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Commit a88a827

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feat: add solutions to lc problem: No.2452 (doocs#2778)
No.2452.Words Within Two Edits of Dictionary
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12 files changed

+163
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‎solution/2300-2399/2395.Find Subarrays With Equal Sum/README_EN.md‎

Lines changed: 7 additions & 1 deletion
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@@ -48,7 +48,13 @@ Note that even though the subarrays have the same content, the two subarrays are
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## Solutions
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51-
### Solution 1
51+
### Solution 1: Hash Table
52+
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We can traverse the array $nums,ドル and use a hash table $vis$ to record the sum of every two adjacent elements in the array. If the sum of the current two elements has already appeared in the hash table, then return `true`. Otherwise, add the sum of the current two elements to the hash table.
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If we finish traversing and haven't found two subarrays that meet the condition, return `false`.
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The time complexity is $O(n),ドル and the space complexity is $O(n),ドル where $n$ is the length of the array $nums$.
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<!-- tabs:start -->
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‎solution/2300-2399/2396.Strictly Palindromic Number/README.md‎

Lines changed: 3 additions & 3 deletions
Original file line numberDiff line numberDiff line change
@@ -46,11 +46,11 @@
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### 方法一:脑筋急转弯
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49-
当 $n=4$ 时,二进制表示为 100ドル,ドル不是回文串;
49+
当 $n = 4$ 时,二进制表示为 100ドル,ドル不是回文串;
5050

51-
当 $n \gt 4$ 时,此时 $n-2$ 的二进制表示为 12ドル,ドル不是回文串。
51+
当 $n \gt 4$ 时,此时 $n - 2$ 进制表示为 12ドル,ドル不是回文串。
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53-
因此,我们直接返回 `false` 即可。
53+
因此,我们可以直接返回 `false`。
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时间复杂度 $O(1),ドル空间复杂度 $O(1)$。
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‎solution/2300-2399/2396.Strictly Palindromic Number/README_EN.md‎

Lines changed: 9 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -43,7 +43,15 @@ Therefore, we return false.
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## Solutions
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46-
### Solution 1
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### Solution 1: Quick Thinking
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When $n = 4,ドル its binary representation is 100ドル,ドル which is not a palindrome;
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When $n \gt 4,ドル its $(n - 2)$-ary representation is 12ドル,ドル which is not a palindrome.
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Therefore, we can directly return `false`.
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The time complexity is $O(1),ドル and the space complexity is $O(1)$.
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<!-- tabs:start -->
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‎solution/2300-2399/2398.Maximum Number of Robots Within Budget/README.md‎

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Original file line numberDiff line numberDiff line change
@@ -50,7 +50,9 @@
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### 方法一:双指针 + 单调队列
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53-
问题实际上是求滑动窗口内的最大值,可以用单调队列来求解。只需要二分枚举窗口 $k$ 的大小,找到一个最大的 $k,ドル使得满足题目要求。
53+
问题实际上是求滑动窗口内的最大值,可以用单调队列来求解。
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55+
我们只需要二分枚举窗口 $k$ 的大小,找到一个最大的 $k,ドル使得满足题目要求。
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实现过程中,实际上不需要进行二分枚举,只需要将固定窗口改为双指针非固定窗口即可。
5658

‎solution/2300-2399/2398.Maximum Number of Robots Within Budget/README_EN.md‎

Lines changed: 9 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -44,7 +44,15 @@ It can be shown that it is not possible to run more than 3 consecutive robots wi
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## Solutions
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47-
### Solution 1
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### Solution 1: Two Pointers + Monotonic Queue
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The problem is essentially finding the maximum value within a sliding window, which can be solved using a monotonic queue.
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We only need to use binary search to enumerate the size of the window $k,ドル and find the largest $k$ that satisfies the problem requirements.
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53+
In the implementation process, we don't actually need to perform binary search enumeration. We just need to change the fixed window to a non-fixed window with double pointers.
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The time complexity is $O(n),ドル and the space complexity is $O(n),ドル where $n$ is the number of robots in the problem.
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<!-- tabs:start -->
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‎solution/2400-2499/2417.Closest Fair Integer/README_EN.md‎

Lines changed: 8 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -39,7 +39,14 @@
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## Solutions
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### Solution 1
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### Solution 1: Case Discussion
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We denote the number of digits of $n$ as $k,ドル and the number of odd and even digits as $a$ and $b$ respectively.
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- If $a = b,ドル then $n$ itself is `fair`, and we can directly return $n$;
47+
- Otherwise, if $k$ is odd, we can find the smallest `fair` number with $k+1$ digits, in the form of `10000111`. If $k$ is even, we can directly brute force `closestFair(n+1)`.
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The time complexity is $O(\sqrt{n} \times \log_{10} n)$.
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<!-- tabs:start -->
4552

‎solution/2400-2499/2452.Words Within Two Edits of Dictionary/README.md‎

Lines changed: 42 additions & 23 deletions
Original file line numberDiff line numberDiff line change
@@ -52,9 +52,9 @@
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### 方法一:暴力枚举
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遍历 `queries` 中的每个单词,对于每个单词,遍历 `dictionary` 中的每个单词,判断两个单词不同字符的位置数是否小于 $3,ドル如果是,则将该单词加入结果集。
55+
我们直接遍历数组 $\text{queries}$ 中的每个单词 $s,ドル再遍历数组 $\text{dictionary}$ 中的每个单词 $t,ドル如果存在一个单词 $t$ 与 $s$ 的编辑距离小于 $3,ドル则将 $s$ 加入答案数组中,然后退出内层循环的遍历。如果不存在这样的单词 $t,ドル则继续遍历下一个单词 $s$。
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时间复杂度 $O(m\times n\times k)$。其中 $m$ 和 $n$ 分别是 `queries` 和 `dictionary` 的长度,而 $k$ 是 `queries` 和 `dictionary` 中单词的长度。
57+
时间复杂度 $O(m\times n\times l),ドル其中 $m$ 和 $n$ 分别是数组 $\text{queries}$ 和 $\text{dictionary}$ 的长度,而 $l$ 是单词的长度。
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<!-- tabs:start -->
6060

@@ -102,7 +102,9 @@ public:
102102
for (auto& s : queries) {
103103
for (auto& t : dictionary) {
104104
int cnt = 0;
105-
for (int i = 0; i < s.size(); ++i) cnt += s[i] != t[i];
105+
for (int i = 0; i < s.size(); ++i) {
106+
cnt += s[i] != t[i];
107+
}
106108
if (cnt < 3) {
107109
ans.emplace_back(s);
108110
break;
@@ -137,15 +139,15 @@ func twoEditWords(queries []string, dictionary []string) (ans []string) {
137139
```ts
138140
function twoEditWords(queries: string[], dictionary: string[]): string[] {
139141
const n = queries[0].length;
140-
return queries.filter(querie => {
141-
for (const s of dictionary) {
142+
return queries.filter(s => {
143+
for (const t of dictionary) {
142144
let diff = 0;
143-
for (let i = 0; i < n; i++) {
144-
if (querie[i] !== s[i]&&++diff>2) {
145-
break;
145+
for (let i = 0; i < n; ++i) {
146+
if (s[i] !== t[i]) {
147+
++diff;
146148
}
147149
}
148-
if (diff <=2) {
150+
if (diff <3) {
149151
return true;
150152
}
151153
}
@@ -157,28 +159,45 @@ function twoEditWords(queries: string[], dictionary: string[]): string[] {
157159
```rust
158160
impl Solution {
159161
pub fn two_edit_words(queries: Vec<String>, dictionary: Vec<String>) -> Vec<String> {
160-
let n = queries[0].len();
161162
queries
162163
.into_iter()
163-
.filter(|querie| {
164-
for s in dictionary.iter() {
165-
let mut diff = 0;
166-
for i in 0..n {
167-
if querie.as_bytes()[i] != s.as_bytes()[i] {
168-
diff += 1;
169-
}
170-
}
171-
if diff <= 2 {
172-
return true;
173-
}
174-
}
175-
false
164+
.filter(|s| {
165+
dictionary.iter().any(|t| {
166+
s
167+
.chars()
168+
.zip(t.chars())
169+
.filter(|&(a, b)| a != b)
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.count() < 3
171+
})
176172
})
177173
.collect()
178174
}
179175
}
180176
```
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178+
```cs
179+
public class Solution {
180+
public IList<string> TwoEditWords(string[] queries, string[] dictionary) {
181+
var ans = new List<string>();
182+
foreach (var s in queries) {
183+
foreach (var t in dictionary) {
184+
int cnt = 0;
185+
for (int i = 0; i < s.Length; i++) {
186+
if (s[i] != t[i]) {
187+
cnt++;
188+
}
189+
}
190+
if (cnt < 3) {
191+
ans.Add(s);
192+
break;
193+
}
194+
}
195+
}
196+
return ans;
197+
}
198+
}
199+
```
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<!-- tabs:end -->
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<!-- end -->

‎solution/2400-2499/2452.Words Within Two Edits of Dictionary/README_EN.md‎

Lines changed: 45 additions & 22 deletions
Original file line numberDiff line numberDiff line change
@@ -47,7 +47,11 @@ Applying any two edits to &quot;yes&quot; cannot make it equal to &quot;not&quot
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## Solutions
4949

50-
### Solution 1
50+
### Solution 1: Brute Force Enumeration
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52+
We directly traverse each word $s$ in the array $\text{queries},ドル and then traverse each word $t$ in the array $\text{dictionary}$. If there exists a word $t$ whose edit distance from $s$ is less than 3ドル,ドル we add $s$ to the answer array and then exit the inner loop. If there is no such word $t,ドル we continue to traverse the next word $s$.
53+
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The time complexity is $O(m \times n \times l),ドル where $m$ and $n$ are the lengths of the arrays $\text{queries}$ and $\text{dictionary}$ respectively, and $l$ is the length of the word.
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<!-- tabs:start -->
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@@ -95,7 +99,9 @@ public:
9599
for (auto& s : queries) {
96100
for (auto& t : dictionary) {
97101
int cnt = 0;
98-
for (int i = 0; i < s.size(); ++i) cnt += s[i] != t[i];
102+
for (int i = 0; i < s.size(); ++i) {
103+
cnt += s[i] != t[i];
104+
}
99105
if (cnt < 3) {
100106
ans.emplace_back(s);
101107
break;
@@ -130,15 +136,15 @@ func twoEditWords(queries []string, dictionary []string) (ans []string) {
130136
```ts
131137
function twoEditWords(queries: string[], dictionary: string[]): string[] {
132138
const n = queries[0].length;
133-
return queries.filter(querie => {
134-
for (const s of dictionary) {
139+
return queries.filter(s => {
140+
for (const t of dictionary) {
135141
let diff = 0;
136-
for (let i = 0; i < n; i++) {
137-
if (querie[i] !== s[i]&&++diff>2) {
138-
break;
142+
for (let i = 0; i < n; ++i) {
143+
if (s[i] !== t[i]) {
144+
++diff;
139145
}
140146
}
141-
if (diff <=2) {
147+
if (diff <3) {
142148
return true;
143149
}
144150
}
@@ -150,28 +156,45 @@ function twoEditWords(queries: string[], dictionary: string[]): string[] {
150156
```rust
151157
impl Solution {
152158
pub fn two_edit_words(queries: Vec<String>, dictionary: Vec<String>) -> Vec<String> {
153-
let n = queries[0].len();
154159
queries
155160
.into_iter()
156-
.filter(|querie| {
157-
for s in dictionary.iter() {
158-
let mut diff = 0;
159-
for i in 0..n {
160-
if querie.as_bytes()[i] != s.as_bytes()[i] {
161-
diff += 1;
162-
}
163-
}
164-
if diff <= 2 {
165-
return true;
166-
}
167-
}
168-
false
161+
.filter(|s| {
162+
dictionary.iter().any(|t| {
163+
s
164+
.chars()
165+
.zip(t.chars())
166+
.filter(|&(a, b)| a != b)
167+
.count() < 3
168+
})
169169
})
170170
.collect()
171171
}
172172
}
173173
```
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175+
```cs
176+
public class Solution {
177+
public IList<string> TwoEditWords(string[] queries, string[] dictionary) {
178+
var ans = new List<string>();
179+
foreach (var s in queries) {
180+
foreach (var t in dictionary) {
181+
int cnt = 0;
182+
for (int i = 0; i < s.Length; i++) {
183+
if (s[i] != t[i]) {
184+
cnt++;
185+
}
186+
}
187+
if (cnt < 3) {
188+
ans.Add(s);
189+
break;
190+
}
191+
}
192+
}
193+
return ans;
194+
}
195+
}
196+
```
197+
175198
<!-- tabs:end -->
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177200
<!-- end -->

‎solution/2400-2499/2452.Words Within Two Edits of Dictionary/Solution.cpp‎

Lines changed: 3 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -5,7 +5,9 @@ class Solution {
55
for (auto& s : queries) {
66
for (auto& t : dictionary) {
77
int cnt = 0;
8-
for (int i = 0; i < s.size(); ++i) cnt += s[i] != t[i];
8+
for (int i = 0; i < s.size(); ++i) {
9+
cnt += s[i] != t[i];
10+
}
911
if (cnt < 3) {
1012
ans.emplace_back(s);
1113
break;
Lines changed: 20 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,20 @@
1+
public class Solution {
2+
public IList<string> TwoEditWords(string[] queries, string[] dictionary) {
3+
var ans = new List<string>();
4+
foreach (var s in queries) {
5+
foreach (var t in dictionary) {
6+
int cnt = 0;
7+
for (int i = 0; i < s.Length; i++) {
8+
if (s[i] != t[i]) {
9+
cnt++;
10+
}
11+
}
12+
if (cnt < 3) {
13+
ans.Add(s);
14+
break;
15+
}
16+
}
17+
}
18+
return ans;
19+
}
20+
}

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