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feat: update solutions to lc problem: No.2744 (doocs#2228)
No.2744.Find Maximum Number of String Pairs
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‎solution/2700-2799/2744.Find Maximum Number of String Pairs/README.md

Lines changed: 22 additions & 24 deletions
Original file line numberDiff line numberDiff line change
@@ -67,11 +67,11 @@
6767

6868
我们可以用哈希表 $cnt$ 来存储数组 $words$ 中每个字符串的反转字符串出现的次数。
6969

70-
遍历数组 $words$，对于每个字符串 $w$，我们直接将 $cnt[w]$ 的值加到答案中，然后将 $w$ 的反转字符串出现的次数加 $1$。
70+
遍历数组 $words$，对于每个字符串 $w$，我们将其反转字符串 $w$ 的出现次数加到答案中，然后将 $w$ 的出现次数加 $1$。
7171

72-
遍历结束后，即可得到最大匹配数目。
72+
最后返回答案。
7373

74-
时间复杂度 $O(L)$，空间复杂度 $O(L)$，其中 $L$ 是数组 $words$ 中字符串的长度之和。
74+
时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是数组 $words$ 的长度。
7575

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

@@ -81,19 +81,20 @@ class Solution:
8181
cnt = Counter()
8282
ans = 0
8383
for w in words:
84-
ans += cnt[w]
85-
cnt[w[::-1]] += 1
84+
ans += cnt[w[::-1]]
85+
cnt[w] += 1
8686
return ans
8787
```
8888

8989
```java
9090
class Solution {
9191
public int maximumNumberOfStringPairs(String[] words) {
92-
Map<String, Integer> cnt = new HashMap<>(words.length);
92+
Map<Integer, Integer> cnt = new HashMap<>();
9393
int ans = 0;
94-
for (String w : words) {
95-
ans += cnt.getOrDefault(w, 0);
96-
cnt.merge(new StringBuilder(w).reverse().toString(), 1, Integer::sum);
94+
for (var w : words) {
95+
int a = w.charAt(0) - 'a', b = w.charAt(1) - 'a';
96+
ans += cnt.getOrDefault(b << 5 | a, 0);
97+
cnt.merge(a << 5 | b, 1, Integer::sum);
9798
}
9899
return ans;
99100
}
@@ -104,12 +105,12 @@ class Solution {
104105
class Solution {
105106
public:
106107
int maximumNumberOfStringPairs(vector<string>& words) {
107-
unordered_map<string, int> cnt;
108+
unordered_map<int, int> cnt;
108109
int ans = 0;
109110
for (auto& w : words) {
110-
ans += cnt[w];
111-
reverse(w.begin(), w.end());
112-
cnt[w]++;
111+
int a = w[0] - 'a', b = w[1] - 'a';
112+
ans += cnt[b << 5 | a];
113+
cnt[a << 5 | b]++;
113114
}
114115
return ans;
115116
}
@@ -118,27 +119,24 @@ public:
118119
119120
```go
120121
func maximumNumberOfStringPairs(words []string) (ans int) {
121-
cnt := map[string]int{}
122+
cnt := map[int]int{}
122123
for _, w := range words {
123-
ans += cnt[w]
124-
s := []byte(w)
125-
for i, j := 0, len(s)-1; i < j; i, j = i+1, j-1 {
126-
s[i], s[j] = s[j], s[i]
127-
}
128-
cnt[string(s)]++
124+
a, b := int(w[0]-'a'), int(w[1]-'a')
125+
ans += cnt[b<<5|a]
126+
cnt[a<<5|b]++
129127
}
130128
return
131129
}
132130
```
133131

134132
```ts
135133
function maximumNumberOfStringPairs(words: string[]): number {
136-
const cnt: Map<string, number> = new Map();
134+
const cnt: { [key: number]: number } = {};
137135
let ans = 0;
138136
for (const w of words) {
139-
ans += cnt.get(w) || 0;
140-
const s = w.split('').reverse().join('');
141-
cnt.set(s, (cnt.get(s) || 0) + 1);
137+
const [a, b] = [w.charCodeAt(0) - 97, w.charCodeAt(w.length - 1) - 97];
138+
ans += cnt[(b << 5) | a] || 0;
139+
cnt[(a << 5) | b] = (cnt[(a << 5) | b] || 0) + 1;
142140
}
143141
return ans;
144142
}

‎solution/2700-2799/2744.Find Maximum Number of String Pairs/README_EN.md

Lines changed: 23 additions & 25 deletions
Original file line numberDiff line numberDiff line change
@@ -60,13 +60,13 @@ It can be proven that 1 is the maximum number of pairs that can be formed.
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6161
### Solution 1: Hash Table
6262

63-
We can use a hash table $cnt$ to store the number of occurrences of each string's reversed string in the array $words$.
63+
We can use a hash table $cnt$ to store the number of occurrences of each reversed string in the array $words$.
6464

65-
Traverse the array $words$, for each string $w$, we directly add the value of $cnt[w]$ to the answer, then increase the occurrence count of $w$'s reversed string by $1$.
65+
We iterate through the array $words$. For each string $w$, we add the number of occurrences of its reversed string to the answer, then increment the count of $w$ by $1$.
6666

67-
After the traversal ends, we can get the maximum number of matches.
67+
Finally, we return the answer.
6868

69-
The time complexity is $O(L)$, and the space complexity is $O(L)$, where $L$ is the sum of the lengths of the strings in the array $words$.
69+
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $words$.
7070

7171
<!-- tabs:start -->
7272

@@ -76,19 +76,20 @@ class Solution:
7676
cnt = Counter()
7777
ans = 0
7878
for w in words:
79-
ans += cnt[w]
80-
cnt[w[::-1]] += 1
79+
ans += cnt[w[::-1]]
80+
cnt[w] += 1
8181
return ans
8282
```
8383

8484
```java
8585
class Solution {
8686
public int maximumNumberOfStringPairs(String[] words) {
87-
Map<String, Integer> cnt = new HashMap<>(words.length);
87+
Map<Integer, Integer> cnt = new HashMap<>();
8888
int ans = 0;
89-
for (String w : words) {
90-
ans += cnt.getOrDefault(w, 0);
91-
cnt.merge(new StringBuilder(w).reverse().toString(), 1, Integer::sum);
89+
for (var w : words) {
90+
int a = w.charAt(0) - 'a', b = w.charAt(1) - 'a';
91+
ans += cnt.getOrDefault(b << 5 | a, 0);
92+
cnt.merge(a << 5 | b, 1, Integer::sum);
9293
}
9394
return ans;
9495
}
@@ -99,12 +100,12 @@ class Solution {
99100
class Solution {
100101
public:
101102
int maximumNumberOfStringPairs(vector<string>& words) {
102-
unordered_map<string, int> cnt;
103+
unordered_map<int, int> cnt;
103104
int ans = 0;
104105
for (auto& w : words) {
105-
ans += cnt[w];
106-
reverse(w.begin(), w.end());
107-
cnt[w]++;
106+
int a = w[0] - 'a', b = w[1] - 'a';
107+
ans += cnt[b << 5 | a];
108+
cnt[a << 5 | b]++;
108109
}
109110
return ans;
110111
}
@@ -113,27 +114,24 @@ public:
113114
114115
```go
115116
func maximumNumberOfStringPairs(words []string) (ans int) {
116-
cnt := map[string]int{}
117+
cnt := map[int]int{}
117118
for _, w := range words {
118-
ans += cnt[w]
119-
s := []byte(w)
120-
for i, j := 0, len(s)-1; i < j; i, j = i+1, j-1 {
121-
s[i], s[j] = s[j], s[i]
122-
}
123-
cnt[string(s)]++
119+
a, b := int(w[0]-'a'), int(w[1]-'a')
120+
ans += cnt[b<<5|a]
121+
cnt[a<<5|b]++
124122
}
125123
return
126124
}
127125
```
128126

129127
```ts
130128
function maximumNumberOfStringPairs(words: string[]): number {
131-
const cnt: Map<string, number> = new Map();
129+
const cnt: { [key: number]: number } = {};
132130
let ans = 0;
133131
for (const w of words) {
134-
ans += cnt.get(w) || 0;
135-
const s = w.split('').reverse().join('');
136-
cnt.set(s, (cnt.get(s) || 0) + 1);
132+
const [a, b] = [w.charCodeAt(0) - 97, w.charCodeAt(w.length - 1) - 97];
133+
ans += cnt[(b << 5) | a] || 0;
134+
cnt[(a << 5) | b] = (cnt[(a << 5) | b] || 0) + 1;
137135
}
138136
return ans;
139137
}

‎solution/2700-2799/2744.Find Maximum Number of String Pairs/Solution.cpp

Lines changed: 4 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -1,12 +1,12 @@
11
class Solution {
22
public:
33
int maximumNumberOfStringPairs(vector<string>& words) {
4-
unordered_map<string, int> cnt;
4+
unordered_map<int, int> cnt;
55
int ans = 0;
66
for (auto& w : words) {
7-
ans += cnt[w];
8-
reverse(w.begin(), w.end());
9-
cnt[w]++;
7+
int a = w[0] - 'a', b = w[1] - 'a';
8+
ans += cnt[b << 5 | a];
9+
cnt[a << 5 | b]++;
1010
}
1111
return ans;
1212
}

‎solution/2700-2799/2744.Find Maximum Number of String Pairs/Solution.go

Lines changed: 4 additions & 7 deletions
Original file line numberDiff line numberDiff line change
@@ -1,12 +1,9 @@
11
func maximumNumberOfStringPairs(words []string) (ans int) {
2-
cnt := map[string]int{}
2+
cnt := map[int]int{}
33
for _, w := range words {
4-
ans += cnt[w]
5-
s := []byte(w)
6-
for i, j := 0, len(s)-1; i < j; i, j = i+1, j-1 {
7-
s[i], s[j] = s[j], s[i]
8-
}
9-
cnt[string(s)]++
4+
a, b := int(w[0]-'a'), int(w[1]-'a')
5+
ans += cnt[b<<5|a]
6+
cnt[a<<5|b]++
107
}
118
return
129
}

‎solution/2700-2799/2744.Find Maximum Number of String Pairs/Solution.java

Lines changed: 5 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -1,10 +1,11 @@
11
class Solution {
22
public int maximumNumberOfStringPairs(String[] words) {
3-
Map<String, Integer> cnt = new HashMap<>(words.length);
3+
Map<Integer, Integer> cnt = new HashMap<>();
44
int ans = 0;
5-
for (String w : words) {
6-
ans += cnt.getOrDefault(w, 0);
7-
cnt.merge(new StringBuilder(w).reverse().toString(), 1, Integer::sum);
5+
for (var w : words) {
6+
int a = w.charAt(0) - 'a', b = w.charAt(1) - 'a';
7+
ans += cnt.getOrDefault(b << 5 | a, 0);
8+
cnt.merge(a << 5 | b, 1, Integer::sum);
89
}
910
return ans;
1011
}

‎solution/2700-2799/2744.Find Maximum Number of String Pairs/Solution.py

Lines changed: 2 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -3,6 +3,6 @@ def maximumNumberOfStringPairs(self, words: List[str]) -> int:
33
cnt = Counter()
44
ans = 0
55
for w in words:
6-
ans += cnt[w]
7-
cnt[w[::-1]] += 1
6+
ans += cnt[w[::-1]]
7+
cnt[w] += 1
88
return ans

‎solution/2700-2799/2744.Find Maximum Number of String Pairs/Solution.ts

Lines changed: 4 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -1,10 +1,10 @@
11
function maximumNumberOfStringPairs(words: string[]): number {
2-
const cnt: Map<string, number> = new Map();
2+
const cnt: { [key: number]: number } = {};
33
let ans = 0;
44
for (const w of words) {
5-
ans += cnt.get(w) || 0;
6-
const s = w.split('').reverse().join('');
7-
cnt.set(s, (cnt.get(s) || 0) + 1);
5+
const [a, b] = [w.charCodeAt(0) - 97, w.charCodeAt(w.length - 1) - 97];
6+
ans += cnt[(b << 5) | a] || 0;
7+
cnt[(a << 5) | b] = (cnt[(a << 5) | b] || 0) + 1;
88
}
99
return ans;
1010
}

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