From 7a8e930a6548df9680214fde10ffff9fd0a9235b Mon Sep 17 00:00:00 2001
From: yanglbme <szuyanglb@outlook.com>
Date: Tue, 24 Sep 2024 09:11:36 +0800
Subject: [PATCH] feat: add solutions to lc problem: No.1239
No.1239.Maximum Length of a Concatenated String with Unique Characters
---
.../README.md | 178 +++++++++++-------
.../README_EN.md | 178 +++++++++++-------
.../Solution.cpp | 34 ++--
.../Solution.go | 35 ++--
.../Solution.java | 36 ++--
.../Solution.py | 26 +--
.../Solution.ts | 36 ++++
7 files changed, 310 insertions(+), 213 deletions(-)
create mode 100644 solution/1200-1299/1239.Maximum Length of a Concatenated String with Unique Characters/Solution.ts
diff --git a/solution/1200-1299/1239.Maximum Length of a Concatenated String with Unique Characters/README.md b/solution/1200-1299/1239.Maximum Length of a Concatenated String with Unique Characters/README.md
index 5c8121c3866f2..dce090b353afa 100644
--- a/solution/1200-1299/1239.Maximum Length of a Concatenated String with Unique Characters/README.md
+++ b/solution/1200-1299/1239.Maximum Length of a Concatenated String with Unique Characters/README.md
@@ -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 -->
@@ -88,23 +94,17 @@ 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
@@ -112,28 +112,26 @@ class Solution:
```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));
+ }
}
}
}
@@ -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));
+ }
}
}
}
@@ -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;
}
```
diff --git a/solution/1200-1299/1239.Maximum Length of a Concatenated String with Unique Characters/README_EN.md b/solution/1200-1299/1239.Maximum Length of a Concatenated String with Unique Characters/README_EN.md
index dc8597040daba..260ee2b0ef79c 100644
--- a/solution/1200-1299/1239.Maximum Length of a Concatenated String with Unique Characters/README_EN.md
+++ b/solution/1200-1299/1239.Maximum Length of a Concatenated String with Unique Characters/README_EN.md
@@ -74,11 +74,17 @@ Maximum length is 4.
<!-- solution:start -->
-### Solution 1: Bit Manipulation + State Compression
+### Solution 1: State Compression + Bit Manipulation
-State compression is used, with a 32-bit number recording the occurrence of letters, and `masks` storing the strings enumerated before.
+Since the problem requires that the characters in the subsequence must not be repeated and all characters are lowercase letters, we can use a binary integer of length $26$ to represent a subsequence. The $i$-th bit being $1$ indicates that the subsequence contains the $i$-th character, and $0$ indicates that it does not contain the $i$-th character.
-The time complexity is $O(2^n + L)$, and the space complexity is $O(2^n)$. Where $n$ and $L$ are the length of the string array and the sum of the lengths of the strings in the array, respectively.
+We can use an array $s$ to store the states of all subsequences that meet the conditions. Initially, $s$ contains only one element $0$.
+
+Then we traverse the array $\textit{arr}$. For each string $t$, we use an integer $x$ to represent the state of $t$. Then we traverse the array $s$. For each state $y$, if $x$ and $y$ have no common characters, we add the union of $x$ and $y$ to $s$ and update the answer.
+
+Finally, we return the answer.
+
+The time complexity is $O(2^n + L)$, and the space complexity is $O(2^n)$. Here, $n$ is the length of the string array, and $L$ is the sum of the lengths of all strings in the array.
<!-- tabs:start -->
@@ -87,23 +93,17 @@ The time complexity is $O(2^n + L)$, and the space complexity is $O(2^n)$. Where
```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
@@ -111,28 +111,26 @@ class Solution:
```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));
+ }
}
}
}
@@ -147,27 +145,25 @@ 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));
+ }
}
}
}
@@ -180,29 +176,69 @@ 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;
}
```
diff --git a/solution/1200-1299/1239.Maximum Length of a Concatenated String with Unique Characters/Solution.cpp b/solution/1200-1299/1239.Maximum Length of a Concatenated String with Unique Characters/Solution.cpp
index 71e2653dcd012..76f4c50b2ac58 100644
--- a/solution/1200-1299/1239.Maximum Length of a Concatenated String with Unique Characters/Solution.cpp
+++ b/solution/1200-1299/1239.Maximum Length of a Concatenated String with Unique Characters/Solution.cpp
@@ -1,30 +1,28 @@
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;
+ x |= 1 << b;
}
- if (mask == 0) {
- continue;
- }
- 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));
+ }
}
}
}
return ans;
}
-};
\ No newline at end of file
+};
diff --git a/solution/1200-1299/1239.Maximum Length of a Concatenated String with Unique Characters/Solution.go b/solution/1200-1299/1239.Maximum Length of a Concatenated String with Unique Characters/Solution.go
index f1f2f02c47eb3..095f0d2809e73 100644
--- a/solution/1200-1299/1239.Maximum Length of a Concatenated String with Unique Characters/Solution.go
+++ b/solution/1200-1299/1239.Maximum Length of a Concatenated String with Unique Characters/Solution.go
@@ -1,25 +1,24 @@
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
-}
\ No newline at end of file
+ return ans
+}
diff --git a/solution/1200-1299/1239.Maximum Length of a Concatenated String with Unique Characters/Solution.java b/solution/1200-1299/1239.Maximum Length of a Concatenated String with Unique Characters/Solution.java
index 47e50759aea30..fee4b6904b825 100644
--- a/solution/1200-1299/1239.Maximum Length of a Concatenated String with Unique Characters/Solution.java
+++ b/solution/1200-1299/1239.Maximum Length of a Concatenated String with Unique Characters/Solution.java
@@ -1,30 +1,28 @@
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));
+ }
}
}
}
return ans;
}
-}
\ No newline at end of file
+}
diff --git a/solution/1200-1299/1239.Maximum Length of a Concatenated String with Unique Characters/Solution.py b/solution/1200-1299/1239.Maximum Length of a Concatenated String with Unique Characters/Solution.py
index 1e63cce9be627..79fd149c270da 100644
--- a/solution/1200-1299/1239.Maximum Length of a Concatenated String with Unique Characters/Solution.py
+++ b/solution/1200-1299/1239.Maximum Length of a Concatenated String with Unique Characters/Solution.py
@@ -1,19 +1,13 @@
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)
diff --git a/solution/1200-1299/1239.Maximum Length of a Concatenated String with Unique Characters/Solution.ts b/solution/1200-1299/1239.Maximum Length of a Concatenated String with Unique Characters/Solution.ts
new file mode 100644
index 0000000000000..15e3c3a6e7269
--- /dev/null
+++ b/solution/1200-1299/1239.Maximum Length of a Concatenated String with Unique Characters/Solution.ts
@@ -0,0 +1,36 @@
+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;
+}