Leetcode 2828: Check if a String Is an Acronym of Words

Input: The input consists of an array `words` of strings and a string `s`.

Example: For example, `words = ['orange', 'pear', 'grape']` and `s = 'opg'`

Constraints:

• 1 <= words.length <= 100

• 1 <= words[i].length <= 10

• 1 <= s.length <= 100

• words[i] and s consist of lowercase English letters.

Output: Return `true` if `s` is an acronym of `words`, otherwise return `false`.

Example: For `words = ['dog', 'cat', 'fish']` and `s = 'dcf'`, the output is `true`.

Constraints:

Goal: The goal is to check if the string `s` matches the concatenation of the first characters from each string in `words`.

Steps:

• Initialize an empty string to store the acronym.

• For each word in `words`, add its first character to the acronym string.

• Compare the acronym with `s` and return `true` if they match, otherwise return `false`.

Goal: The input will consist of a list of strings and a target acronym string.

Steps:

• The length of `words` is between 1 and 100.

• Each word in `words` has a length between 1 and 10.

• The length of `s` is between 1 and 100.

Assumptions:

• The input strings only contain lowercase English letters.

• The solution should work within the constraints for both the `words` array and string `s`.

• Input: For `words = ['orange', 'pear', 'grape']` and `s = 'opg'`

• Explanation: The first characters of the words 'orange', 'pear', and 'grape' are 'o', 'p', and 'g', respectively. Hence, `s = 'opg'` is the acronym.

Approach: The solution involves creating the acronym by concatenating the first characters of the strings in `words` and comparing it with `s`.

Observations:

• The task is simple: extract the first letter of each word and compare it with the given string `s`.

• By iterating over the list of words and constructing the acronym, we can efficiently check if `s` matches.

Steps:

• Create an empty string to store the acronym.

• Iterate through each word in `words` and append its first character to the acronym string.

• Compare the final acronym with `s`. Return `true` if they match, otherwise return `false`.

Empty Inputs:

• An empty input for `words` would not happen due to the problem constraints.

Large Inputs:

• Ensure that the algorithm can handle the maximum input size of `words`.

Special Values:

• Handle cases where `s` has fewer characters than the total number of words.

Constraints:

• The solution should run efficiently within the time and space constraints for the given problem.

bool isAcronym(vector<string>& words, string s) {
    string res = "";
    for(string w: words)
        res += w[0];
    return s == res;
}

1 : Function Definition

bool isAcronym(vector<string>& words, string s) {

The function 'isAcronym' is defined, taking a vector of strings 'words' and a string 's'. It checks whether 's' is an acronym formed by the first letter of each word in 'words'.

2 : Variable Initialization

    string res = "";

A string variable 'res' is initialized as an empty string. This variable will store the concatenated first letters of each word in the 'words' vector.

3 : Loop Start

    for(string w: words)

A for loop is initiated, iterating over each word 'w' in the 'words' vector. This loop will extract the first character of each word.

4 : Concatenate First Letter

        res += w[0];

The first character of each word 'w' is concatenated to the 'res' string, effectively forming the acronym.

5 : Return Statement

    return s == res;

The function compares the formed acronym 'res' with the input string 's'. If they are equal, it returns true; otherwise, false.

Best Case: O(n)

Average Case: O(n)

Worst Case: O(n)

Description: The time complexity is O(n), where n is the length of the `words` array.

Best Case: O(1)

Worst Case: O(n)

Description: The space complexity is O(n), where n is the number of words, as we store the acronym string.

Leetcode 2828: Check if a String Is an Acronym of Words

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