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Commit cc4815a

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feat: add solutions to lc problem: No.3412 (doocs#3930)
No.3412.Find Mirror Score of a String
1 parent c644d65 commit cc4815a

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7 files changed

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7 files changed

+319
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‎solution/3400-3499/3412.Find Mirror Score of a String/README.md‎

Lines changed: 110 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -77,32 +77,138 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3412.Fi
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<!-- solution:start -->
7979

80-
### 方法一
80+
### 方法一:哈希表
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我们可以用一个哈希表 $\textit{d}$ 来存储每个未标记的字符的下标列表,其中键是字符,值是下标列表。
83+
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我们遍历字符串 $\textit{s},ドル对于每个字符 $\textit{x},ドル我们找到其镜像字符 $\textit{y},ドル如果 $\textit{d}$ 中存在 $\textit{y},ドル我们就取出 $\textit{y}$ 对应的下标列表 $\textit{ls},ドル取出 $\textit{ls}$ 的最后一个元素 $\textit{j},ドル并将 $\textit{j}$ 从 $\textit{ls}$ 中移除。如果 $\textit{ls}$ 变为空,我们就将 $\textit{y}$ 从 $\textit{d}$ 中移除。此时,我们就找到了一个满足条件的下标对 $(\textit{j}, \textit{i}),ドル并将 $\textit{i} - \textit{j}$ 加到答案中。否则,我们将 $\textit{x}$ 加入到 $\textit{d}$ 中。
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时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 为字符串 $\textit{s}$ 的长度。
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<!-- tabs:start -->
8389

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#### Python3
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```python
87-
93+
class Solution:
94+
def calculateScore(self, s: str) -> int:
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d = defaultdict(list)
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ans = 0
97+
for i, x in enumerate(s):
98+
y = chr(ord("a") + ord("z") - ord(x))
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if d[y]:
100+
j = d[y].pop()
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ans += i - j
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else:
103+
d[x].append(i)
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return ans
88105
```
89106

90107
#### Java
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92109
```java
93-
110+
class Solution {
111+
public long calculateScore(String s) {
112+
Map<Character, List<Integer>> d = new HashMap<>(26);
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int n = s.length();
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long ans = 0;
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for (int i = 0; i < n; ++i) {
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char x = s.charAt(i);
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char y = (char) ('a' + 'z' - x);
118+
if (d.containsKey(y)) {
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var ls = d.get(y);
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int j = ls.remove(ls.size() - 1);
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if (ls.isEmpty()) {
122+
d.remove(y);
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}
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ans += i - j;
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} else {
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d.computeIfAbsent(x, k -> new ArrayList<>()).add(i);
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}
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}
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return ans;
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}
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}
94132
```
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#### C++
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```cpp
99-
137+
class Solution {
138+
public:
139+
long long calculateScore(string s) {
140+
unordered_map<char, vector<int>> d;
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int n = s.length();
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long long ans = 0;
143+
for (int i = 0; i < n; ++i) {
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char x = s[i];
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char y = 'a' + 'z' - x;
146+
if (d.contains(y)) {
147+
vector<int>& ls = d[y];
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int j = ls.back();
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ls.pop_back();
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if (ls.empty()) {
151+
d.erase(y);
152+
}
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ans += i - j;
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} else {
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d[x].push_back(i);
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}
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}
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return ans;
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}
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};
100161
```
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#### Go
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```go
166+
func calculateScore(s string) (ans int64) {
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d := make(map[rune][]int)
168+
for i, x := range s {
169+
y := 'a' + 'z' - x
170+
if ls, ok := d[y]; ok {
171+
j := ls[len(ls)-1]
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d[y] = ls[:len(ls)-1]
173+
if len(d[y]) == 0 {
174+
delete(d, y)
175+
}
176+
ans += int64(i - j)
177+
} else {
178+
d[x] = append(d[x], i)
179+
}
180+
}
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return
182+
}
183+
```
105184

185+
#### TypeScript
186+
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```ts
188+
function calculateScore(s: string): number {
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const d: Map<string, number[]> = new Map();
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const n = s.length;
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let ans = 0;
192+
for (let i = 0; i < n; i++) {
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const x = s[i];
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const y = String.fromCharCode('a'.charCodeAt(0) + 'z'.charCodeAt(0) - x.charCodeAt(0));
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if (d.has(y)) {
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const ls = d.get(y)!;
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const j = ls.pop()!;
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if (ls.length === 0) {
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d.delete(y);
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}
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ans += i - j;
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} else {
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if (!d.has(x)) {
205+
d.set(x, []);
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}
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d.get(x)!.push(i);
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}
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}
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return ans;
211+
}
106212
```
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<!-- tabs:end -->

‎solution/3400-3499/3412.Find Mirror Score of a String/README_EN.md‎

Lines changed: 110 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -75,32 +75,138 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3412.Fi
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<!-- solution:start -->
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78-
### Solution 1
78+
### Solution 1: Hash Table
79+
80+
We can use a hash table $\textit{d}$ to store the index list of each unmarked character, where the key is the character and the value is the list of indices.
81+
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We traverse the string $\textit{s},ドル and for each character $\textit{x},ドル we find its mirror character $\textit{y}$. If $\textit{d}$ contains $\textit{y},ドル we take out the index list $\textit{ls}$ corresponding to $\textit{y},ドル take out the last element $\textit{j}$ from $\textit{ls},ドル and remove $\textit{j}$ from $\textit{ls}$. If $\textit{ls}$ becomes empty, we remove $\textit{y}$ from $\textit{d}$. At this point, we have found a pair of indices $(\textit{j}, \textit{i})$ that meet the condition, and we add $\textit{i} - \textit{j}$ to the answer. Otherwise, we add $\textit{x}$ to $\textit{d}$.
83+
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The time complexity is $O(n),ドル and the space complexity is $O(n)$. Here, $n$ is the length of the string $\textit{s}$.
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<!-- tabs:start -->
8187

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#### Python3
8389

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```python
85-
91+
class Solution:
92+
def calculateScore(self, s: str) -> int:
93+
d = defaultdict(list)
94+
ans = 0
95+
for i, x in enumerate(s):
96+
y = chr(ord("a") + ord("z") - ord(x))
97+
if d[y]:
98+
j = d[y].pop()
99+
ans += i - j
100+
else:
101+
d[x].append(i)
102+
return ans
86103
```
87104

88105
#### Java
89106

90107
```java
91-
108+
class Solution {
109+
public long calculateScore(String s) {
110+
Map<Character, List<Integer>> d = new HashMap<>(26);
111+
int n = s.length();
112+
long ans = 0;
113+
for (int i = 0; i < n; ++i) {
114+
char x = s.charAt(i);
115+
char y = (char) ('a' + 'z' - x);
116+
if (d.containsKey(y)) {
117+
var ls = d.get(y);
118+
int j = ls.remove(ls.size() - 1);
119+
if (ls.isEmpty()) {
120+
d.remove(y);
121+
}
122+
ans += i - j;
123+
} else {
124+
d.computeIfAbsent(x, k -> new ArrayList<>()).add(i);
125+
}
126+
}
127+
return ans;
128+
}
129+
}
92130
```
93131

94132
#### C++
95133

96134
```cpp
97-
135+
class Solution {
136+
public:
137+
long long calculateScore(string s) {
138+
unordered_map<char, vector<int>> d;
139+
int n = s.length();
140+
long long ans = 0;
141+
for (int i = 0; i < n; ++i) {
142+
char x = s[i];
143+
char y = 'a' + 'z' - x;
144+
if (d.contains(y)) {
145+
vector<int>& ls = d[y];
146+
int j = ls.back();
147+
ls.pop_back();
148+
if (ls.empty()) {
149+
d.erase(y);
150+
}
151+
ans += i - j;
152+
} else {
153+
d[x].push_back(i);
154+
}
155+
}
156+
return ans;
157+
}
158+
};
98159
```
99160
100161
#### Go
101162
102163
```go
164+
func calculateScore(s string) (ans int64) {
165+
d := make(map[rune][]int)
166+
for i, x := range s {
167+
y := 'a' + 'z' - x
168+
if ls, ok := d[y]; ok {
169+
j := ls[len(ls)-1]
170+
d[y] = ls[:len(ls)-1]
171+
if len(d[y]) == 0 {
172+
delete(d, y)
173+
}
174+
ans += int64(i - j)
175+
} else {
176+
d[x] = append(d[x], i)
177+
}
178+
}
179+
return
180+
}
181+
```
103182

183+
#### TypeScript
184+
185+
```ts
186+
function calculateScore(s: string): number {
187+
const d: Map<string, number[]> = new Map();
188+
const n = s.length;
189+
let ans = 0;
190+
for (let i = 0; i < n; i++) {
191+
const x = s[i];
192+
const y = String.fromCharCode('a'.charCodeAt(0) + 'z'.charCodeAt(0) - x.charCodeAt(0));
193+
194+
if (d.has(y)) {
195+
const ls = d.get(y)!;
196+
const j = ls.pop()!;
197+
if (ls.length === 0) {
198+
d.delete(y);
199+
}
200+
ans += i - j;
201+
} else {
202+
if (!d.has(x)) {
203+
d.set(x, []);
204+
}
205+
d.get(x)!.push(i);
206+
}
207+
}
208+
return ans;
209+
}
104210
```
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106212
<!-- tabs:end -->
Lines changed: 24 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,24 @@
1+
class Solution {
2+
public:
3+
long long calculateScore(string s) {
4+
unordered_map<char, vector<int>> d;
5+
int n = s.length();
6+
long long ans = 0;
7+
for (int i = 0; i < n; ++i) {
8+
char x = s[i];
9+
char y = 'a' + 'z' - x;
10+
if (d.contains(y)) {
11+
vector<int>& ls = d[y];
12+
int j = ls.back();
13+
ls.pop_back();
14+
if (ls.empty()) {
15+
d.erase(y);
16+
}
17+
ans += i - j;
18+
} else {
19+
d[x].push_back(i);
20+
}
21+
}
22+
return ans;
23+
}
24+
};
Lines changed: 17 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,17 @@
1+
func calculateScore(s string) (ans int64) {
2+
d := make(map[rune][]int)
3+
for i, x := range s {
4+
y := 'a' + 'z' - x
5+
if ls, ok := d[y]; ok {
6+
j := ls[len(ls)-1]
7+
d[y] = ls[:len(ls)-1]
8+
if len(d[y]) == 0 {
9+
delete(d, y)
10+
}
11+
ans += int64(i - j)
12+
} else {
13+
d[x] = append(d[x], i)
14+
}
15+
}
16+
return
17+
}
Lines changed: 22 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,22 @@
1+
class Solution {
2+
public long calculateScore(String s) {
3+
Map<Character, List<Integer>> d = new HashMap<>(26);
4+
int n = s.length();
5+
long ans = 0;
6+
for (int i = 0; i < n; ++i) {
7+
char x = s.charAt(i);
8+
char y = (char) ('a' + 'z' - x);
9+
if (d.containsKey(y)) {
10+
var ls = d.get(y);
11+
int j = ls.remove(ls.size() - 1);
12+
if (ls.isEmpty()) {
13+
d.remove(y);
14+
}
15+
ans += i - j;
16+
} else {
17+
d.computeIfAbsent(x, k -> new ArrayList<>()).add(i);
18+
}
19+
}
20+
return ans;
21+
}
22+
}
Lines changed: 12 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,12 @@
1+
class Solution:
2+
def calculateScore(self, s: str) -> int:
3+
d = defaultdict(list)
4+
ans = 0
5+
for i, x in enumerate(s):
6+
y = chr(ord("a") + ord("z") - ord(x))
7+
if d[y]:
8+
j = d[y].pop()
9+
ans += i - j
10+
else:
11+
d[x].append(i)
12+
return ans
Lines changed: 24 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,24 @@
1+
function calculateScore(s: string): number {
2+
const d: Map<string, number[]> = new Map();
3+
const n = s.length;
4+
let ans = 0;
5+
for (let i = 0; i < n; i++) {
6+
const x = s[i];
7+
const y = String.fromCharCode('a'.charCodeAt(0) + 'z'.charCodeAt(0) - x.charCodeAt(0));
8+
9+
if (d.has(y)) {
10+
const ls = d.get(y)!;
11+
const j = ls.pop()!;
12+
if (ls.length === 0) {
13+
d.delete(y);
14+
}
15+
ans += i - j;
16+
} else {
17+
if (!d.has(x)) {
18+
d.set(x, []);
19+
}
20+
d.get(x)!.push(i);
21+
}
22+
}
23+
return ans;
24+
}

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