Update abbreviation.rs

Author: AliAlimohammadi

src/dynamic_programming/abbreviation.rs

@@ -1,14 +1,14 @@
 //! Abbreviation Problem Solution
 //!
-//! This module solves the abbreviation problem: determining if string `a` can be 
-//! transformed into string `b` by capitalizing zero or more lowercase letters and 
+//! This module solves the abbreviation problem: determining if string `a` can be
+//! transformed into string `b` by capitalizing zero or more lowercase letters and
 //! deleting all remaining lowercase letters.
 
 /// Determines if string `a` can be transformed into string `b` by:
 /// 1. Capitalizing zero or more lowercase letters in `a`
 /// 2. Deleting all remaining lowercase letters
 ///
-/// The solution uses dynamic programming where `dp[i][j]` represents whether 
+/// The solution uses dynamic programming where `dp[i][j]` represents whether
 /// the first `i` characters of `a` can form the first `j` characters of `b`.
 ///
 /// # Arguments
@@ -25,7 +25,7 @@
 /// # Examples
 /// ```
 /// use the_algorithms_rust::dynamic_programming::abbreviation;
-/// 
+///
 /// assert_eq!(abbreviation("daBcd", "ABC"), true);
 /// assert_eq!(abbreviation("dBcd", "ABC"), false);
 /// ```
@@ -34,36 +34,36 @@ pub fn abbreviation(a: &str, b: &str) -> bool {
     let b_chars: Vec<char> = b.chars().collect();
     let n = a_chars.len();
     let m = b_chars.len();
-    
+
     // dp[i][j] represents whether first i chars of a can form first j chars of b
     let mut dp = vec![vec![false; m + 1]; n + 1];
-    
+
     // Base case: empty string a can form empty string b
     dp[0][0] = true;
-    
+
     // Fill the first column: we can form empty b by deleting all lowercase letters
     for i in 0..n {
         if a_chars[i].is_lowercase() {
             dp[i + 1][0] = dp[i][0];
         }
     }
-    
+
     for i in 0..n {
         for j in 0..=m {
             if dp[i][j] {
                 // If we can match current position, check next characters
                 if j < m && a_chars[i].to_ascii_uppercase() == b_chars[j] {
                     dp[i + 1][j + 1] = true;
                 }
-                
+
                 // If current character in a is lowercase, we can delete it
                 if a_chars[i].is_lowercase() {
                     dp[i + 1][j] = true;
                 }
             }
         }
     }
-    
+
     dp[n][m]
 }
 
@@ -86,6 +86,8 @@ mod tests {
         // Original test cases from the problem
         test_daBcd_ABC: ("daBcd", "ABC") => true,
         test_dBcd_ABC: ("dBcd", "ABC") => false,
+
+        // Additional test cases
         test_AbcE_ABE: ("AbcE", "ABE") => true,
         test_AbcE_ABC: ("AbcE", "ABC") => false,
         test_abcde_ABCDE: ("abcde", "ABCDE") => true,
@@ -94,15 +96,15 @@ mod tests {
         test_ABCDE_ABCD: ("ABCDE", "ABCD") => false,
         test_aBcDe_ABCDE: ("aBcDe", "ABCDE") => true,
         test_aBcDe_ABCD: ("aBcDe", "ABCD") => true,
-        
+
         // Edge test cases
         test_empty_both: ("", "") => true,
         test_empty_a: ("", "ABC") => false,
         test_empty_b: ("abc", "") => true,
         test_only_lowercase: ("abc", "ABC") => true,
         test_only_uppercase: ("ABC", "ABC") => true,
         test_mismatched_uppercase: ("ABD", "ABC") => false,
-        
+
         // Complex cases from HackerRank
         test_complex_1: ("LLZOSYAMQRMBTZXTQMQcKGLR", "LLZOSYAMBTZXMQKLR") => false,
         test_complex_2: ("MGYXKOVSMAHKOLAZZKWXKS", "MGXKOVSAHKOLZKKDP") => false,