From 921708acaafca386e1f8df8bcbddf167cf196805 Mon Sep 17 00:00:00 2001 From: yanglbme Date: Tue, 18 Mar 2025 11:08:15 +0800 Subject: [PATCH] feat: add solutions to lc problems: No.2712,2716 * No.2712.Minimum Cost to Make All Characters Equal * No.2716.Minimize String Length --- .../README.md | 18 ++++++++++++++++++ .../README_EN.md | 18 ++++++++++++++++++ .../Solution.rs | 13 +++++++++++++ .../2716.Minimize String Length/README.md | 15 ++++++++++++--- .../2716.Minimize String Length/README_EN.md | 17 +++++++++++++---- .../2716.Minimize String Length/Solution.cpp | 5 ++--- .../2716.Minimize String Length/Solution.cs | 5 +++++ 7 files changed, 81 insertions(+), 10 deletions(-) create mode 100644 solution/2700-2799/2712.Minimum Cost to Make All Characters Equal/Solution.rs create mode 100644 solution/2700-2799/2716.Minimize String Length/Solution.cs diff --git a/solution/2700-2799/2712.Minimum Cost to Make All Characters Equal/README.md b/solution/2700-2799/2712.Minimum Cost to Make All Characters Equal/README.md index 744cd2e6fbaab..bb2932f430b2c 100644 --- a/solution/2700-2799/2712.Minimum Cost to Make All Characters Equal/README.md +++ b/solution/2700-2799/2712.Minimum Cost to Make All Characters Equal/README.md @@ -156,6 +156,24 @@ function minimumCost(s: string): number { } ``` +#### Rust + +```rust +impl Solution { + pub fn minimum_cost(s: String) -> i64 { + let mut ans = 0; + let n = s.len(); + let s = s.as_bytes(); + for i in 1..n { + if s[i] != s[i - 1] { + ans += i.min(n - i); + } + } + ans as i64 + } +} +``` + diff --git a/solution/2700-2799/2712.Minimum Cost to Make All Characters Equal/README_EN.md b/solution/2700-2799/2712.Minimum Cost to Make All Characters Equal/README_EN.md index 4fdc8b5d7b20c..39e07f3e335db 100644 --- a/solution/2700-2799/2712.Minimum Cost to Make All Characters Equal/README_EN.md +++ b/solution/2700-2799/2712.Minimum Cost to Make All Characters Equal/README_EN.md @@ -155,6 +155,24 @@ function minimumCost(s: string): number { } ``` +#### Rust + +```rust +impl Solution { + pub fn minimum_cost(s: String) -> i64 { + let mut ans = 0; + let n = s.len(); + let s = s.as_bytes(); + for i in 1..n { + if s[i] != s[i - 1] { + ans += i.min(n - i); + } + } + ans as i64 + } +} +``` + diff --git a/solution/2700-2799/2712.Minimum Cost to Make All Characters Equal/Solution.rs b/solution/2700-2799/2712.Minimum Cost to Make All Characters Equal/Solution.rs new file mode 100644 index 0000000000000..6899f03efe1a2 --- /dev/null +++ b/solution/2700-2799/2712.Minimum Cost to Make All Characters Equal/Solution.rs @@ -0,0 +1,13 @@ +impl Solution { + pub fn minimum_cost(s: String) -> i64 { + let mut ans = 0; + let n = s.len(); + let s = s.as_bytes(); + for i in 1..n { + if s[i] != s[i - 1] { + ans += i.min(n - i); + } + } + ans as i64 + } +} diff --git a/solution/2700-2799/2716.Minimize String Length/README.md b/solution/2700-2799/2716.Minimize String Length/README.md index 86079bba90735..05254fb0682f3 100644 --- a/solution/2700-2799/2716.Minimize String Length/README.md +++ b/solution/2700-2799/2716.Minimize String Length/README.md @@ -72,7 +72,7 @@ tags: 题目实际上可以转化为求字符串中不同字符的个数，因此，我们只需要统计字符串中不同字符的个数即可。 -时间复杂度 $O(n)$，空间复杂度 $O(C)$。其中 $n$ 是字符串的长度；而 $C$ 是字符集的大小，本题中字符集为小写英文字母，因此 $C=26$。 +时间复杂度 $O(n)$，其中 $n$ 是字符串 $\textit{s}$ 的长度。空间复杂度 $O(|\Sigma|)$，其中 $\Sigma$ 是字符集，这里是小写英文字母，因此 $|\Sigma|=26$。 @@ -104,8 +104,7 @@ class Solution { class Solution { public: int minimizedStringLength(string s) { - unordered_set ss(s.begin(), s.end()); - return ss.size(); + return unordered_set(s.begin(), s.end()).size(); } }; ``` @@ -143,6 +142,16 @@ impl Solution { } ``` +#### C# + +```cs +public class Solution { + public int MinimizedStringLength(string s) { + return new HashSet(s).Count; + } +} +``` + diff --git a/solution/2700-2799/2716.Minimize String Length/README_EN.md b/solution/2700-2799/2716.Minimize String Length/README_EN.md index d3fd7ea4de41b..7cf23ab0fa6c1 100644 --- a/solution/2700-2799/2716.Minimize String Length/README_EN.md +++ b/solution/2700-2799/2716.Minimize String Length/README_EN.md @@ -100,9 +100,9 @@ tags: ### Solution 1: Hash Table -The problem can actually be transformed into finding the number of different characters in the string. Therefore, we only need to count the number of different characters in the string. +The problem can actually be transformed into finding the number of distinct characters in the string. Therefore, we only need to count the number of distinct characters in the string. -The time complexity is $O(n)$, and the space complexity is $O(C)$. Here, $n$ is the length of the string, and $C$ is the size of the character set. In this problem, the character set is lowercase English letters, so $C=26$. +The time complexity is $O(n)$, where $n$ is the length of the string $\textit{s}$. The space complexity is $O(|\Sigma|)$, where $\Sigma$ is the character set. In this case, it's lowercase English letters, so $|\Sigma|=26$. @@ -134,8 +134,7 @@ class Solution { class Solution { public: int minimizedStringLength(string s) { - unordered_set ss(s.begin(), s.end()); - return ss.size(); + return unordered_set(s.begin(), s.end()).size(); } }; ``` @@ -173,6 +172,16 @@ impl Solution { } ``` +#### C# + +```cs +public class Solution { + public int MinimizedStringLength(string s) { + return new HashSet(s).Count; + } +} +``` + diff --git a/solution/2700-2799/2716.Minimize String Length/Solution.cpp b/solution/2700-2799/2716.Minimize String Length/Solution.cpp index 4826683b74eff..23f78810d5f7a 100644 --- a/solution/2700-2799/2716.Minimize String Length/Solution.cpp +++ b/solution/2700-2799/2716.Minimize String Length/Solution.cpp @@ -1,7 +1,6 @@ class Solution { public: int minimizedStringLength(string s) { - unordered_set ss(s.begin(), s.end()); - return ss.size(); + return unordered_set(s.begin(), s.end()).size(); } -}; \ No newline at end of file +}; diff --git a/solution/2700-2799/2716.Minimize String Length/Solution.cs b/solution/2700-2799/2716.Minimize String Length/Solution.cs new file mode 100644 index 0000000000000..8c3b7ebe8acdb --- /dev/null +++ b/solution/2700-2799/2716.Minimize String Length/Solution.cs @@ -0,0 +1,5 @@ +public class Solution { + public int MinimizedStringLength(string s) { + return new HashSet(s).Count; + } +}