Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Build confidence with an intuition-first walkthrough focused on array fundamentals.
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] and s consist of lowercase English letters.Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array
["alice","bob","charlie"] "abc"
["an","apple"] "a"
["never","gonna","give","up","on","you"] "ngguoy"
word-abbreviation)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2828: Check if a String Is an Acronym of Words
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);
}
}
// Accepted solution for LeetCode #2828: Check if a String Is an Acronym of Words
func isAcronym(words []string, s string) bool {
t := []byte{}
for _, w := range words {
t = append(t, w[0])
}
return string(t) == s
}
# Accepted solution for LeetCode #2828: Check if a String Is an Acronym of Words
class Solution:
def isAcronym(self, words: List[str], s: str) -> bool:
return "".join(w[0] for w in words) == s
// Accepted solution for LeetCode #2828: Check if a String Is an Acronym of Words
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
}
}
// Accepted solution for LeetCode #2828: Check if a String Is an Acronym of Words
function isAcronym(words: string[], s: string): boolean {
return words.map(w => w[0]).join('') === s;
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.
Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.