In an age where artificial intelligence (AI) is rapidly evolving, the essence of programming remains irreplaceable. It’s a common misconception that the rise of AI will render the need for human coders obsolete. However, the reality is quite the contrary. As AI continues to shape our digital landscape, the demand for skilled programmers who can understand, debug, and enhance these intelligent systems has never been higher. It’s not just about instructing a computer to perform tasks anymore; it’s about crafting the very intellect that drives future innovations.
Embarking on the journey of mastering algorithms and data structures is not merely an academic endeavor; it’s a pathway to becoming a key player in the technological evolution. These foundational skills are the building blocks for solving complex problems, not just in software development but also in the realms of AI and machine learning. There’s no magic wand — achieving proficiency requires consistent effort and dedication.
If you’re aspiring to contribute to the tech revolutions of tomorrow, remember that the journey of a thousand codes begins with a single algorithm. While the golden era of programming began decades ago, today presents the next best opportunity to dive into the world of programming. Let’s secure your place in the future of technology, starting with a seemingly simple but profoundly insightful LeetCode problem: deciphering acronyms from arrays of words.
https://leetcode.com/problems/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.
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 <= 100