# Given an integer, return the integer with reversed digits.# Note: The integer could be either positive or negative.
def solution(x):    string = str(x)
    if string[0] == '-':        return int('-'+string[:0:-1])    else:        return int(string[::-1])
print(solution(-231))print(solution(345))

# For a given sentence, return the average word length. # Note: Remember to remove punctuation first.
sentence1 = "Hi all, my name is Tom...I am originally from Australia."sentence2 = "I need to work very hard to learn more about algorithms in Python!"
def solution(sentence):    for p in "!?',;.":        sentence = sentence.replace(p, '')    words = sentence.split()    return round(sum(len(word) for word in words)/len(words),2)
print(solution(sentence1))print(solution(sentence2))

# Given two non-negative integers num1 and num2 represented as string, return the sum of num1 and num2.# You must not use any built-in BigInteger library or convert the inputs to integer directly.
#Notes:#Both num1 and num2 contains only digits 0-9.#Both num1 and num2 does not contain any leading zero.
num1 = '364'num2 = '1836'
# Approach 1: def solution(num1,num2):    eval(num1) + eval(num2)    return str(eval(num1) + eval(num2))
print(solution(num1,num2))
#Approach2 #Given a string of length one, the ord() function returns an integer representing the Unicode code point of the character #when the argument is a unicode object, or the value of the byte when the argument is an 8-bit string.
def solution(num1, num2):    n1, n2 = 0, 0    m1, m2 = 10**(len(num1)-1), 10**(len(num2)-1)
    for i in num1:        n1 += (ord(i) - ord("0")) * m1         m1 = m1//10        
    for i in num2:        n2 += (ord(i) - ord("0")) * m2        m2 = m2//10
    return str(n1 + n2)print(solution(num1, num2))

# Given a string, find the first non-repeating character in it and return its index. # If it doesn't exist, return -1. # Note: all the input strings are already lowercase.
#Approach 1def solution(s):    frequency = {}    for i in s:        if i not in frequency:            frequency[i] = 1        else:            frequency[i] +=1    for i in range(len(s)):        if frequency[s[i]] == 1:            return i    return -1
print(solution('alphabet'))print(solution('barbados'))print(solution('crunchy'))
print('###')
#Approach 2import collections
def solution(s):    # build hash map : character and how often it appears    count = collections.Counter(s) # <-- gives back a dictionary with words occurrence count                                          #Counter({'l': 1, 'e': 3, 't': 1, 'c': 1, 'o': 1, 'd': 1})    # find the index    for idx, ch in enumerate(s):        if count[ch] == 1:            return idx         return -1
print(solution('alphabet'))print(solution('barbados'))print(solution('crunchy'))

# Given a non-empty string s, you may delete at most one character. Judge whether you can make it a palindrome.# The string will only contain lowercase characters a-z.
s = 'radkar'def solution(s):    for i in range(len(s)):        t = s[:i] + s[i+1:]        if t == t[::-1]: return True
    return s == s[::-1]
solution(s)

# Given an array of integers, determine whether the array is monotonic or not.A = [6, 5, 4, 4] B = [1,1,1,3,3,4,3,2,4,2]C = [1,1,2,3,7]
def solution(nums):     return (all(nums[i] <= nums[i + 1] for i in range(len(nums) - 1)) or             all(nums[i] >= nums[i + 1] for i in range(len(nums) - 1))) 
print(solution(A)) print(solution(B)) print(solution(C))

#Given an array nums, write a function to move all zeroes to the end of it while maintaining the relative order of #the non-zero elements.
array1 = [0,1,0,3,12]array2 = [1,7,0,0,8,0,10,12,0,4]
def solution(nums):    for i in nums:        if 0 in nums:            nums.remove(0)            nums.append(0)    return nums
solution(array1)solution(array2)

# Given an array containing None values fill in the None values with most recent # non None value in the array array1 = [1,None,2,3,None,None,5,None]
def solution(array):    valid = 0                res = []                     for i in nums:         if i is not None:                res.append(i)            valid = i        else:            res.append(valid)    return res
solution(array1)

#Given two sentences, return an array that has the words that appear in one sentence and not#the other and an array with the words in common. 
sentence1 = 'We are really pleased to meet you in our city'sentence2 = 'The city was hit by a really heavy storm'
def solution(sentence1, sentence2):    set1 = set(sentence1.split())    set2 = set(sentence2.split())
    return sorted(list(set1^set2)), sorted(list(set1&set2)) # ^ A.symmetric_difference(B), & A.intersection(B)
print(solution(sentence1, sentence2))

# Given k numbers which are less than n, return the set of prime number among them# Note: The task is to write a program to print all Prime numbers in an Interval.# Definition: A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. 
n = 35def solution(n):    prime_nums = []    for num in range(n):        if num > 1: # all prime numbers are greater than 1            for i in range(2, num):                if (num % i) == 0: # if the modulus == 0 is means that the number can be divided by a number preceding it                    break            else:                prime_nums.append(num)    return prime_numssolution(n)