From 1c39bcee9f8f30f359ef98f46431e89a85392c03 Mon Sep 17 00:00:00 2001
From: jackfiled <xcrenchangjun@outlook.com>
Date: Wed, 20 Dec 2023 10:42:38 +0800
Subject: [PATCH] 20231220 Finished

---
 src/solution/mod.rs                           |  1 +
 ...heck_if_a_string_is_an_acronym_of_words.rs | 78 +++++++++++++++++++
 2 files changed, 79 insertions(+)
 create mode 100644 src/solution/s2828_check_if_a_string_is_an_acronym_of_words.rs

diff --git a/src/solution/mod.rs b/src/solution/mod.rs
index 5e2b42e..bf37cc0 100644
--- a/src/solution/mod.rs
+++ b/src/solution/mod.rs
@@ -4,3 +4,4 @@ mod s0020_valid_parentheses;
 mod s2697_lexicographically_smallest_palindrome;
 mod s0002_add_two_numbers;
 mod s0003_longest_substring_without_repeating_characters;
+mod s2828_check_if_a_string_is_an_acronym_of_words;
diff --git a/src/solution/s2828_check_if_a_string_is_an_acronym_of_words.rs b/src/solution/s2828_check_if_a_string_is_an_acronym_of_words.rs
new file mode 100644
index 0000000..eaecc18
--- /dev/null
+++ b/src/solution/s2828_check_if_a_string_is_an_acronym_of_words.rs
@@ -0,0 +1,78 @@
+/**
+ * [2828] Check if a String Is an Acronym of Words
+ *
+ * Given an array of strings words and a string s, determine if s is an acronym of words.
+ * The string s is considered an acronym of words if it can be formed by concatenating the first character of each string in words in order. For example, "ab" can be formed from ["apple", "banana"], but it can't be formed from ["bear", "aardvark"].
+ * Return true if s is an acronym of words, and false otherwise. 
+ *  
+ * <strong class="example">Example 1:
+ *
+ * Input: words = ["alice","bob","charlie"], s = "abc"
+ * Output: true
+ * Explanation: The first character in the words "alice", "bob", and "charlie" are 'a', 'b', and 'c', respectively. Hence, s = "abc" is the acronym. 
+ *
+ * <strong class="example">Example 2:
+ *
+ * Input: words = ["an","apple"], s = "a"
+ * Output: false
+ * Explanation: The first character in the words "an" and "apple" are 'a' and 'a', respectively. 
+ * The acronym formed by concatenating these characters is "aa". 
+ * Hence, s = "a" is not the acronym.
+ *
+ * <strong class="example">Example 3:
+ *
+ * Input: words = ["never","gonna","give","up","on","you"], s = "ngguoy"
+ * Output: true
+ * Explanation: By concatenating the first character of the words in the array, we get the string "ngguoy". 
+ * Hence, s = "ngguoy" is the acronym.
+ *
+ *  
+ * Constraints:
+ *
+ * 	1 <= words.length <= 100
+ * 	1 <= words[i].length <= 10
+ * 	1 <= s.length <= 100
+ * 	words[i] and s consist of lowercase English letters.
+ *
+ */
+pub struct Solution {}
+
+// problem: https://leetcode.cn/problems/check-if-a-string-is-an-acronym-of-words/
+// discuss: https://leetcode.cn/problems/check-if-a-string-is-an-acronym-of-words/discuss/?currentPage=1&orderBy=most_votes&query=
+
+// submission codes start here
+
+impl Solution {
+    pub fn is_acronym(words: Vec<String>, s: String) -> bool {
+        if words.len() != s.len() {
+            return false;
+        }
+
+        for (index, value) in s.chars().enumerate() {
+            if !words[index].starts_with(value) {
+                return false;
+            }
+        }
+
+        true
+    }
+}
+
+// submission codes end
+
+#[cfg(test)]
+mod tests {
+    use super::*;
+
+    #[test]
+    fn test_2828() {
+        assert_eq!(true, Solution::is_acronym(vec![
+            String::from("alice"),
+            String::from("bob"),
+            String::from("charlie")], String::from("abc")));
+
+        assert_eq!(false, Solution::is_acronym(vec![
+            String::from("an"),
+            String::from("apple")], String::from("a")));
+    }
+}