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2828. Check if a String Is an Acronym of Words
Description
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.
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.
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.
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 <= 1001 <= words[i].length <= 101 <= s.length <= 100words[i]andsconsist of lowercase English letters.
Solutions
Solution 1: Simulation
We can iterate over each string in the array $words$, concatenate their first letters to form a new string $t$, and then check if $t$ is equal to $s$.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $words$.
Solution 2: Simulation (Space Optimization)
First, we check if the number of strings in $words$ is equal to the length of $s$. If not, $s$ is definitely not an acronym of the first letters of $words$, and we directly return $false$.
Then, we iterate over each character in $s$, checking if it is equal to the first letter of the corresponding string in $words$. If not, $s$ is definitely not an acronym of the first letters of $words$, and we directly return $false$.
After the iteration, if we haven’t returned $false$, then $s$ is an acronym of the first letters of $words$, and we return $true$.
The time complexity is $O(n)$, where $n$ is the length of the array $words$. The space complexity is $O(1)$.
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class Solution { public boolean isAcronym(List<String> words, String s) { StringBuilder t = new StringBuilder(); for (var w : words) { t.append(w.charAt(0)); } return t.toString().equals(s); } } -
class Solution { public: bool isAcronym(vector<string>& words, string s) { string t; for (auto& w : words) { t += w[0]; } return t == s; } }; -
class Solution: def isAcronym(self, words: List[str], s: str) -> bool: return "".join(w[0] for w in words) == s -
func isAcronym(words []string, s string) bool { t := []byte{} for _, w := range words { t = append(t, w[0]) } return string(t) == s } -
function isAcronym(words: string[], s: string): boolean { return words.map(w => w[0]).join('') === s; } -
impl Solution { pub fn is_acronym(words: Vec<String>, s: String) -> bool { words .iter() .map(|w| w.chars().next().unwrap_or_default()) .collect::<String>() == s } }