Check if a String Is an Acronym of Words - Practice Coding Problems

Check if a String Is an Acronym of Words - Problem

Array String Easy

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.

Input & Output

Example 1 — Basic Valid Acronym

$ Input: words = ["alice","bob","charlie"], s = "abc"

› Output: true

💡 Note: Taking first letters: 'a' from "alice", 'b' from "bob", 'c' from "charlie" gives "abc" which matches s

Example 2 — Invalid Acronym

$ Input: words = ["an","apple"], s = "a"

› Output: false

💡 Note: First letters are 'a' and 'a', forming "aa", but s = "a" has different length

Example 3 — Single Character Match

$ Input: words = ["never"], s = "n"

› Output: true

💡 Note: Single word "never" has first letter 'n' which matches s = "n"

Constraints

1 ≤ words.length ≤ 100
1 ≤ words[i].length ≤ 10
1 ≤ s.length ≤ 100
words[i] and s consist of lowercase English letters

Visualization

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Understanding the Visualization

Input

Array of words and target string

Extract

Take first character of each word

Compare

Check if extracted characters match target string

Key Takeaway

🎯 Key Insight: Simply compare first character of each word with corresponding position in target string

Asked in

G Google 25 a Amazon 20 M Microsoft 15

The key insight is to check if the first character of each word in sequence forms the target string. The optimal approach is direct character comparison with O(n) time and O(1) space complexity.

Common Approaches

Approach	Time	Space	Notes
✓ Direct Character Comparison	O(n)	O(1)	Compare each word's first character with corresponding character in s
String Building Approach	O(n)	O(n)	Build acronym string from first characters, then compare with target

Direct Character Comparison — Algorithm Steps

Step 1: Check if s.length equals words.length
Step 2: Loop through each word and compare first character with s[i]

Visualization

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Step-by-Step Walkthrough

Length Check

Verify words array length matches string length

Character Match

Compare first letter of each word with corresponding character

Result

Return true if all characters match

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>

bool solution(char words[][100], int wordsCount, char* s) {
    int sLen = strlen(s);
    if (wordsCount != sLen) {
        return false;
    }
    
    for (int i = 0; i < wordsCount; i++) {
        if (words[i][0] != s[i]) {
            return false;
        }
    }
    
    return true;
}

int main() {
    char line[1000];
    char words[100][100];
    int wordsCount = 0;
    
    fgets(line, sizeof(line), stdin);
    
    // Parse JSON array manually
    char* ptr = line + 1; // Skip [
    char* token = strtok(ptr, ",]");
    while (token != NULL) {
        // Remove quotes and whitespace
        while (*token == ' ' || *token == '"') token++;
        int len = strlen(token);
        while (len > 0 && (token[len-1] == ' ' || token[len-1] == '"' || token[len-1] == '\n')) {
            token[len-1] = '\0';
            len--;
        }
        strcpy(words[wordsCount], token);
        wordsCount++;
        token = strtok(NULL, ",]");
    }
    
    char s[100];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0; // Remove newline
    
    bool result = solution(words, wordsCount, s);
    printf(result ? "true\n" : "false\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single loop through n words, each comparison is O(1)

✓ Linear Growth

Space Complexity

O(1)

Only using variables, no extra data structures

✓ Linear Space

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Medium Frequency

~8 min Avg. Time

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Input & Output

Constraints

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Related Problems

Common Approaches

Direct Character Comparison — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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