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feat: add solutions to lc problem: No.1239 #3555

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Sep 24, 2024
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feat: add solutions to lc problem: No.1239 #3555

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178 changes: 107 additions & 71 deletions ...9/1239.Maximum Length of a Concatenated String with Unique Characters/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -75,11 +75,17 @@ tags:

<!-- solution:start -->

### 方法一：位运算 + 状态压缩
### 方法一：状态压缩 + 位运算

状态压缩，用一个 $32$ 位数记录字母的出现情况，`masks` 存储之前枚举的字符串。
由于题目要求子序列的字符不能重复，字符都是小写字母，因此，我们可以用一个长度为 $26$ 的二进制整数来表示一个子序列，其中第 $i$ 位为 $1$ 表示子序列中含有滴 $i$ 个字符，为 $0$ 表示不含有第 $i$ 个字符。

时间复杂度 $O(2^n + L)$，空间复杂度 $O(2^n)$。其中 $n$ 和 $L$ 分别是字符串数组的长度和字符串数组中字符串的长度之和。
我们可以用一个数组 $s$ 来存储所有满足条件的子序列的状态，初始时 $s$ 中只有一个元素 $0$。

然后我们遍历数组 $\textit{arr}$，对于每个字符串 $t$，我们用一个整数 $x$ 来表示 $t$ 的状态，然后我们遍历数组 $s$，对于每个状态 $y$，如果 $x$ 和 $y$ 之间没有相同的字符，那么我们将 $x$ 和 $y$ 的并集加入到 $s$ 中，并更新答案。

最后我们返回答案即可。

时间复杂度 $O(2^n + L)$，空间复杂度 $O(2^n)$，其中 $n$ 是字符串数组的长度，而 $L$ 是字符串数组中所有字符串的长度之和。

<!-- tabs:start -->

Expand All @@ -88,52 +94,44 @@ tags:
```python
class Solution:
def maxLength(self, arr: List[str]) -> int:
ans = 0
masks = [0]
for s in arr:
mask = 0
for c in s:
i = ord(c) - ord('a')
if mask >> i & 1:
mask = 0
s = [0]
for t in arr:
x = 0
for b in map(lambda c: ord(c) - 97, t):
if x >> b & 1:
x = 0
break
mask |= 1 << i
if mask == 0:
continue
for m in masks:
if m & mask == 0:
masks.append(m | mask)
ans = max(ans, (m | mask).bit_count())
return ans
x |= 1 << b
if x:
s.extend((x | y) for y in s if (x & y) == 0)
return max(x.bit_count() for x in s)
```

#### Java

```java
class Solution {
public int maxLength(List<String> arr) {
List<Integer> s = new ArrayList<>();
s.add(0);
int ans = 0;
List<Integer> masks = new ArrayList<>();
masks.add(0);
for (var s : arr) {
int mask = 0;
for (int i = 0; i < s.length(); ++i) {
int j = s.charAt(i) - 'a';
if (((mask >> j) & 1) == 1) {
mask = 0;
for (var t : arr) {
int x = 0;
for (char c : t.toCharArray()) {
int b = c - 'a';
if ((x >> b & 1) == 1) {
x = 0;
break;
}
mask |= 1 << j;
x |= 1 << b;
}
if (mask == 0) {
continue;
}
int n = masks.size();
for (int i = 0; i < n; ++i) {
int m = masks.get(i);
if ((m & mask) == 0) {
masks.add(m | mask);
ans = Math.max(ans, Integer.bitCount(m | mask));
if (x > 0) {
for (int i = s.size() - 1; i >= 0; --i) {
int y = s.get(i);
if ((x & y) == 0) {
s.add(x | y);
ans = Math.max(ans, Integer.bitCount(x | y));
}
}
}
}
Expand All @@ -148,27 +146,25 @@ class Solution {
class Solution {
public:
int maxLength(vector<string>& arr) {
vector<int> s = {0};
int ans = 0;
vector<int> masks = {0};
for (auto& s : arr) {
int mask = 0;
for (auto& c : s) {
int i = c - 'a';
if (mask >> i & 1) {
mask = 0;
for (const string& t : arr) {
int x = 0;
for (char c : t) {
int b = c - 'a';
if ((x >> b & 1) == 1) {
x = 0;
break;
}
mask |= 1 << i;
}
if (mask == 0) {
continue;
x |= 1 << b;
}
int n = masks.size();
for (int i = 0; i < n; ++i) {
int m = masks[i];
if ((m & mask) == 0) {
masks.push_back(m | mask);
ans = max(ans, __builtin_popcount(m | mask));
if (x > 0) {
for (int i = s.size() - 1; i >= 0; --i) {
int y = s[i];
if ((x & y) == 0) {
s.push_back(x | y);
ans = max(ans, __builtin_popcount(x | y));
}
}
}
}
Expand All @@ -181,29 +177,69 @@ public:

```go
func maxLength(arr []string) (ans int) {
masks := []int{0}
for _, s := range arr {
mask := 0
for _, c := range s {
i := int(c - 'a')
if mask>>i&1 == 1 {
mask = 0
s := []int{0}
for _, t := range arr {
x := 0
for _, c := range t {
b := int(c - 'a')
if (x>>b)&1 == 1 {
x = 0
break
}
mask |= 1 << i
x |= 1 << b
}
if mask == 0 {
continue
}
n := len(masks)
for _, m := range masks[:n] {
if m&mask == 0 {
masks = append(masks, m|mask)
ans = max(ans, bits.OnesCount(uint(m|mask)))
if x > 0 {
for i := len(s) - 1; i >= 0; i-- {
y := s[i]
if (x & y) == 0 {
s = append(s, x|y)
ans = max(ans, bits.OnesCount(uint(x|y)))
}
}
}
}
return
return ans
}
```

#### TypeScript

```ts
function maxLength(arr: string[]): number {
const s: number[] = [0];
let ans = 0;
for (const t of arr) {
let x = 0;
for (const c of t) {
const b = c.charCodeAt(0) - 97;
if ((x >> b) & 1) {
x = 0;
break;
}
x |= 1 << b;
}

if (x > 0) {
for (let i = s.length - 1; ~i; --i) {
const y = s[i];
if ((x & y) === 0) {
s.push(x | y);
ans = Math.max(ans, bitCount(x | y));
}
}
}
}

return ans;
}

function bitCount(i: number): number {
i = i - ((i >>> 1) & 0x55555555);
i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);
i = (i + (i >>> 4)) & 0x0f0f0f0f;
i = i + (i >>> 8);
i = i + (i >>> 16);
return i & 0x3f;
}
```

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