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feat: add solutions to lc problem: No.2306 (doocs#1992)
No.2306.Naming a Company
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‎solution/2300-2399/2306.Naming a Company/README.md

+43-11
Original file line numberDiff line numberDiff line change
@@ -62,7 +62,15 @@
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**方法一：枚举计数**
6464

65-
$f[i][j]$ 表示有多少个首字母为 $i$ 的字符串，将首字符换成 $j$ 后，未在 $ideas$ 中出现过的字符串个数。
65+
我们定义 $f[i][j]$ 表示 $ideas$ 中以第 $i$ 个字母开头，替换为第 $j$ 个字母后，不在 $ideas$ 中的字符串的个数。初始时 $f[i][j] = 0$。另外，用一个哈希表 $s$ 记录 $ideas$ 中的字符串，方便我们开快速判断某个字符串是否在 $ideas$ 中。
66+
67+
接下来，我们遍历 $ideas$ 中字符串，对于当前遍历到的字符串 $v$，我们枚举替换后的第一个字母 $j$，如果 $v$ 替换后的字符串不在 $ideas$ 中，那么我们就更新 $f[i][j] = f[i][j] + 1$。
68+
69+
最后，我们再次遍历 $ideas$ 中字符串，对于当前遍历到的字符串 $v$，我们枚举替换后的第一个字母 $j$，如果 $v$ 替换后的字符串不在 $ideas$ 中，那么我们就更新答案 $ans = ans + f[j][i]$。
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最终答案即为 $ans$。
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时间复杂度 $O(n \times m \times |\Sigma|)$，空间复杂度 $O(|\Sigma|^2)$。其中 $n$ 和 $m$ 分别是 $ideas$ 中字符串的个数和字符串的最大长度，而 $|\Sigma|$ 是字符串中出现的字符集，本题中 $|\Sigma| \leq 26$。
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<!-- tabs:start -->
6876

@@ -138,7 +146,7 @@ class Solution {
138146
public:
139147
long long distinctNames(vector<string>& ideas) {
140148
unordered_set<string> s(ideas.begin(), ideas.end());
141-
vector<vector<int>> f(26, vector<int>(26));
149+
int f[26][26]{};
142150
for (auto v : ideas) {
143151
int i = v[0] - 'a';
144152
for (int j = 0; j < 26; ++j) {
@@ -149,7 +157,7 @@ public:
149157
}
150158
}
151159
long long ans = 0;
152-
for (auto v : ideas) {
160+
for (auto& v : ideas) {
153161
int i = v[0] - 'a';
154162
for (int j = 0; j < 26; ++j) {
155163
v[0] = j + 'a';
@@ -166,15 +174,12 @@ public:
166174
### **Go**
167175
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```go
169-
func distinctNames(ideas []string) int64 {
177+
func distinctNames(ideas []string) (ans int64) {
170178
s := map[string]bool{}
171179
for _, v := range ideas {
172180
s[v] = true
173181
}
174-
f := make([][]int, 26)
175-
for i := range f {
176-
f[i] = make([]int, 26)
177-
}
182+
f := [26][26]int{}
178183
for _, v := range ideas {
179184
i := int(v[0] - 'a')
180185
t := []byte(v)
@@ -185,7 +190,7 @@ func distinctNames(ideas []string) int64 {
185190
}
186191
}
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}
188-
var ans int64
193+
189194
for _, v := range ideas {
190195
i := int(v[0] - 'a')
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t := []byte(v)
@@ -196,14 +201,41 @@ func distinctNames(ideas []string) int64 {
196201
}
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}
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}
199-
return ans
204+
return
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}
201206
```
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203208
### **TypeScript**
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205210
```ts
206-
211+
function distinctNames(ideas: string[]): number {
212+
const s = new Set(ideas);
213+
const f: number[][] = Array(26)
214+
.fill(0)
215+
.map(() => Array(26).fill(0));
216+
for (const v of s) {
217+
const i = v.charCodeAt(0) - 'a'.charCodeAt(0);
218+
const t = [...v];
219+
for (let j = 0; j < 26; ++j) {
220+
t[0] = String.fromCharCode('a'.charCodeAt(0) + j);
221+
if (!s.has(t.join(''))) {
222+
f[i][j]++;
223+
}
224+
}
225+
}
226+
let ans = 0;
227+
for (const v of s) {
228+
const i = v.charCodeAt(0) - 'a'.charCodeAt(0);
229+
const t = [...v];
230+
for (let j = 0; j < 26; ++j) {
231+
t[0] = String.fromCharCode('a'.charCodeAt(0) + j);
232+
if (!s.has(t.join(''))) {
233+
ans += f[j][i];
234+
}
235+
}
236+
}
237+
return ans;
238+
}
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```
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### **...**

‎solution/2300-2399/2306.Naming a Company/README_EN.md

+46-10
Original file line numberDiff line numberDiff line change
@@ -56,6 +56,18 @@ The following are some examples of invalid selections:
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## Solutions
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59+
**Solution 1: Enumeration Counting**
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We define $f[i][j]$ to represent the number of strings in $ideas$ that start with the $i$th letter and are not in $ideas$ after being replaced with the $j$th letter. Initially, $f[i][j] = 0$. Additionally, we use a hash table $s$ to record the strings in $ideas$, which allows us to quickly determine whether a string is in $ideas$.
62+
63+
Next, we traverse the strings in $ideas$. For the current string $v$, we enumerate the first letter $j$ after replacement. If the string after $v$ is replaced is not in $ideas$, then we update $f[i][j] = f[i][j] + 1$.
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65+
Finally, we traverse the strings in $ideas$ again. For the current string $v$, we enumerate the first letter $j$ after replacement. If the string after $v$ is replaced is not in $ideas$, then we update the answer $ans = ans + f[j][i]$.
66+
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The final answer is $ans$.
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The time complexity is $O(n \times m \times |\Sigma|)$, and the space complexity is $O(|\Sigma|^2)$. Here, $n$ and $m$ are the number of strings in $ideas$ and the maximum length of the strings, respectively, and $|\Sigma|$ is the character set that appears in the string. In this problem, $|\Sigma| \leq 26$.
70+
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<!-- tabs:start -->
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6173
### **Python3**
@@ -126,7 +138,7 @@ class Solution {
126138
public:
127139
long long distinctNames(vector<string>& ideas) {
128140
unordered_set<string> s(ideas.begin(), ideas.end());
129-
vector<vector<int>> f(26, vector<int>(26));
141+
int f[26][26]{};
130142
for (auto v : ideas) {
131143
int i = v[0] - 'a';
132144
for (int j = 0; j < 26; ++j) {
@@ -137,7 +149,7 @@ public:
137149
}
138150
}
139151
long long ans = 0;
140-
for (auto v : ideas) {
152+
for (auto& v : ideas) {
141153
int i = v[0] - 'a';
142154
for (int j = 0; j < 26; ++j) {
143155
v[0] = j + 'a';
@@ -154,15 +166,12 @@ public:
154166
### **Go**
155167
156168
```go
157-
func distinctNames(ideas []string) int64 {
169+
func distinctNames(ideas []string) (ans int64) {
158170
s := map[string]bool{}
159171
for _, v := range ideas {
160172
s[v] = true
161173
}
162-
f := make([][]int, 26)
163-
for i := range f {
164-
f[i] = make([]int, 26)
165-
}
174+
f := [26][26]int{}
166175
for _, v := range ideas {
167176
i := int(v[0] - 'a')
168177
t := []byte(v)
@@ -173,7 +182,7 @@ func distinctNames(ideas []string) int64 {
173182
}
174183
}
175184
}
176-
var ans int64
185+
177186
for _, v := range ideas {
178187
i := int(v[0] - 'a')
179188
t := []byte(v)
@@ -184,14 +193,41 @@ func distinctNames(ideas []string) int64 {
184193
}
185194
}
186195
}
187-
return ans
196+
return
188197
}
189198
```
190199

191200
### **TypeScript**
192201

193202
```ts
194-
203+
function distinctNames(ideas: string[]): number {
204+
const s = new Set(ideas);
205+
const f: number[][] = Array(26)
206+
.fill(0)
207+
.map(() => Array(26).fill(0));
208+
for (const v of s) {
209+
const i = v.charCodeAt(0) - 'a'.charCodeAt(0);
210+
const t = [...v];
211+
for (let j = 0; j < 26; ++j) {
212+
t[0] = String.fromCharCode('a'.charCodeAt(0) + j);
213+
if (!s.has(t.join(''))) {
214+
f[i][j]++;
215+
}
216+
}
217+
}
218+
let ans = 0;
219+
for (const v of s) {
220+
const i = v.charCodeAt(0) - 'a'.charCodeAt(0);
221+
const t = [...v];
222+
for (let j = 0; j < 26; ++j) {
223+
t[0] = String.fromCharCode('a'.charCodeAt(0) + j);
224+
if (!s.has(t.join(''))) {
225+
ans += f[j][i];
226+
}
227+
}
228+
}
229+
return ans;
230+
}
195231
```
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197233
### **...**

‎solution/2300-2399/2306.Naming a Company/Solution.cpp

+26-26
Original file line numberDiff line numberDiff line change
@@ -1,27 +1,27 @@
1-
class Solution {
2-
public:
3-
long long distinctNames(vector<string>& ideas) {
4-
unordered_set<string> s(ideas.begin(), ideas.end());
5-
vector<vector<int>> f(26, vector<int>(26));
6-
for (auto v : ideas) {
7-
int i = v[0] - 'a';
8-
for (int j = 0; j < 26; ++j) {
9-
v[0] = j + 'a';
10-
if (!s.count(v)) {
11-
++f[i][j];
12-
}
13-
}
14-
}
15-
long long ans = 0;
16-
for (auto v : ideas) {
17-
int i = v[0] - 'a';
18-
for (int j = 0; j < 26; ++j) {
19-
v[0] = j + 'a';
20-
if (!s.count(v)) {
21-
ans += f[j][i];
22-
}
23-
}
24-
}
25-
return ans;
26-
}
1+
class Solution {
2+
public:
3+
long long distinctNames(vector<string>& ideas) {
4+
unordered_set<string> s(ideas.begin(), ideas.end());
5+
int f[26][26]{};
6+
for (auto v : ideas) {
7+
int i = v[0] - 'a';
8+
for (int j = 0; j < 26; ++j) {
9+
v[0] = j + 'a';
10+
if (!s.count(v)) {
11+
++f[i][j];
12+
}
13+
}
14+
}
15+
long long ans = 0;
16+
for (auto& v : ideas) {
17+
int i = v[0] - 'a';
18+
for (int j = 0; j < 26; ++j) {
19+
v[0] = j + 'a';
20+
if (!s.count(v)) {
21+
ans += f[j][i];
22+
}
23+
}
24+
}
25+
return ans;
26+
}
2727
};

‎solution/2300-2399/2306.Naming a Company/Solution.go

+4-7
Original file line numberDiff line numberDiff line change
@@ -1,12 +1,9 @@
1-
func distinctNames(ideas []string) int64 {
1+
func distinctNames(ideas []string) (ans int64) {
22
s := map[string]bool{}
33
for _, v := range ideas {
44
s[v] = true
55
}
6-
f := make([][]int, 26)
7-
for i := range f {
8-
f[i] = make([]int, 26)
9-
}
6+
f := [26][26]int{}
107
for _, v := range ideas {
118
i := int(v[0] - 'a')
129
t := []byte(v)
@@ -17,7 +14,7 @@ func distinctNames(ideas []string) int64 {
1714
}
1815
}
1916
}
20-
var ans int64
17+
2118
for _, v := range ideas {
2219
i := int(v[0] - 'a')
2320
t := []byte(v)
@@ -28,5 +25,5 @@ func distinctNames(ideas []string) int64 {
2825
}
2926
}
3027
}
31-
return ans
28+
return
3229
}

‎solution/2300-2399/2306.Naming a Company/Solution.ts

+28
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,28 @@
1+
function distinctNames(ideas: string[]): number {
2+
const s = new Set(ideas);
3+
const f: number[][] = Array(26)
4+
.fill(0)
5+
.map(() => Array(26).fill(0));
6+
for (const v of s) {
7+
const i = v.charCodeAt(0) - 'a'.charCodeAt(0);
8+
const t = [...v];
9+
for (let j = 0; j < 26; ++j) {
10+
t[0] = String.fromCharCode('a'.charCodeAt(0) + j);
11+
if (!s.has(t.join(''))) {
12+
f[i][j]++;
13+
}
14+
}
15+
}
16+
let ans = 0;
17+
for (const v of s) {
18+
const i = v.charCodeAt(0) - 'a'.charCodeAt(0);
19+
const t = [...v];
20+
for (let j = 0; j < 26; ++j) {
21+
t[0] = String.fromCharCode('a'.charCodeAt(0) + j);
22+
if (!s.has(t.join(''))) {
23+
ans += f[j][i];
24+
}
25+
}
26+
}
27+
return ans;
28+
}

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