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feat: add solutions to lc problem: No.2068
No.2068.Check Whether Two Strings are Almost Equivalent
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‎solution/2000-2099/2068.Check Whether Two Strings are Almost Equivalent/README.md

+67-27
Original file line numberDiff line numberDiff line change
@@ -60,7 +60,11 @@
6060

6161
<!-- 这里可写通用的实现逻辑 -->
6262

63-
**方法一：哈希表计数**
63+
**方法一：计数**
64+
65+
我们可以创建一个长度为 $26$ 的数组 $cnt$，记录两个字符串中每个字母出现的次数之差。最后遍历 $cnt$，如果有任意一个字母出现的次数之差大于 $3$，则返回 `false`，否则返回 `true`。
66+
67+
时间复杂度 $O(n)$，空间复杂度 $O(C)$。其中 $n$ 是字符串的长度；而 $C$ 是字符集的大小，本题中 $C = 26$。
6468

6569
<!-- tabs:start -->
6670

@@ -71,12 +75,10 @@
7175
```python
7276
class Solution:
7377
def checkAlmostEquivalent(self, word1: str, word2: str) -> bool:
74-
counter = defaultdict(int)
75-
for c in word1:
76-
counter[c] += 1
78+
cnt = Counter(word1)
7779
for c in word2:
78-
counter[c] -= 1
79-
return all(abs(x) <= 3 for x in counter.values())
80+
cnt[c] -= 1
81+
return all(abs(x) <= 3 for x in cnt.values())
8082
```
8183

8284
### **Java**
@@ -86,15 +88,15 @@ class Solution:
8688
```java
8789
class Solution {
8890
public boolean checkAlmostEquivalent(String word1, String word2) {
89-
int[] counter = new int[26];
90-
for (char c : word1.toCharArray()) {
91-
++counter[c - 'a'];
91+
int[] cnt = new int[26];
92+
for (int i = 0; i < word1.length(); ++i) {
93+
++cnt[word1.charAt(i) - 'a'];
9294
}
93-
for (char c : word2.toCharArray()) {
94-
--counter[c - 'a'];
95+
for (int i = 0; i < word2.length(); ++i) {
96+
--cnt[word2.charAt(i) - 'a'];
9597
}
96-
for (int i = 0; i < 26; ++i) {
97-
if (Math.abs(counter[i]) > 3) {
98+
for (int x : cnt) {
99+
if (Math.abs(x) > 3) {
98100
return false;
99101
}
100102
}
@@ -109,13 +111,19 @@ class Solution {
109111
class Solution {
110112
public:
111113
bool checkAlmostEquivalent(string word1, string word2) {
112-
vector<int> counter(26);
113-
for (char& c : word1) ++counter[c - 'a'];
114-
for (char& c : word2) --counter[c - 'a'];
115-
for (int i = 0; i < 26; ++i)
116-
if (abs(counter[i]) > 3)
117-
return 0;
118-
return 1;
114+
int cnt[26]{};
115+
for (char& c : word1) {
116+
++cnt[c - 'a'];
117+
}
118+
for (char& c : word2) {
119+
--cnt[c - 'a'];
120+
}
121+
for (int i = 0; i < 26; ++i) {
122+
if (abs(cnt[i]) > 3) {
123+
return false;
124+
}
125+
}
126+
return true;
119127
}
120128
};
121129
```
@@ -124,22 +132,54 @@ public:
124132
125133
```go
126134
func checkAlmostEquivalent(word1 string, word2 string) bool {
127-
counter := make([]int, 26)
128-
for i := range word1 {
129-
counter[word1[i]-'a']++
135+
cnt := [26]int{}
136+
for _, c := range word1 {
137+
cnt[c-'a']++
130138
}
131-
for i := range word2 {
132-
counter[word2[i]-'a']--
139+
for _, c := range word2 {
140+
cnt[c-'a']--
133141
}
134-
for _, v := range counter {
135-
if v > 3 || -v > 3 {
142+
for _, x := range cnt {
143+
if x > 3 || x < -3 {
136144
return false
137145
}
138146
}
139147
return true
140148
}
141149
```
142150

151+
### **TypeScript**
152+
153+
```ts
154+
function checkAlmostEquivalent(word1: string, word2: string): boolean {
155+
const cnt: number[] = new Array(26).fill(0);
156+
for (const c of word1) {
157+
++cnt[c.charCodeAt(0) - 97];
158+
}
159+
for (const c of word2) {
160+
--cnt[c.charCodeAt(0) - 97];
161+
}
162+
return cnt.every(x => Math.abs(x) <= 3);
163+
}
164+
```
165+
166+
### **C#**
167+
168+
```cs
169+
public class Solution {
170+
public bool CheckAlmostEquivalent(string word1, string word2) {
171+
int[] cnt = new int[26];
172+
foreach (var c in word1) {
173+
cnt[c - 'a']++;
174+
}
175+
foreach (var c in word2) {
176+
cnt[c - 'a']--;
177+
}
178+
return cnt.All(x => Math.Abs(x) <= 3);
179+
}
180+
}
181+
```
182+
143183
### **PHP**
144184

145185
```php

‎solution/2000-2099/2068.Check Whether Two Strings are Almost Equivalent/README_EN.md

+68-26
Original file line numberDiff line numberDiff line change
@@ -57,35 +57,39 @@ The difference is 4, which is more than the allowed 3.
5757

5858
## Solutions
5959

60+
**Approach 1: Counting**
61+
62+
We can create an array $cnt$ of length $26$ to record the difference in the number of times each letter appears in the two strings. Then we traverse $cnt$, if any letter appears the difference in the number of times greater than $3$, then return `false`, otherwise return `true`.
63+
64+
The time complexity is $O(n)$ and the space complexity is $O(C)$. Where $n$ is the length of the string, and $C$ is the size of the character set, and in this question $C = 26$.
65+
6066
<!-- tabs:start -->
6167

6268
### **Python3**
6369

6470
```python
6571
class Solution:
6672
def checkAlmostEquivalent(self, word1: str, word2: str) -> bool:
67-
counter = defaultdict(int)
68-
for c in word1:
69-
counter[c] += 1
73+
cnt = Counter(word1)
7074
for c in word2:
71-
counter[c] -= 1
72-
return all(abs(x) <= 3 for x in counter.values())
75+
cnt[c] -= 1
76+
return all(abs(x) <= 3 for x in cnt.values())
7377
```
7478

7579
### **Java**
7680

7781
```java
7882
class Solution {
7983
public boolean checkAlmostEquivalent(String word1, String word2) {
80-
int[] counter = new int[26];
81-
for (char c : word1.toCharArray()) {
82-
++counter[c - 'a'];
84+
int[] cnt = new int[26];
85+
for (int i = 0; i < word1.length(); ++i) {
86+
++cnt[word1.charAt(i) - 'a'];
8387
}
84-
for (char c : word2.toCharArray()) {
85-
--counter[c - 'a'];
88+
for (int i = 0; i < word2.length(); ++i) {
89+
--cnt[word2.charAt(i) - 'a'];
8690
}
87-
for (int i = 0; i < 26; ++i) {
88-
if (Math.abs(counter[i]) > 3) {
91+
for (int x : cnt) {
92+
if (Math.abs(x) > 3) {
8993
return false;
9094
}
9195
}
@@ -100,13 +104,19 @@ class Solution {
100104
class Solution {
101105
public:
102106
bool checkAlmostEquivalent(string word1, string word2) {
103-
vector<int> counter(26);
104-
for (char& c : word1) ++counter[c - 'a'];
105-
for (char& c : word2) --counter[c - 'a'];
106-
for (int i = 0; i < 26; ++i)
107-
if (abs(counter[i]) > 3)
108-
return 0;
109-
return 1;
107+
int cnt[26]{};
108+
for (char& c : word1) {
109+
++cnt[c - 'a'];
110+
}
111+
for (char& c : word2) {
112+
--cnt[c - 'a'];
113+
}
114+
for (int i = 0; i < 26; ++i) {
115+
if (abs(cnt[i]) > 3) {
116+
return false;
117+
}
118+
}
119+
return true;
110120
}
111121
};
112122
```
@@ -115,22 +125,54 @@ public:
115125
116126
```go
117127
func checkAlmostEquivalent(word1 string, word2 string) bool {
118-
counter := make([]int, 26)
119-
for i := range word1 {
120-
counter[word1[i]-'a']++
128+
cnt := [26]int{}
129+
for _, c := range word1 {
130+
cnt[c-'a']++
121131
}
122-
for i := range word2 {
123-
counter[word2[i]-'a']--
132+
for _, c := range word2 {
133+
cnt[c-'a']--
124134
}
125-
for _, v := range counter {
126-
if v > 3 || -v > 3 {
135+
for _, x := range cnt {
136+
if x > 3 || x < -3 {
127137
return false
128138
}
129139
}
130140
return true
131141
}
132142
```
133143

144+
### **TypeScript**
145+
146+
```ts
147+
function checkAlmostEquivalent(word1: string, word2: string): boolean {
148+
const cnt: number[] = new Array(26).fill(0);
149+
for (const c of word1) {
150+
++cnt[c.charCodeAt(0) - 97];
151+
}
152+
for (const c of word2) {
153+
--cnt[c.charCodeAt(0) - 97];
154+
}
155+
return cnt.every(x => Math.abs(x) <= 3);
156+
}
157+
```
158+
159+
### **C#**
160+
161+
```cs
162+
public class Solution {
163+
public bool CheckAlmostEquivalent(string word1, string word2) {
164+
int[] cnt = new int[26];
165+
foreach (var c in word1) {
166+
cnt[c - 'a']++;
167+
}
168+
foreach (var c in word2) {
169+
cnt[c - 'a']--;
170+
}
171+
return cnt.All(x => Math.Abs(x) <= 3);
172+
}
173+
}
174+
```
175+
134176
### **PHP**
135177

136178
```php

‎solution/2000-2099/2068.Check Whether Two Strings are Almost Equivalent/Solution.cpp

+13-7
Original file line numberDiff line numberDiff line change
@@ -1,12 +1,18 @@
11
class Solution {
22
public:
33
bool checkAlmostEquivalent(string word1, string word2) {
4-
vector<int> counter(26);
5-
for (char& c : word1) ++counter[c - 'a'];
6-
for (char& c : word2) --counter[c - 'a'];
7-
for (int i = 0; i < 26; ++i)
8-
if (abs(counter[i]) > 3)
9-
return 0;
10-
return 1;
4+
int cnt[26]{};
5+
for (char& c : word1) {
6+
++cnt[c - 'a'];
7+
}
8+
for (char& c : word2) {
9+
--cnt[c - 'a'];
10+
}
11+
for (int i = 0; i < 26; ++i) {
12+
if (abs(cnt[i]) > 3) {
13+
return false;
14+
}
15+
}
16+
return true;
1117
}
1218
};

‎solution/2000-2099/2068.Check Whether Two Strings are Almost Equivalent/Solution.cs

+12
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,12 @@
1+
public class Solution {
2+
public bool CheckAlmostEquivalent(string word1, string word2) {
3+
int[] cnt = new int[26];
4+
foreach (var c in word1) {
5+
cnt[c - 'a']++;
6+
}
7+
foreach (var c in word2) {
8+
cnt[c - 'a']--;
9+
}
10+
return cnt.All(x => Math.Abs(x) <= 3);
11+
}
12+
}

‎solution/2000-2099/2068.Check Whether Two Strings are Almost Equivalent/Solution.go

+7-7
Original file line numberDiff line numberDiff line change
@@ -1,13 +1,13 @@
11
func checkAlmostEquivalent(word1 string, word2 string) bool {
2-
counter := make([]int, 26)
3-
for i := range word1 {
4-
counter[word1[i]-'a']++
2+
cnt := [26]int{}
3+
for _, c := range word1 {
4+
cnt[c-'a']++
55
}
6-
for i := range word2 {
7-
counter[word2[i]-'a']--
6+
for _, c := range word2 {
7+
cnt[c-'a']--
88
}
9-
for _, v := range counter {
10-
if v > 3 || -v > 3 {
9+
for _, x := range cnt {
10+
if x > 3 || x < -3 {
1111
return false
1212
}
1313
}

‎solution/2000-2099/2068.Check Whether Two Strings are Almost Equivalent/Solution.java

+7-7
Original file line numberDiff line numberDiff line change
@@ -1,14 +1,14 @@
11
class Solution {
22
public boolean checkAlmostEquivalent(String word1, String word2) {
3-
int[] counter = new int[26];
4-
for (char c : word1.toCharArray()) {
5-
++counter[c - 'a'];
3+
int[] cnt = new int[26];
4+
for (int i = 0; i < word1.length(); ++i) {
5+
++cnt[word1.charAt(i) - 'a'];
66
}
7-
for (char c : word2.toCharArray()) {
8-
--counter[c - 'a'];
7+
for (int i = 0; i < word2.length(); ++i) {
8+
--cnt[word2.charAt(i) - 'a'];
99
}
10-
for (int i = 0; i < 26; ++i) {
11-
if (Math.abs(counter[i]) > 3) {
10+
for (int x : cnt) {
11+
if (Math.abs(x) > 3) {
1212
return false;
1313
}
1414
}

‎solution/2000-2099/2068.Check Whether Two Strings are Almost Equivalent/Solution.py

+3-5
Original file line numberDiff line numberDiff line change
@@ -1,8 +1,6 @@
11
class Solution:
22
def checkAlmostEquivalent(self, word1: str, word2: str) -> bool:
3-
counter = defaultdict(int)
4-
for c in word1:
5-
counter[c] += 1
3+
cnt = Counter(word1)
64
for c in word2:
7-
counter[c] -= 1
8-
return all(abs(x) <= 3 for x in counter.values())
5+
cnt[c] -= 1
6+
return all(abs(x) <= 3 for x in cnt.values())

‎solution/2000-2099/2068.Check Whether Two Strings are Almost Equivalent/Solution.ts

+10
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,10 @@
1+
function checkAlmostEquivalent(word1: string, word2: string): boolean {
2+
const cnt: number[] = new Array(26).fill(0);
3+
for (const c of word1) {
4+
++cnt[c.charCodeAt(0) - 97];
5+
}
6+
for (const c of word2) {
7+
--cnt[c.charCodeAt(0) - 97];
8+
}
9+
return cnt.every(x => Math.abs(x) <= 3);
10+
}

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