ruby-problems

Solving algorithms with Ruby!

Reverse String

Write a function to reverse a string

def reverse(string)
  # your magic
end

Palindrome

Write a function to check if a given string is a palindrome

def isPalindrome(string)
  # your magic
end

Prime Number

Write a function to check if a number is a prime number

def PrimeNumber?(number)
  # your magic
end

Prime Factors

Write a function to find all prime factors of a number

def primeFactors(number)
  # your magic
end

Missing Number

You have an array of numbers 1 to 100 in an array. Only one number is missing in the array. The array is unsorted. Find the missing number.

def missingNumber(array)
  # your magic
end

Caesar Cipher

Write a function to encrypt a message (by shifting char code up or down). Write a function to decrypt a message

Useful methods:

• String::ord
• String::chr

def encrypt(message)
  # your magic
end

def decrypt(message)
  # your magic
end

Number Persistence

In mathematics, the persistence of a number is the number of times one must apply a given operation to an integer before reaching a fixed point at which the operation no longer alters the number.

The additive persistence of 2718 is 2: first we find that 2 + 7 + 1 + 8 = 18, and then that 1 + 8 = 9. Since 9 is just a single number, we stop here.

Write a function to find the additive persistence of a number

def additivePersistence(number)
  # your magic
end

Triple Double

Have the function TripleDouble(num1,num2) take both parameters being passed, and return 1 if there is a straight triple of a number at any place in num1 and also a straight double of the same number in num2.

For example: if num1 equals 451999277 and num2 equals 41177722899, then return 1 because in the first parameter you have the straight triple 999 and you have a straight double, 99, of the same number in the second parameter. If this isn't the case, return 0.

def tripleDouble(triple,double)
  # your magic
end

Multiplicative Persistence

In mathematics, the persistence of a number is the number of times one must apply a given operation to an integer before reaching a fixed point at which the operation no longer alters the number.

The multiplicative persistence of 39 is 3, because it takes three steps to reduce 39 to a single digit: 39 → 27 → 14 → 4.

Write a function to find the multiplicative persistence of a number. You must use recursion.

def multiplicative_persistence()
  # your magic
end