GitHub Repo: https://github.com/ImBIOS/common-char-extractor-ts

In the realm of coding interviews, algorithmic challenges serve not just as tests of technical knowledge but as windows into the problem-solving acumen of candidates. One such intriguing problem, often encountered, revolves around string manipulation—a fundamental skill in a developer's arsenal. Today, we dive into a captivating problem: identifying common characters across multiple strings, and how we approached its solution using TypeScript.

The Challenge Unveiled

The task at hand was seemingly straightforward yet deceptively complex: given an array of strings, we were to write a function that extracts and returns an array of characters that appear in the same position across all strings. For example, given the input ["abcd", "bcd", "cde", "cdef"], the expected output would be ["c", "d"], signifying that 'c' and 'd' are the common characters shared among all strings in the corresponding positions.

Crafting the Solution

Our journey began with the choice of TypeScript for the solution—a decision driven by TypeScript's robust type system, enhancing code reliability and developer productivity. The initial step involved iterating over each character of the first string, using it as a reference to compare with the corresponding characters in the subsequent strings.

/**
 * Extracts and returns an array of characters that appear in the same position
 * across all strings within the input array.
 *
 * @param {string[]} words - An array of strings to be analyzed.
 * @returns {string[]} An array of characters that are common across all provided strings,
 * maintaining their order of appearance in the first string.
 */
export function extractCommonLetters(words: string[]): string[] {
  if (words.length === 0) {
    return [];
  }

  const copiedWords = [...words];
  const result: string[] = [];
  const referenceWord = copiedWords[0];

  // Iterate over each character in the first word
  for (let i = 0; i < referenceWord.length; i++) {
    const currentChar = referenceWord[i];
    let isCharCommonAcrossWords = true;

    // Check if the current character exists in each of the other words
    for (let j = 1; j < copiedWords.length; j++) {
      const word = copiedWords[j];
      const charIndex = word.indexOf(currentChar);

      // If the character is not found, break and set flag to false
      if (charIndex === -1) {
        isCharCommonAcrossWords = false;
        break;
      } else {
        // Remove the found character from the word to handle duplicates
        copiedWords[j] = word.slice(0, charIndex) + word.slice(charIndex + 1);
      }
    }

    // If the character is common across all words, add it to the result
    if (isCharCommonAcrossWords) {
      result.push(currentChar);
    }
  }

  return result;
}

This snippet encapsulates the essence of our approach, emphasizing readability and straightforward logic to tackle the problem.

Venturing Beyond: Testing the Solution

To validate our solution, we employed a suite of test cases covering various scenarios—from basic cases to more complex ones involving special characters and case sensitivity. This thorough testing ensured our algorithm's robustness and correctness across a wide range of inputs.

Analyzing Space and Time Complexity

Upon reflection, our solution's time complexity is primarily determined by the number of characters in the first string (let's denote this as N) and the total number of strings (denote this as M). For each character in the first string, we check its occurrence in the corresponding position across all other strings, leading to a time complexity of O(N*M).

The space complexity of our solution is O(N), as we store the common characters in an array. In the worst-case scenario, where all characters are common across all strings, the size of this array would be proportional to the length of the first string.

Pathways to Improvement

While our solution stands efficient for a moderate number of strings and characters, there's always room for optimization. Here are a few strategies to enhance its performance further:

1. Early Termination: If at any point a character does not match in the corresponding position across all strings, we can break early from the loop, saving unnecessary comparisons.

2. Optimizing for Short Strings: Starting with the shortest string as the reference could potentially reduce the number of iterations, as it minimizes the maximum number of characters to be checked.

3. Parallel Processing: For large datasets, leveraging parallel processing techniques to compare characters across strings concurrently could significantly reduce execution time.

4. Hashing Techniques: Utilizing a hash map to keep track of character positions and occurrences might offer a more sophisticated way to identify common characters, especially when dealing with extensive sets of strings.

Concluding Thoughts

The common characters challenge serves as a remarkable testament to the elegance of string manipulation problems, inviting developers to delve into the nuances of algorithms and data structures. Through TypeScript, we not only found a solution but also embraced a journey of clarity, type safety, and efficient problem-solving.

As we wrap up, it's essential to remember that the journey through coding challenges is as much about the path taken as it is about the destination reached. Each problem, including this one, offers a unique opportunity to refine our skills, rethink our approaches, and, most importantly, continue learning.