LC 2000. Reverse Prefix of Word (Rust)

Author: raul-sauco

README.md

@@ -588,6 +588,7 @@ Solutions to LeetCode problems. The first column links to the problem in LeetCod
 | [1980. Find Unique Binary String][lc1980]                                                     | 🟠 Medium  | [![rust](res/rs.png)][lc1980rs]                                                                       |
 | [1991. Find the Middle Index in Array][lc1991]                                                | 🟢 Easy    | [![python](res/py.png)][lc1991py]                                                                     |
 | [1996. The Number of Weak Characters in the Game][lc1996]                                     | 🟠 Medium  | [![python](res/py.png)][lc1996py]                                                                     |
+| [2000. Reverse Prefix of Word][lc2000]                                                        | 🟢 Easy    | [![rust](res/rs.png)][lc2000rs]                                                                       |
 | [2007. Find Original Array From Doubled Array][lc2007]                                        | 🟠 Medium  | [![python](res/py.png)][lc2007py]                                                                     |
 | [2009. Minimum Number of Operations to Make Array Continuous][lc2009]                         | 🔴 Hard    | [![rust](res/rs.png)][lc2009rs]                                                                       |
 | [2013. Detect Squares][lc2013]                                                                | 🟠 Medium  | [![python](res/py.png)][lc2013py]                                                                     |
@@ -2003,6 +2004,8 @@ Solutions to LeetCode problems. The first column links to the problem in LeetCod
 [lc1991py]: leetcode/find-the-middle-index-in-array.py
 [lc1996]: https://leetcode.com/problems/the-number-of-weak-characters-in-the-game/
 [lc1996py]: leetcode/the-number-of-weak-characters-in-the-game.py
+[lc2000]: https://leetcode.com/problems/reverse-prefix-of-word/
+[lc2000rs]: leetcode/reverse-prefix-of-word.rs
 [lc2007]: https://leetcode.com/problems/find-original-array-from-doubled-array/
 [lc2007py]: leetcode/find-original-array-from-doubled-array.py
 [lc2009]: https://leetcode.com/problems/minimum-number-of-operations-to-make-array-continuous/

leetcode/reverse-prefix-of-word.rs

@@ -0,0 +1,65 @@
+// 2000. Reverse Prefix of Word
+// 🟢 Easy
+//
+// https://leetcode.com/problems/reverse-prefix-of-word/
+//
+// Tags: Two Pointers - String
+
+struct Solution;
+impl Solution {
+    /// Check if the character is found in the input, if found reverse the characters up to and
+    /// including that index, then append the remaining characters and return that.
+    ///
+    /// Time complexity: O(n) - We iterate over the characters once to find the first index of, if
+    /// found, we iterate in reverse over the characters up to that index then forward over the
+    /// characters after that index, building the result.
+    /// Space complexity: O(1) - Unless iterators are internally using extra memory, which I don't
+    /// think is the case, then constant extra memory.
+    ///
+    /// Runtime 0 ms Beats 100%
+    /// Memory 2.07 MB Beats 90%
+    pub fn reverse_prefix(word: String, ch: char) -> String {
+        match word.chars().position(|c| c == ch) {
+            Some(i) => word[..=i]
+                .chars()
+                .rev()
+                .chain(word[i + 1..].chars())
+                .collect(),
+            None => word,
+        }
+    }
+}
+
+// Tests.
+fn main() {
+    let tests = [
+        ("abcdefd", 'd', "dcbaefd"),
+        ("xyxzxe", 'z', "zxyxxe"),
+        ("abcd", 'z', "abcd"),
+    ];
+    println!("\n\x1b[92m» Running {} tests...\x1b[0m", tests.len());
+    let mut success = 0;
+    for (i, t) in tests.iter().enumerate() {
+        let res = Solution::reverse_prefix(t.0.to_string(), t.1);
+        if res == t.2 {
+            success += 1;
+            println!("\x1b[92m✔\x1b[95m Test {} passed!\x1b[0m", i);
+        } else {
+            println!(
+                "\x1b[31mx\x1b[95m Test {} failed expected: {:?} but got {}!!\x1b[0m",
+                i, t.2, res
+            );
+        }
+    }
+    println!();
+    if success == tests.len() {
+        println!("\x1b[30;42m✔ All tests passed!\x1b[0m")
+    } else if success == 0 {
+        println!("\x1b[31mx \x1b[41;37mAll tests failed!\x1b[0m")
+    } else {
+        println!(
+            "\x1b[31mx\x1b[95m {} tests failed!\x1b[0m",
+            tests.len() - success
+        )
+    }
+}