top of page
Search
  • Writer's pictureMarko Reljić

A Guide Through Initial Coding Interviews: Exploring Three Key Questions



Embarking on the coding interview journey can be a daunting task for many aspiring programmers. It's a crucial step towards landing a programming job, where your skills in problem-solving, logic, and coding are put to the test. This blog aims to shed light on the coding interview process by dissecting three common yet intriguing questions that you might encounter early on in your programming career.


Each question, distinct in its challenge, serves as an excellent platform for honing your coding abilities. We'll delve into the specifics of these problems, unraveling the thought processes and strategies needed to find effective solutions. This exploration is not just about cracking codes but about understanding the essence of problem-solving in software development.


Whether you're a newcomer to the world of programming or preparing for your next job interview, this guide is designed to equip you with the insights and confidence needed to tackle similar challenges. By walking you through these examples, we hope to inspire a problem-solving mindset and enhance your coding interview preparedness.

Let's dive into the first of these challenges, breaking down the approach and solution, to kickstart your journey with a solid foundation in tackling coding interview questions.


Challenge Overview: Identifying the Longest Word

In this programming challenge, the goal is to develop a function named `LongestWord` that receives a string parameter, `sen`, and returns the most extended word within the string. Should there be multiple words of identical maximum length, the function should return the first occurrence. The string may contain punctuation, which should not affect the evaluation, and words can include numbers, making them eligible as potential longest words.


Complete Solution

Here's a concise and effective solution to tackle this challenge:

function LongestWord(sen) { 
  // Remove punctuation from the string and normalize spaces
  const words = sen.replace(/[^\w\s]|_/g, "").replace(/\s{2,}/g, " ").split(" ");
  let longestWord = "";
  // Iterate through each word to find the longest
  words.forEach((word) => {
    if (word.length > longestWord.length) {
      longestWord = word;
    }
  })
  return longestWord; 
}


Detailed Explanation


The Task

The function `LongestWord` is designed to sift through an input string to find and return the longest word it contains. It treats both letters and numbers as part of words and excludes punctuation from consideration.


Initial String Processing

To ensure punctuation does not skew our word length calculations, we first remove these characters from `sen`. This cleanup is achieved with:

sen.replace(/[^\w\s]|_/g, "")

where `[^\w\s]|_` is a regular expression pattern that matches any character not a word (`\w`) or whitespace (`\s`), along with underscores (`_`). This effectively strips out punctuation. Following this, we tackle any instances of multiple spaces using `.replace(/\s{2,}/g, " ")`, ensuring words are uniformly separated by a single space.


Splitting the String into Words

With a cleaned string, we split `sen` into an array of words based on spaces:

.split(" ")

This array allows us to individually examine each word's length.


Finding and Returning the Longest Word

We then iterate through the array of words, comparing the lengths of current and previously stored words to find the longest one. This process is straightforward: if the current word's length exceeds that of the `longestWord`, we update `longestWord` to hold the current word.

By the end of the iteration, `longestWord` contains the longest word found in the input string, which is then returned as the function's output.


Conclusion

This solution to the "Identifying the Longest Word" challenge highlights key programming skills: string manipulation, regular expressions, and iteration through arrays. Through a step-by-step breakdown, we see how each part of the process contributes to efficiently solving the problem, showcasing practical techniques for string processing and conditional logic in JavaScript.


Challenge Overview: Identifying Common Elements

This programming challenge involves creating a function called `FindIntersection` that takes an array of strings, `strArr`, as its input. This array contains two elements, each a list of comma-separated numbers in ascending order. The function's purpose is to determine the numbers that appear in both lists and return them as a comma-separated string in sorted order. If there are no common elements between the two lists, the function should return the string "false."


Complete Solution

Here's a straightforward solution to address this challenge:

function FindIntersection(strArr) { 
  // Split the first and second elements of strArr into arrays of numbers
  const arrayOne = strArr[0].split(",").map(number => parseInt(number.trim()));
  const arrayTwo = strArr[1].split(",").map(number => parseInt(number.trim()));
  let response = [];
  // Create a Set from the second array for efficient lookup
  const array2Set = new Set(arrayTwo);
  // Iterate through the first array to find common elements
  for(let i = 0; i < arrayOne.length; i++) {
      if(array2Set.has(arrayOne[i])) {
        response.push(arrayOne[i])
      }
  }
  // Return the result as a comma-separated string or false if no intersection
  return response.length === 0 ? "false" : response.join(", ");
}


Detailed Explanation


The Task

The `FindIntersection` function is designed to identify numbers present in both lists provided as string elements within an array, `strArr`. This function must handle string manipulation, conversion to integers, and comparison between two sets of numbers.


Preparing the Input

The first step involves converting the comma-separated strings in `strArr` into arrays of integers for easy manipulation. This conversion is necessary because the inputs are strings, and we need to perform numerical comparisons. The `.split(",")` method divides the string into an array of substrings, and `.map(number => parseInt(number.trim()))` converts each substring into an integer, ensuring any spaces are removed with `.trim()`.


Comparing the Lists

To efficiently find common elements, we convert the second list of numbers into a `Set`, `array2Set`. Sets in JavaScript are collections of unique values and provide an efficient way to check for the presence of an element using the `.has()` method. This approach reduces the problem to a simple lookup for each element of the first array in the second array's set, significantly optimizing performance.


Generating the Output

As we iterate through the first array, we collect common elements in the `response` array. If `response` remains empty, it indicates no common elements were found, and we return "false." Otherwise, we join the elements of `response` into a comma-separated string, which is then returned as the function's output.


Conclusion

This solution to the "Identifying Common Elements" challenge demonstrates essential skills in handling arrays, strings, and set operations in JavaScript. By breaking down the process into clear steps—input preparation, comparison, and output formatting—we can effectively solve the problem with an emphasis on readability and efficiency. Through such problems, programmers can sharpen their ability to transform complex requirements into practical, executable code.



Challenge Overview: Optimizing Classroom Seating

In this intriguing challenge, you are tasked with developing a function called `SeatingStudents` that evaluates an array of integers, `arr`, representing classroom seating arrangements. The first element, `K`, denotes the total number of desks, arranged in two columns, with the subsequent elements (`r1, r2, r3, ...`) indicating which desks are already occupied. The goal is to calculate the number of possible ways two students can be seated next to each other—either side-by-side in the same row or one above the other in the same column.


Complete Solution

Here is an effective solution for calculating the seating arrangements:

function SeatingStudents(arr) {
  const numberOfDesks = arr.shift();
  let numberOfWays = 0;
  for (let i = 1; i <= numberOfDesks; i++) {
    // Check for seating to the right and below for odd-numbered desks
    if (i % 2 !== 0) {
      const tableToTheRight = i + 1;
      const tableToTheBottom = i + 2;
      // Increment ways if right and bottom positions are not occupied
      if (!arr.includes(i) && !arr.includes(tableToTheRight) && tableToTheRight <= numberOfDesks) {
        numberOfWays++;
      }
      if (!arr.includes(i) && !arr.includes(tableToTheBottom) && tableToTheBottom <= numberOfDesks) {
        numberOfWays++;
      }
    } else {
      // For even-numbered desks, only check below
      const tableToTheBottom = i + 2;
      if (!arr.includes(i) && !arr.includes(tableToTheBottom) && tableToTheBottom <= numberOfDesks) {
        numberOfWays++;
      }
    }
  }
  return numberOfWays;
}


Detailed Explanation


Understanding the Task

This function seeks to identify possible seating configurations in a classroom setup where desks are organized in two columns. Considering the constraints of already occupied desks, the challenge lies in calculating how many pairs of adjacent (either side-by-side or vertically aligned) desks remain available.


Algorithm Logic

The core of the solution revolves around iterating through the list of desks, using `K` to understand the classroom's size and layout. For each desk (`i`), the algorithm checks potential neighboring seats—directly to the right (`i + 1`) for odd-numbered desks and directly below (`i + 2`) for all desks. This method accounts for the classroom's two-column structure.

1. Handling Occupied Desks: By utilizing `arr.includes()`, the solution efficiently filters out any scenarios where either the current desk or the neighboring desk is occupied, ensuring only viable seating arrangements are considered.

2. Calculating Possible Seatings: The iteration process incrementally builds the total number of feasible seating combinations by checking adjacent seat availability and ensuring these checks stay within the bounds of `K`, the total desk count.


Result Compilation

After completing the iteration, the algorithm returns the cumulative count of seating arrangements, offering a precise answer to the seating configuration challenge.


Conclusion

This problem emphasizes not just algorithmic thinking but also a keen understanding of spatial arrangements and constraints handling. It beautifully illustrates how even simple arithmetic and logical operations can solve seemingly complex real-world problems. Through this challenge, we gain insights into practical problem-solving techniques, especially in optimizing space and managing constraints—skills invaluable in both programming and beyond.


47 views0 comments

Recent Posts

See All

Comments


bottom of page