if the difference between the indices of each letter equals the corresponding value in the `distance` array. If any discrepancy is found, return `false`. If the traversal completes without discrepancies, return `true`.
+
+The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(|\Sigma|)$, where $\Sigma$ is the character set, which in this case is the set of lowercase letters.
 
 <!-- tabs:start -->
 
@@ -79,10 +83,11 @@ Because distance[0] = 1, s is not a well-spaced string.
 class Solution:
     def checkDistances(self, s: str, distance: List[int]) -> bool:
         d = defaultdict(int)
-        for i, c in enumerate(s, 1):
-            if d[c] and i - d[c] - 1 != distance[ord(c) - ord('a')]:
+        for i, c in enumerate(map(ord, s), 1):
+            j = c - ord("a")
+            if d[j] and i - d[j] - 1 != distance[j]:
                 return False
-            d[c] = i
+            d[j] = i
         return True
 ```
 
@@ -92,7 +97,7 @@ class Solution:
 class Solution {
     public boolean checkDistances(String s, int[] distance) {
         int[] d = new int[26];
-        for (int i = 1, n = s.length(); i <= n; ++i) {
+        for (int i = 1; i <= s.length(); ++i) {
             int j = s.charAt(i - 1) - 'a';
             if (d[j] > 0 && i - d[j] - 1 != distance[j]) {
                 return false;
@@ -143,8 +148,8 @@ func checkDistances(s string, distance []int) bool {
 
 ```ts
 function checkDistances(s: string, distance: number[]): boolean {
+    const d: number[] = Array(26).fill(0);
     const n = s.length;
-    const d: number[] = new Array(26).fill(0);
     for (let i = 1; i <= n; ++i) {
         const j = s.charCodeAt(i - 1) - 97;
         if (d[j] && i - d[j] - 1 !== distance[j]) {

solution/2300-2399/2399.Check Distances Between Same Letters/Solution.java

@@ -1,7 +1,7 @@
 class Solution {
     public boolean checkDistances(String s, int[] distance) {
         int[] d = new int[26];
-        for (int i = 1, n = s.length(); i <= n; ++i) {
+        for (int i = 1; i <= s.length(); ++i) {
             int j = s.charAt(i - 1) - 'a';
             if (d[j] > 0 && i - d[j] - 1 != distance[j]) {
                 return false;
@@ -10,4 +10,4 @@ public boolean checkDistances(String s, int[] distance) {
         }
         return true;
     }
-}
+}

solution/2300-2399/2399.Check Distances Between Same Letters/Solution.py

@@ -1,8 +1,9 @@
 class Solution:
     def checkDistances(self, s: str, distance: List[int]) -> bool:
         d = defaultdict(int)
-        for i, c in enumerate(s, 1):
-            if d[c] and i - d[c] - 1 != distance[ord(c) - ord('a')]:
+        for i, c in enumerate(map(ord, s), 1):
+            j = c - ord("a")
+            if d[j] and i - d[j] - 1 != distance[j]:
                 return False
-            d[c] = i
+            d[j] = i
         return True

solution/2300-2399/2399.Check Distances Between Same Letters/Solution.ts

@@ -1,6 +1,6 @@
 function checkDistances(s: string, distance: number[]): boolean {
+    const d: number[] = Array(26).fill(0);
     const n = s.length;
-    const d: number[] = new Array(26).fill(0);
     for (let i = 1; i <= n; ++i) {
         const j = s.charCodeAt(i - 1) - 97;
         if (d[j] && i - d[j] - 1 !== distance[j]) {

solution/2400-2499/2400.Number of Ways to Reach a Position After Exactly k Steps/README.md

@@ -134,7 +134,7 @@ public:
         const int mod = 1e9 + 7;
         int f[k + 1][k + 1];
         memset(f, -1, sizeof(f));
-        function<int(int, int)> dfs = [&](int i, int j) -> int {
+        auto dfs = [&](auto&& dfs, int i, int j) -> int {
             if (i > j || j < 0) {
                 return 0;
             }
@@ -144,10 +144,10 @@ public:
             if (f[i][j] != -1) {
                 return f[i][j];
             }
-            f[i][j] = (dfs(i + 1, j - 1) + dfs(abs(i - 1), j - 1)) % mod;
+            f[i][j] = (dfs(dfs, i + 1, j - 1) + dfs(dfs, abs(i - 1), j - 1)) % mod;
             return f[i][j];
         };
-        return dfs(abs(startPos - endPos), k);
+        return dfs(dfs, abs(startPos - endPos), k);
     }
 };
 ```

solution/2400-2499/2400.Number of Ways to Reach a Position After Exactly k Steps/README_EN.md

@@ -135,7 +135,7 @@ public:
         const int mod = 1e9 + 7;
         int f[k + 1][k + 1];
         memset(f, -1, sizeof(f));
-        function<int(int, int)> dfs = [&](int i, int j) -> int {
+        auto dfs = [&](auto&& dfs, int i, int j) -> int {
             if (i > j || j < 0) {
                 return 0;
             }
@@ -145,10 +145,10 @@ public:
             if (f[i][j] != -1) {
                 return f[i][j];
             }
-            f[i][j] = (dfs(i + 1, j - 1) + dfs(abs(i - 1), j - 1)) % mod;
+            f[i][j] = (dfs(dfs, i + 1, j - 1) + dfs(dfs, abs(i - 1), j - 1)) % mod;
             return f[i][j];
         };
-        return dfs(abs(startPos - endPos), k);
+        return dfs(dfs, abs(startPos - endPos), k);
     }
 };
 ```

solution/2400-2499/2400.Number of Ways to Reach a Position After Exactly k Steps/Solution.cpp

@@ -4,7 +4,7 @@ class Solution {
         const int mod = 1e9 + 7;
         int f[k + 1][k + 1];
         memset(f, -1, sizeof(f));
-        function<int(int, int)> dfs = [&](int i, int j) -> int {
+        auto dfs = [&](auto&& dfs, int i, int j) -> int {
             if (i > j || j < 0) {
                 return 0;
             }
@@ -14,9 +14,9 @@ class Solution {
             if (f[i][j] != -1) {
                 return f[i][j];
             }
-            f[i][j] = (dfs(i + 1, j - 1) + dfs(abs(i - 1), j - 1)) % mod;
+            f[i][j] = (dfs(dfs, i + 1, j - 1) + dfs(dfs, abs(i - 1), j - 1)) % mod;
             return f[i][j];
         };
-        return dfs(abs(startPos - endPos), k);
+        return dfs(dfs, abs(startPos - endPos), k);
     }
-};
+};