Merge branch 'jk-palindrome' of https://github.com/jsca-kwok/Ruby into jk-palindrome

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Jessica Kwok 2021-06-29 14:04:08 -07:00
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* [Almost Palindrome Checker](https://github.com/TheAlgorithms/Ruby/blob/master/data_structures/arrays/strings/almost_palindrome_checker.rb)
* [Anagram Checker](https://github.com/TheAlgorithms/Ruby/blob/master/data_structures/arrays/strings/anagram_checker.rb)
* [Jewels And Stones](https://github.com/TheAlgorithms/Ruby/blob/master/data_structures/arrays/strings/jewels_and_stones.rb)
* [Palindrome](https://github.com/TheAlgorithms/Ruby/blob/master/data_structures/arrays/strings/palindrome.rb)
* [Remove Vowels](https://github.com/TheAlgorithms/Ruby/blob/master/data_structures/arrays/strings/remove_vowels.rb)
* [Two Sum](https://github.com/TheAlgorithms/Ruby/blob/master/data_structures/arrays/two_sum.rb)
* [Two Sum Ii](https://github.com/TheAlgorithms/Ruby/blob/master/data_structures/arrays/two_sum_ii.rb)

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# Challenge name: Valid Palindrome
#
# Given a string s, determine if it is a palindrome,
# considering only alphanumeric characters and ignoring cases.
#
# Example 1:
# Input: s = "A man, a plan, a canal: Panama"
# Output: true
# Explanation: "amanaplanacanalpanama" is a palindrome.
#
# Example 2:
# Input: s = "race a car"
# Output: false
# Explanation: "raceacar" is not a palindrome.
#
# Constraints:
# 1 <= s.length <= 2 * 105
# s consists only of printable ASCII characters.
# @param {String} s
# @return {Boolean}
#
# Approach 1: Using Ruby method .reverse
#
# Time Complexity: O(n)
#
def is_palindrome(s)
letters_only = s.downcase.gsub(/[^0-9a-z]/i, '')
letters_only.reverse == letters_only
end
s = 'A man, a plan, a canal: Panama'
puts is_palindrome(s)
# Output: true
# Explanation: "amanaplanacanalpanama" is a palindrome.
s = 'race a car'
puts is_palindrome(s)
# Output: false
# Explanation: "raceacar" is not a palindrome.
s = 'ab_a'
puts is_palindrome(s)
# Output: true
# Explanation: "aba" is a palindrome.
#
# Approach 2: Reversed array
#
# Complexity Analysis:
#
# Time Complexity: O(n), in length n of the string.
#
# We need to iterate through the string: When we filter out non-alphanumeric characters and convert the remaining
# characters to lower-case. When we reverse the string. When we compare the original and the reversed strings.
# Each iteration runs linearly in time (since each character operation completes in constant time).
# Thus, the effective run-time complexity is linear.
#
# Space Complexity: O(n), in length n of the string. We need O(n) additional
# space to store the filtered string and the reversed string.
#
def is_palindrome(s)
letters_only_array = s.downcase.gsub(/[^0-9a-z]/i, '').split('')
reversed_array = []
letters_only_array.each do |letter|
reversed_array.unshift(letter)
end
letters_only_array == reversed_array
end
s = 'A man, a plan, a canal: Panama'
puts is_palindrome(s)
# Output: true
# Explanation: "amanaplanacanalpanama" is a palindrome.
s = 'race a car'
puts is_palindrome(s)
# Output: false
# Explanation: "raceacar" is not a palindrome.
s = 'ab_a'
puts is_palindrome(s)
# Output: true
# Explanation: "aba" is a palindrome.
#
# Approach 2: Two Pointers
#
#
# Complexity Analysis:
#
# Time Complexity: O(n), in length n of the string. We traverse over each
# character at most once until the two pointers meet in the middle, or when
# we break and return early.
# Space Complexity: O(1). No extra space required, at all.
#
def is_palindrome(s)
letters_only = s.downcase.gsub(/[^0-9a-z]/i, '')
p1 = 0
p2 = letters_only.length - 1
while p1 < p2
if letters_only[p1] == letters_only[p2]
p1 += 1
p2 -= 1
else
return false
end
end
true
end
s = 'A man, a plan, a canal: Panama'
puts is_palindrome(s)
# Output: true
# Explanation: "amanaplanacanalpanama" is a palindrome.
s = 'race a car'
puts is_palindrome(s)
# Output: false
# Explanation: "raceacar" is not a palindrome.
s = 'ab_a'
puts is_palindrome(s)
# Output: true
# Explanation: "aba" is a palindrome.