An Anagram Detection Example

• One string is an anagram of another if the second is simply a rearrangement of the first.
• We check to see that each character in the first string actually occurs in the second. 

Python code:

def anagramSolution1(s1,s2):

s1= s1.replace(" ", "").lower()
s2= s2.replace(" ", "").lower()

#edge cases
if len(s1) != len(s2):
return False
# create a dictonary
count = {}

for letter in s1:
if letter in count:
count[letter] += 1

else:
count[letter] = 1

for letter in s2:
if letter in count:
count[letter] -= 1
else:
count[letter] = 1

for k in count:
if count[k] != 0:
return False

return True

print(anagramSolution1('abcd','dcba'))

Day 3: 10001st prime

Question:
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.

What is the 10 001st prime number?

def isPrime(n):    

    if n < 2: return "Neither prime, nor composite"
    for i in range(2, n):       

    # even division so its false      

         if n % i == 0:            

            return False    

    return True

def nthPrime(n):   

    numberOfPrimes = 0    

    prime = 1
    while numberOfPrimes < n:       

          prime += 1        

          if isPrime(prime):            

              numberOfPrimes += 1    

    return prime

print(nthPrime(10001))

Picture:

Day 2: Sum square difference

Question:

The sum of the squares of the first ten natural numbers is,

1² + 2² + ... + 10² = 385

The square of the sum of the first ten natural numbers is,

(1 + 2 + ... + 10)² = 55² = 3025

Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640.

Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

Solution:

The sum of natural numbers can be expressed as

The square of the sum:                                                                                                

Python code:

product =0

for i in range(1,101):    

    square = i* i #suare of sum    #sum of square    

    product +=square

print("The sum of the squares of the first ten natural numbers is,",product)

# The square of the sum of the first ten natural numbers is,

num =0

for i in range(1,101):   

    num += i
# squaring the number

result = (num)*(num)

print("The square of the sum of the first ten natural numbers i",result)
print ("The difference between the sum of the squares of the first ten 

        \natural numbers and the square of the sum is",result - product)

Day 1: largest palindrome

Question 1)

A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.

Find the largest palindrome made from the product of two 3-digit numbers.

Answer:

The largest palindrome can be written as: abccba Which then simpifies to: 100000a + 10000b + 1000c + 100c + 10b + a And then: 100001a + 10010b + 1100c Factoring out 11, you get: 11(9091a + 910b + 100c) So the palindrome must be divisible by 11. Seeing as 11 is prime, at least one of the numbers must be divisible by 11.

#Create a sequence from 100 to 999

three_dig = range(100,1000)
# and an empty list for the palindromes

list_palindromes = []
# loop twice over the number sequence

for n1 in three_dig:    

for n2 in three_dig:       

 # Get the product       

 number =  n1 * n2
 # Transform it in string      

 number = str(number)
 # Check the whole string to test      

  # if it's a palindrome       

 if number == number[::-1]:         

   # Add the palindrome to the list        

 list_palindromes.append(int(number))

print max(list_palindromes)