From 3cc14f7977a56c1240a9290f16a685ad322be7a7 Mon Sep 17 00:00:00 2001 From: Libin YANG Date: Wed, 11 Jun 2025 20:50:28 +0800 Subject: [PATCH 1/2] feat: add solutions to lc problem: No.3579 No.3579.Minimum Steps to Convert String with Operations --- .../README.md | 12 +++++++++++- .../README_EN.md | 12 +++++++++++- 2 files changed, 22 insertions(+), 2 deletions(-) diff --git a/solution/3500-3599/3579.Minimum Steps to Convert String with Operations/README.md b/solution/3500-3599/3579.Minimum Steps to Convert String with Operations/README.md index a5717cc4a6e3b..2a6c4191e378d 100644 --- a/solution/3500-3599/3579.Minimum Steps to Convert String with Operations/README.md +++ b/solution/3500-3599/3579.Minimum Steps to Convert String with Operations/README.md @@ -144,7 +144,17 @@ tags: -### 方法一 +### 方法一：贪心 + 动态规划 + +我们定义 $f[i]$ 表示将 $\textit{word1}$ 的前 $i$ 个字符转换为 $\textit{word2}$ 的前 $i$ 个字符所需的最小操作数。那么答案为 $f[n]$，其中 $n$ 是 $\textit{word1}$ 和 $\textit{word2}$ 的长度。 + +我们可以通过遍历所有可能的分割点来计算 $f[i]$。对于每个分割点 $j$，我们需要计算将 $\textit{word1}[j:i]$ 转换为 $\textit{word2}[j:i]$ 所需的最小操作数。 + +我们可以使用一个辅助函数 $\text{calc}(l, r, \text{rev})$ 来计算从 $\textit{word1}[l:r]$ 转换为 $\textit{word2}[l:r]$ 所需的最小操作数，其中 $\text{rev}$ 表示是否需要反转子串。由于反转前后进行其它操作的结果是一样的，所以我们可以考虑不反转，以及首先进行一次反转后再进行其它操作。因此有 $f[i] = \min_{j < i} (f[j] + \min(\text{calc}(j, i-1, \text{false}), 1 + \text{calc}(j, i-1, \text{true})))$。 + +接下来我们需要实现 $\text{calc}(l, r, \text{rev})$ 函数。我们用一个二维数组 $cnt$ 来记录 $\textit{word1}$ 和 $\textit{word2}$ 中字符的配对情况。对于每个字符对 $(a, b)$，如果 $a \neq b$，我们需要检查 $cnt[b][a]$ 是否大于 $0$。如果是，我们可以将其配对，减少一次操作；否则，我们需要增加一次操作，并将 $cnt[a][b]$ 加 $1$。 + +时间复杂度 $O(n^3 + |\Sigma|^2)$，空间复杂度 $O(n + |\Sigma|^2)$，其中 $n$ 是字符串的长度，而 $|\Sigma|$ 是字符集大小（本题中为 $26$）。 diff --git a/solution/3500-3599/3579.Minimum Steps to Convert String with Operations/README_EN.md b/solution/3500-3599/3579.Minimum Steps to Convert String with Operations/README_EN.md index 3a479ed5ab9a9..6f2cc39109b9f 100644 --- a/solution/3500-3599/3579.Minimum Steps to Convert String with Operations/README_EN.md +++ b/solution/3500-3599/3579.Minimum Steps to Convert String with Operations/README_EN.md @@ -139,7 +139,17 @@ tags: -### Solution 1 +### Solution 1: Greedy + Dynamic Programming + +We define $f[i]$ as the minimum number of operations required to convert the first $i$ characters of $\textit{word1}$ to the first $i$ characters of $\textit{word2}$. The answer is $f[n]$, where $n$ is the length of both $\textit{word1}$ and $\textit{word2}$. + +We can compute $f[i]$ by enumerating all possible split points. For each split point $j$, we need to calculate the minimum number of operations required to convert $\textit{word1}[j:i]$ to $\textit{word2}[j:i]$. + +We can use a helper function $\text{calc}(l, r, \text{rev})$ to compute the minimum number of operations needed to convert $\textit{word1}[l:r]$ to $\textit{word2}[l:r]$, where $\text{rev}$ indicates whether to reverse the substring. Since the result of performing other operations before or after a reversal is the same, we only need to consider not reversing, and reversing once before other operations. Therefore, $f[i] = \min_{j < i} (f[j] + \min(\text{calc}(j, i-1, \text{false}), 1 + \text{calc}(j, i-1, \text{true})))$. + +Next, we need to implement the $\text{calc}(l, r, \text{rev})$ function. We use a 2D array $cnt$ to record the pairing status of characters between $\textit{word1}$ and $\textit{word2}$. For each character pair $(a, b)$, if $a \neq b$, we check whether $cnt[b][a] > 0$. If so, we can pair them and reduce one operation; otherwise, we need to add one operation and increment $cnt[a][b]$ by $1$. + +The time complexity is $O(n^3 + |\Sigma|^2)$ and the space complexity is $O(n + |\Sigma|^2)$, where $n$ is the length of the string and $|\Sigma|$ is the size of the character set (which is $26$ in this problem). From 75fe9ed066832b81eba78a9cb2f5145461148f8b Mon Sep 17 00:00:00 2001 From: Libin YANG Date: Wed, 11 Jun 2025 20:53:55 +0800 Subject: [PATCH 2/2] feat: add solution to lc problem --- .../README.md | 47 +++++++++++++++++++ .../README_EN.md | 47 +++++++++++++++++++ .../Solution.rs | 42 +++++++++++++++++ 3 files changed, 136 insertions(+) create mode 100644 solution/3500-3599/3579.Minimum Steps to Convert String with Operations/Solution.rs diff --git a/solution/3500-3599/3579.Minimum Steps to Convert String with Operations/README.md b/solution/3500-3599/3579.Minimum Steps to Convert String with Operations/README.md index 2a6c4191e378d..eb1fcab8cccbc 100644 --- a/solution/3500-3599/3579.Minimum Steps to Convert String with Operations/README.md +++ b/solution/3500-3599/3579.Minimum Steps to Convert String with Operations/README.md @@ -367,6 +367,53 @@ function minOperations(word1: string, word2: string): number { } ``` +#### Rust + +```rust +impl Solution { + pub fn min_operations(word1: String, word2: String) -> i32 { + let n = word1.len(); + let word1 = word1.as_bytes(); + let word2 = word2.as_bytes(); + let mut f = vec![i32::MAX; n + 1]; + f[0] = 0; + + for i in 1..=n { + for j in 0..i { + let a = Self::calc(word1, word2, j, i - 1, false); + let b = 1 + Self::calc(word1, word2, j, i - 1, true); + let t = a.min(b); + f[i] = f[i].min(f[j] + t); + } + } + + f[n] + } + + fn calc(word1: &[u8], word2: &[u8], l: usize, r: usize, rev: bool) -> i32 { + let mut cnt = [[0i32; 26]; 26]; + let mut res = 0; + + for i in l..=r { + let j = if rev { r - (i - l) } else { i }; + let a = (word1[j] - b'a') as usize; + let b = (word2[i] - b'a') as usize; + + if a != b { + if cnt[b][a] > 0 { + cnt[b][a] -= 1; + } else { + cnt[a][b] += 1; + res += 1; + } + } + } + + res + } +} +``` + diff --git a/solution/3500-3599/3579.Minimum Steps to Convert String with Operations/README_EN.md b/solution/3500-3599/3579.Minimum Steps to Convert String with Operations/README_EN.md index 6f2cc39109b9f..5ec617958ee38 100644 --- a/solution/3500-3599/3579.Minimum Steps to Convert String with Operations/README_EN.md +++ b/solution/3500-3599/3579.Minimum Steps to Convert String with Operations/README_EN.md @@ -362,6 +362,53 @@ function minOperations(word1: string, word2: string): number { } ``` +#### Rust + +```rust +impl Solution { + pub fn min_operations(word1: String, word2: String) -> i32 { + let n = word1.len(); + let word1 = word1.as_bytes(); + let word2 = word2.as_bytes(); + let mut f = vec![i32::MAX; n + 1]; + f[0] = 0; + + for i in 1..=n { + for j in 0..i { + let a = Self::calc(word1, word2, j, i - 1, false); + let b = 1 + Self::calc(word1, word2, j, i - 1, true); + let t = a.min(b); + f[i] = f[i].min(f[j] + t); + } + } + + f[n] + } + + fn calc(word1: &[u8], word2: &[u8], l: usize, r: usize, rev: bool) -> i32 { + let mut cnt = [[0i32; 26]; 26]; + let mut res = 0; + + for i in l..=r { + let j = if rev { r - (i - l) } else { i }; + let a = (word1[j] - b'a') as usize; + let b = (word2[i] - b'a') as usize; + + if a != b { + if cnt[b][a] > 0 { + cnt[b][a] -= 1; + } else { + cnt[a][b] += 1; + res += 1; + } + } + } + + res + } +} +``` + diff --git a/solution/3500-3599/3579.Minimum Steps to Convert String with Operations/Solution.rs b/solution/3500-3599/3579.Minimum Steps to Convert String with Operations/Solution.rs new file mode 100644 index 0000000000000..8fc0efb9cacf0 --- /dev/null +++ b/solution/3500-3599/3579.Minimum Steps to Convert String with Operations/Solution.rs @@ -0,0 +1,42 @@ +impl Solution { + pub fn min_operations(word1: String, word2: String) -> i32 { + let n = word1.len(); + let word1 = word1.as_bytes(); + let word2 = word2.as_bytes(); + let mut f = vec![i32::MAX; n + 1]; + f[0] = 0; + + for i in 1..=n { + for j in 0..i { + let a = Self::calc(word1, word2, j, i - 1, false); + let b = 1 + Self::calc(word1, word2, j, i - 1, true); + let t = a.min(b); + f[i] = f[i].min(f[j] + t); + } + } + + f[n] + } + + fn calc(word1: &[u8], word2: &[u8], l: usize, r: usize, rev: bool) -> i32 { + let mut cnt = [[0i32; 26]; 26]; + let mut res = 0; + + for i in l..=r { + let j = if rev { r - (i - l) } else { i }; + let a = (word1[j] - b'a') as usize; + let b = (word2[i] - b'a') as usize; + + if a != b { + if cnt[b][a] > 0 { + cnt[b][a] -= 1; + } else { + cnt[a][b] += 1; + res += 1; + } + } + } + + res + } +}