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Commit 62ac0a2

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feat: update lc problems (doocs#3346)
1 parent 9d87743 commit 62ac0a2

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34 files changed

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-179
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34 files changed

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‎solution/0600-0699/0614.Second Degree Follower/README.md

Lines changed: 1 addition & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -26,7 +26,7 @@ tags:
2626
| follower | varchar |
2727
+-------------+---------+
2828
(followee, follower) 是该表的主键(具有唯一值的列的组合)。
29-
该表的每一行表示关注者关注了社交网络上的关注者。
29+
该表的每一行表示关注者关注了社交网络上的被关注者。
3030
不会有用户关注他们自己。
3131
</pre>
3232

‎solution/0800-0899/0833.Find And Replace in String/README.md

Lines changed: 1 addition & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -4,6 +4,7 @@ difficulty: 中等
44
edit_url: https://github.com/doocs/leetcode/edit/main/solution/0800-0899/0833.Find%20And%20Replace%20in%20String/README.md
55
tags:
66
- 数组
7+
- 哈希表
78
- 字符串
89
- 排序
910
---

‎solution/0800-0899/0833.Find And Replace in String/README_EN.md

Lines changed: 1 addition & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -4,6 +4,7 @@ difficulty: Medium
44
edit_url: https://github.com/doocs/leetcode/edit/main/solution/0800-0899/0833.Find%20And%20Replace%20in%20String/README_EN.md
55
tags:
66
- Array
7+
- Hash Table
78
- String
89
- Sorting
910
---

‎solution/1000-1099/1001.Grid Illumination/README.md

Lines changed: 3 additions & 3 deletions
Original file line numberDiff line numberDiff line change
@@ -80,15 +80,15 @@ tags:
8080

8181
假设一盏灯的坐标为 $(x, y),ドル那么它所在的行的数值为 $x,ドル列的数值为 $y,ドル正对角线的数值为 $x-y,ドル反对角线的数值为 $x+y$。确定某一直线的唯一数值标识后,我们就可以通过哈希表来记录某一直线所拥有的灯的数目。
8282

83-
我们遍历数组 $lamps,ドル将当前遍历到的灯所在的行、列和正、反对角线拥有灯的数目分别加 1ドル$。
83+
我们遍历数组 $\textit{lamps},ドル将当前遍历到的灯所在的行、列和正、反对角线拥有灯的数目分别加 1ドル$。
8484

85-
注意,在处理 $lamps$ 时,需要进行去重,因为我们将重复的灯看作同一盏灯。
85+
注意,在处理 $\textit{lamps}$ 时,需要进行去重,因为我们将重复的灯看作同一盏灯。
8686

8787
接下来,我们遍历 queries,判断当前查询点所在的行,列和正、反对角线是否有灯,如果有,则置 1ドル,ドル即该点在查询时是被照亮的。然后进行关闭操作,查找查询点所在的八近邻点及它本身是否有灯,如果有,将该点所在的行、列和正、反对角线的灯数目分别减 1ドル,ドル并且将灯从网格中去掉。
8888

8989
最后,返回答案数组即可。
9090

91-
时间复杂度 $O(m + q),ドル其中 $m$ 和 $q$ 分别为数组 $lamps$ 和 $queries$ 的长度。
91+
时间复杂度 $O(m + q),ドル其中 $m$ 和 $q$ 分别为数组 $\textit{lamps}$ 和 $\textit{queries}$ 的长度。
9292

9393
<!-- tabs:start -->
9494

‎solution/1000-1099/1001.Grid Illumination/README_EN.md

Lines changed: 13 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -75,7 +75,19 @@ The 1<sup>st</sup>&nbsp;query asks if the lamp at grid[1][0] is illuminated or n
7575

7676
<!-- solution:start -->
7777

78-
### Solution 1
78+
### Solution 1: Hash Table
79+
80+
Suppose the coordinates of a lamp are $(x, y)$. Then, the row value is $x,ドル the column value is $y,ドル the main diagonal value is $x-y,ドル and the anti-diagonal value is $x+y$. Once we determine the unique value identifier for a line, we can use a hash table to record the number of lamps on that line.
81+
82+
We traverse the array $\textit{lamps},ドル and for each lamp, we increment the count of lamps in its row, column, main diagonal, and anti-diagonal by 1ドル$.
83+
84+
Note that when processing $\textit{lamps},ドル we need to remove duplicates because we treat repeated lamps as the same lamp.
85+
86+
Next, we traverse the queries and check if there are lamps in the row, column, main diagonal, or anti-diagonal of the current query point. If there are, we set the value to 1ドル,ドル indicating that the point is illuminated during the query. Then, we perform the turn-off operation by checking the eight neighboring points of the query point and the point itself to see if there are any lamps. If there are, we decrement the count of lamps in the corresponding row, column, main diagonal, and anti-diagonal by 1ドル$ and remove the lamp from the grid.
87+
88+
Finally, we return the answer array.
89+
90+
The time complexity is $O(m + q),ドル where $m$ and $q$ are the lengths of the arrays $\textit{lamps}$ and $\textit{queries},ドル respectively.
7991

8092
<!-- tabs:start -->
8193

‎solution/1000-1099/1002.Find Common Characters/README.md

Lines changed: 38 additions & 47 deletions
Original file line numberDiff line numberDiff line change
@@ -58,7 +58,7 @@ tags:
5858

5959
我们用一个长度为 26ドル$ 的数组 $cnt$ 记录每个字符在所有字符串中出现的最小次数,最后遍历 $cnt$ 数组,将出现次数大于 0ドル$ 的字符加入答案即可。
6060

61-
时间复杂度 $O(n \sum w_i),ドル空间复杂度 $O(C)$。其中 $n$ 为字符串数组 $words$ 的长度,而 $w_i$ 为字符串数组 $words$ 中第 $i$ 个字符串的长度,另外 $C$ 为字符集的大小,本题中 $C = 26$。
61+
时间复杂度 $O(n \sum w_i),ドル空间复杂度 $O(|\Sigma|)$。其中 $n$ 为字符串数组 $words$ 的长度,而 $w_i$ 为字符串数组 $words$ 中第 $i$ 个字符串的长度,另外 $|\Sigma|$ 为字符集的大小,本题中 $|\Sigma| = 26$。
6262

6363
<!-- tabs:start -->
6464

@@ -69,13 +69,10 @@ class Solution:
6969
def commonChars(self, words: List[str]) -> List[str]:
7070
cnt = Counter(words[0])
7171
for w in words:
72-
ccnt = Counter(w)
73-
for c in cnt.keys():
74-
cnt[c] = min(cnt[c], ccnt[c])
75-
ans = []
76-
for c, v in cnt.items():
77-
ans.extend([c] * v)
78-
return ans
72+
t = Counter(w)
73+
for c in cnt:
74+
cnt[c] = min(cnt[c], t[c])
75+
return list(cnt.elements())
7976
```
8077

8178
#### Java
@@ -84,21 +81,19 @@ class Solution:
8481
class Solution {
8582
public List<String> commonChars(String[] words) {
8683
int[] cnt = new int[26];
87-
Arrays.fill(cnt, 10000);
88-
for (String w : words) {
89-
int[] ccnt = new int[26];
84+
Arrays.fill(cnt, 20000);
85+
for (var w : words) {
86+
int[] t = new int[26];
9087
for (int i = 0; i < w.length(); ++i) {
91-
++ccnt[w.charAt(i) - 'a'];
88+
++t[w.charAt(i) - 'a'];
9289
}
9390
for (int i = 0; i < 26; ++i) {
94-
cnt[i] = Math.min(cnt[i], ccnt[i]);
91+
cnt[i] = Math.min(cnt[i], t[i]);
9592
}
9693
}
9794
List<String> ans = new ArrayList<>();
9895
for (int i = 0; i < 26; ++i) {
99-
while (cnt[i]-- > 0) {
100-
ans.add(String.valueOf((char) (i + 'a')));
101-
}
96+
ans.addAll(Collections.nCopies(cnt[i], String.valueOf((char) ('a' + i))));
10297
}
10398
return ans;
10499
}
@@ -111,21 +106,20 @@ class Solution {
111106
class Solution {
112107
public:
113108
vector<string> commonChars(vector<string>& words) {
114-
int cnt[26];
115-
memset(cnt, 0x3f, sizeof(cnt));
116-
for (auto& w : words) {
117-
int ccnt[26]{};
118-
for (char& c : w) {
119-
++ccnt[c - 'a'];
109+
vector<int> cnt(26, 20000);
110+
for (const auto& w : words) {
111+
vector<int> t(26, 0);
112+
for (char c : w) {
113+
++t[c - 'a'];
120114
}
121115
for (int i = 0; i < 26; ++i) {
122-
cnt[i] = min(cnt[i], ccnt[i]);
116+
cnt[i] = min(cnt[i], t[i]);
123117
}
124118
}
125119
vector<string> ans;
126120
for (int i = 0; i < 26; ++i) {
127-
while (cnt[i]--) {
128-
ans.emplace_back(1, i + 'a');
121+
for (int j = 0; j < cnt[i]; ++j) {
122+
ans.push_back(string(1, 'a' + i));
129123
}
130124
}
131125
return ans;
@@ -137,49 +131,46 @@ public:
137131
138132
```go
139133
func commonChars(words []string) (ans []string) {
140-
cnt := [26]int{}
134+
cnt := make([]int, 26)
141135
for i := range cnt {
142-
cnt[i] = 1 << 30
136+
cnt[i] = 20000
143137
}
144138
for _, w := range words {
145-
ccnt := [26]int{}
139+
t := make([]int, 26)
146140
for _, c := range w {
147-
ccnt[c-'a']++
141+
t[c-'a']++
148142
}
149-
for i, v := range cnt {
150-
cnt[i] = min(v, ccnt[i])
143+
for i:= 0; i < 26; i++ {
144+
cnt[i] = min(cnt[i], t[i])
151145
}
152146
}
153-
for i, v := range cnt {
154-
for v > 0 {
155-
ans = append(ans, string(i+'a'))
156-
v--
147+
for i := 0; i < 26; i++ {
148+
for j := 0; j < cnt[i]; j++ {
149+
ans = append(ans, string('a'+rune(i)))
157150
}
158151
}
159-
return
152+
return ans
160153
}
161154
```
162155

163156
#### TypeScript
164157

165158
```ts
166159
function commonChars(words: string[]): string[] {
167-
const freq:number[] = Array(26).fill(Number.POSITIVE_INFINITY);
160+
const cnt = Array(26).fill(20000);
168161
const aCode = 'a'.charCodeAt(0);
169-
for (const word of words) {
170-
const t:number[] = Array(26).fill(0);
171-
for (const c of word) {
172-
++t[c.charCodeAt(0) - aCode];
162+
for (const w of words) {
163+
const t = Array(26).fill(0);
164+
for (const c of w) {
165+
t[c.charCodeAt(0) - aCode]++;
173166
}
174-
for (let i = 0; i < 26; ++i) {
175-
freq[i] = Math.min(freq[i], t[i]);
167+
for (let i = 0; i < 26; i++) {
168+
cnt[i] = Math.min(cnt[i], t[i]);
176169
}
177170
}
178171
const ans: string[] = [];
179-
for (let i = 0; i < 26; ++i) {
180-
if (freq[i]) {
181-
ans.push(...String.fromCharCode(i + aCode).repeat(freq[i]));
182-
}
172+
for (let i = 0; i < 26; i++) {
173+
cnt[i] && ans.push(...String.fromCharCode(i + aCode).repeat(cnt[i]));
183174
}
184175
return ans;
185176
}

‎solution/1000-1099/1002.Find Common Characters/README_EN.md

Lines changed: 42 additions & 47 deletions
Original file line numberDiff line numberDiff line change
@@ -45,7 +45,11 @@ tags:
4545

4646
<!-- solution:start -->
4747

48-
### Solution 1
48+
### Solution 1: Counting
49+
50+
We use an array $cnt$ of length 26ドル$ to record the minimum number of times each character appears in all strings. Finally, we traverse the $cnt$ array and add characters with a count greater than 0ドル$ to the answer.
51+
52+
The time complexity is $O(n \sum w_i),ドル and the space complexity is $O(|\Sigma|)$. Here, $n$ is the length of the string array $words,ドル $w_i$ is the length of the $i$-th string in the array $words,ドル and $|\Sigma|$ is the size of the character set, which is 26ドル$ in this problem.
4953

5054
<!-- tabs:start -->
5155

@@ -56,13 +60,10 @@ class Solution:
5660
def commonChars(self, words: List[str]) -> List[str]:
5761
cnt = Counter(words[0])
5862
for w in words:
59-
ccnt = Counter(w)
60-
for c in cnt.keys():
61-
cnt[c] = min(cnt[c], ccnt[c])
62-
ans = []
63-
for c, v in cnt.items():
64-
ans.extend([c] * v)
65-
return ans
63+
t = Counter(w)
64+
for c in cnt:
65+
cnt[c] = min(cnt[c], t[c])
66+
return list(cnt.elements())
6667
```
6768

6869
#### Java
@@ -71,21 +72,19 @@ class Solution:
7172
class Solution {
7273
public List<String> commonChars(String[] words) {
7374
int[] cnt = new int[26];
74-
Arrays.fill(cnt, 10000);
75-
for (String w : words) {
76-
int[] ccnt = new int[26];
75+
Arrays.fill(cnt, 20000);
76+
for (var w : words) {
77+
int[] t = new int[26];
7778
for (int i = 0; i < w.length(); ++i) {
78-
++ccnt[w.charAt(i) - 'a'];
79+
++t[w.charAt(i) - 'a'];
7980
}
8081
for (int i = 0; i < 26; ++i) {
81-
cnt[i] = Math.min(cnt[i], ccnt[i]);
82+
cnt[i] = Math.min(cnt[i], t[i]);
8283
}
8384
}
8485
List<String> ans = new ArrayList<>();
8586
for (int i = 0; i < 26; ++i) {
86-
while (cnt[i]-- > 0) {
87-
ans.add(String.valueOf((char) (i + 'a')));
88-
}
87+
ans.addAll(Collections.nCopies(cnt[i], String.valueOf((char) ('a' + i))));
8988
}
9089
return ans;
9190
}
@@ -98,21 +97,20 @@ class Solution {
9897
class Solution {
9998
public:
10099
vector<string> commonChars(vector<string>& words) {
101-
int cnt[26];
102-
memset(cnt, 0x3f, sizeof(cnt));
103-
for (auto& w : words) {
104-
int ccnt[26]{};
105-
for (char& c : w) {
106-
++ccnt[c - 'a'];
100+
vector<int> cnt(26, 20000);
101+
for (const auto& w : words) {
102+
vector<int> t(26, 0);
103+
for (char c : w) {
104+
++t[c - 'a'];
107105
}
108106
for (int i = 0; i < 26; ++i) {
109-
cnt[i] = min(cnt[i], ccnt[i]);
107+
cnt[i] = min(cnt[i], t[i]);
110108
}
111109
}
112110
vector<string> ans;
113111
for (int i = 0; i < 26; ++i) {
114-
while (cnt[i]--) {
115-
ans.emplace_back(1, i + 'a');
112+
for (int j = 0; j < cnt[i]; ++j) {
113+
ans.push_back(string(1, 'a' + i));
116114
}
117115
}
118116
return ans;
@@ -124,49 +122,46 @@ public:
124122
125123
```go
126124
func commonChars(words []string) (ans []string) {
127-
cnt := [26]int{}
125+
cnt := make([]int, 26)
128126
for i := range cnt {
129-
cnt[i] = 1 << 30
127+
cnt[i] = 20000
130128
}
131129
for _, w := range words {
132-
ccnt := [26]int{}
130+
t := make([]int, 26)
133131
for _, c := range w {
134-
ccnt[c-'a']++
132+
t[c-'a']++
135133
}
136-
for i, v := range cnt {
137-
cnt[i] = min(v, ccnt[i])
134+
for i:= 0; i < 26; i++ {
135+
cnt[i] = min(cnt[i], t[i])
138136
}
139137
}
140-
for i, v := range cnt {
141-
for v > 0 {
142-
ans = append(ans, string(i+'a'))
143-
v--
138+
for i := 0; i < 26; i++ {
139+
for j := 0; j < cnt[i]; j++ {
140+
ans = append(ans, string('a'+rune(i)))
144141
}
145142
}
146-
return
143+
return ans
147144
}
148145
```
149146

150147
#### TypeScript
151148

152149
```ts
153150
function commonChars(words: string[]): string[] {
154-
const freq:number[] = Array(26).fill(Number.POSITIVE_INFINITY);
151+
const cnt = Array(26).fill(20000);
155152
const aCode = 'a'.charCodeAt(0);
156-
for (const word of words) {
157-
const t:number[] = Array(26).fill(0);
158-
for (const c of word) {
159-
++t[c.charCodeAt(0) - aCode];
153+
for (const w of words) {
154+
const t = Array(26).fill(0);
155+
for (const c of w) {
156+
t[c.charCodeAt(0) - aCode]++;
160157
}
161-
for (let i = 0; i < 26; ++i) {
162-
freq[i] = Math.min(freq[i], t[i]);
158+
for (let i = 0; i < 26; i++) {
159+
cnt[i] = Math.min(cnt[i], t[i]);
163160
}
164161
}
165162
const ans: string[] = [];
166-
for (let i = 0; i < 26; ++i) {
167-
if (freq[i]) {
168-
ans.push(...String.fromCharCode(i + aCode).repeat(freq[i]));
169-
}
163+
for (let i = 0; i < 26; i++) {
164+
cnt[i] && ans.push(...String.fromCharCode(i + aCode).repeat(cnt[i]));
170165
}
171166
return ans;
172167
}

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