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Commit b9568fa

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feat: update solutions to lc problem: No.3003 (#4789)
1 parent cf8d64c commit b9568fa

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‎solution/3000-3099/3003.Maximize the Number of Partitions After Operations/README.md‎

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<!-- solution:start -->
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### 方法一
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### 方法一:记忆化搜索
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我们设计一个函数 $\textit{dfs}(i, \textit{cur}, t)$ 表示当前处理到字符串 $s$ 的下标 $i,ドル当前前缀中已经包含的字符集合为 $\textit{cur},ドル并且还可以修改 $t$ 次字符时,能够得到的最大分割数量。那么答案即为 $\textit{dfs}(0, 0, 1)$。
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函数 $\textit{dfs}(i, \textit{cur}, t)$ 的执行逻辑如下:
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1. 如果 $i \geq n,ドル说明已经处理完字符串 $s,ドル返回 1。
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2. 计算当前字符 $s[i]$ 对应的位掩码 $v = 1 \ll (s[i] - 'a'),ドル并计算更新后的字符集合 $\textit{nxt} = \textit{cur} \mid v$。
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3. 如果 $\textit{nxt}$ 中的位数超过 $k,ドル说明当前前缀已经包含超过 $k$ 个不同字符,我们需要进行一次分割,此时分割数量加 1,并递归调用 $\textit{dfs}(i + 1, v, t)$;否则,继续递归调用 $\textit{dfs}(i + 1, \textit{nxt}, t)$。
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4. 如果 $t > 0,ドル说明我们还可以修改一次字符。我们尝试将当前字符 $s[i]$ 修改为任意一个小写字母(共 26 种选择),对于每个选择,计算更新后的字符集合 $\textit{nxt} = \textit{cur} \mid (1 \ll j),ドル并根据是否超过 $k$ 个不同字符,选择相应的递归调用方式,更新最大分割数量。
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5. 使用哈希表缓存已经计算过的状态,避免重复计算。
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时间复杂度 $O(n \times |\Sigma| \times k),ドル空间复杂度 $O(n \times |\Sigma| \times k)$。其中 $n$ 为字符串 $s$ 的长度,而 $|\Sigma|$ 为字符集大小。
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<!-- tabs:start -->
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int maxPartitionsAfterOperations(string s, int k) {
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int n = s.size();
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unordered_map<long long, int> f;
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function<int(int, int, int)> dfs = [&](int i, int cur, int t) {
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auto dfs = [&](this auto&& dfs, int i, int cur, int t) -> int {
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if (i >= n) {
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return 1;
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}

‎solution/3000-3099/3003.Maximize the Number of Partitions After Operations/README_EN.md‎

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Memoized Search
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We design a function $\textit{dfs}(i, \textit{cur}, t)$ that represents the maximum number of partitions we can obtain when currently processing index $i$ of string $s,ドル the current prefix already contains the character set $\textit{cur},ドル and we can still modify $t$ characters. Then the answer is $\textit{dfs}(0, 0, 1)$.
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The execution logic of function $\textit{dfs}(i, \textit{cur}, t)$ is as follows:
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1. If $i \geq n,ドル it means we have finished processing string $s,ドル return 1.
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2. Calculate the bitmask $v = 1 \ll (s[i] - 'a')$ corresponding to the current character $s[i],ドル and calculate the updated character set $\textit{nxt} = \textit{cur} \mid v$.
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3. If the number of bits in $\textit{nxt}$ exceeds $k,ドル it means the current prefix already contains more than $k$ distinct characters. We need to make a partition, increment the partition count by 1, and recursively call $\textit{dfs}(i + 1, v, t)$. Otherwise, continue recursively calling $\textit{dfs}(i + 1, \textit{nxt}, t)$.
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4. If $t > 0,ドル it means we can still modify a character once. We try to change the current character $s[i]$ to any lowercase letter (26 choices in total). For each choice, calculate the updated character set $\textit{nxt} = \textit{cur} \mid (1 \ll j),ドル and based on whether it exceeds $k$ distinct characters, choose the corresponding recursive call method to update the maximum partition count.
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5. Use a hash table to cache already computed states to avoid redundant calculations.
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The time complexity is $O(n \times |\Sigma| \times k)$ and the space complexity is $O(n \times |\Sigma| \times k),ドル where $n$ is the length of string $s,ドル and $|\Sigma|$ is the size of the character set.
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<!-- tabs:start -->
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int maxPartitionsAfterOperations(string s, int k) {
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int n = s.size();
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unordered_map<long long, int> f;
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function<int(int, int, int)> dfs = [&](int i, int cur, int t) {
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auto dfs = [&](this auto&& dfs, int i, int cur, int t) -> int {
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if (i >= n) {
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return 1;
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}

‎solution/3000-3099/3003.Maximize the Number of Partitions After Operations/Solution.cpp‎

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int maxPartitionsAfterOperations(string s, int k) {
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int n = s.size();
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unordered_map<long long, int> f;
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function<int(int, int, int)> dfs = [&](int i, int cur, int t) {
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autodfs = [&](thisauto&& dfs, int i, int cur, int t) -> int {
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if (i >= n) {
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return 1;
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}
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};
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return dfs(0, 0, 1);
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}
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};
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};

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