Day: May 2, 2024

Common patterns: iteration, hashing, two-pointer, basic transformations. Try each yourself first – ask for solutions/hints per question when ready.

1. Reverse a string without using .reverse
reverse_string("hello") => "olleh"
2. Palindrome check
   palindrome?("racecar") => true
palindrome?("hello") => false
3. Count vowels
   count_vowels("programming") => 3
4. Find max in array without .max
find_max([3, 7, 2, 9, 4]) => 9
5. Remove duplicates from array, preserve order
   remove_duplicates([1,2,2,3,1,4]) => [1,2,3,4]
6. FizzBuzz (1 to n)
   fizzbuzz(15) => ["1","2","Fizz","4","Buzz",...,"FizzBuzz"]
7. Anagram check
   anagram?("listen", "silent") => true
8. Sum of array, no .sum
array_sum([1,2,3,4]) => 10
9. Capitalize each word (title case), no .capitalize on whole string
   title_case("the ruby language") => "The Ruby Language"
10. Find second largest number
    second_largest([4, 1, 9, 7, 9]) => 7

1. Reverse a string without using .reverse

Concept

1. finding the last string character index to find the last string character first

2. then decreasing the index to find the upto the first character

def reverse_string(str)
  last_str_index = str.length - 1
  result = ""
  
  while last_str_index >= 0
    result << str[last_str_index]

    last_str_index -= 1
  end

  result
end

puts reverse_string("hello")
puts reverse_string("programming")

Solution 2: Another way without using Index variables

def reverse_string(str)
  str.each_char.reduce("") { |result, char| char + result }
end

Concept

Prepend each character to an accumulator instead of appending – that flips the order without touching any index. each_char + reduce replaces the while-loop/counter entirely.

2. Palindrome check

def palindrome?(str)
  first_char_index = 0
  last_char_index = str.length - 1

  while first_char_index < last_char_index
    if str[first_char_index] != str[last_char_index]
      return false 
    end

    first_char_index += 1
    last_char_index -= 1
  end

  return true
end

p palindrome?("ala")
p palindrome?("alla")
p palindrome?("racecar")
p palindrome?("car")

Concept

1. Two pointer approach

Two indices start at opposite ends of the string and move toward each other, comparing elements pairwise:

• first_char_index starts at 0, last_char_index starts at length - 1
• Each iteration compares str[first] vs str[last] – if they ever mismatch, it can’t be a palindrome, so return immediately
• Otherwise, both pointers move inward (first += 1, last -= 1) until they meet or cross (first < last becomes false)
• If the loop finishes without a mismatch, all mirrored pairs matched → palindrome

Why this pattern in general: it’s the go-to when you need to compare elements from both ends of a sequence without extra space – palindromes, reversing in-place, “sorted array pair sum” problems, container/water-trapping problems all reuse this exact skeleton (two indices, converge or diverge, one comparison per step).

One edge case worth saying out loud in an interview: this correctly handles even-length (abba) and odd-length (aba, middle char never gets compared to itself) without any special-casing – that’s often a follow-up question.

Count vowels

p count_vowels("programming")