Given two numbers: n: number of consecutive letters beginning at a k: string length Output all possible k-length possible strings containing the first n letters of the alphabet. Letter repeats are allowed. Sample Run: # of letters: 4 length: 3 aaa aab aac aad aba abb abc abd aca acb acc acd ada adb adc add baa bab bac bad bba bbb bbc bbd bca bcb bcc bcd bda bdb bdc bdd caa cab cac cad cba cbb cbc cbd cca ccb ccc ccd cda cdb cdc cdd daa dab dac dad dba dbb dbc dbd dca dcb dcc dcd dda ddb ddc ddd
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Given two numbers:
n: number of consecutive letters beginning at a
k: string length
Output all possible k-length possible strings containing the first n letters of the alphabet. Letter repeats are allowed.
Sample Run:
# of letters: 4
length: 3
aaa
aab
aac
aad
aba
abb
abc
abd
aca
acb
acc
acd
ada
adb
adc
add
baa
bab
bac
bad
bba
bbb
bbc
bbd
bca
bcb
bcc
bcd
bda
bdb
bdc
bdd
caa
cab
cac
cad
cba
cbb
cbc
cbd
cca
ccb
ccc
ccd
cda
cdb
cdc
cdd
daa
dab
dac
dad
dba
dbb
dbc
dbd
dca
dcb
dcc
dcd
dda
ddb
ddc
ddd
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- Up for the count def counting_series (n): The Champernowme word 1234567891011121314151617181920212223., also known as the counting series, is an infinitely long string of digits made up of all positive integers written out in ascending order without any separators. This function should return the digit at position n of the Champernowne word. Position counting again starts from zero for us budding computer scientists. Of course, the automated tester will throw at your function values of n huge enough that those who construct the Champernowne word as an explicit string will run out of both time and space long before receiving the answer. Instead, note how the fractal structure of this infinite sequence starts with 9 single-digit numbers, followed by 90 two-digit numbers, followed by 900 three-digit numbers, and so on. This observation gifts you a pair of seven league boots that allow their wearer skip prefixes of this series in exponential leaps and bounds, instead of having to crawl…CodeWorkout Gym Course Search exercises... Q Search kola shreya@columbusstate.edu X176: count7 Given a non-negative int n, return the count of the occurrences of 7 as a digit, so for example 717 yields 2. (no loops). Note that mod (%) by 10 yields the rightmost digit (126 % 10 is 6), while divide () by 10 removes the rightmost digit (126/10 is 12). Your Answer: 1 public int count7(int n) 4} Check my answer! Reset Next exercise Feedback CodeWorkout © Virginia Tech About License Privacy Contact an 123 45Generate a random list of 150 integers. Calculate the Standard Deviation of the List. Ask the user for an integer and determine if its exceptional or not.
- TAKE A LIST OF STRING FROM THE USER AND COUNT THE NUMBER OF UNIQUE ELEMENTS WITHOUT USING THE FOR, WHILE, Do-WHILE LOOPS. PREFERRED PROGRAMMING LANGUAGE: JAVA/C++In python, The function count_contains_x takes a list of strings (wordList) and returns the count of how many strings in the list contain at least one upper- or lower-case 'x'. Hint: Use the accumulator pattern to accumulate the count. Hint: You might want to use the string method count. For example: Test Result strlist = ["xerox", "OXEN", "whiffleball", "XOXOX"] print(count_contains_x(strlist)) 3 fish = ["angel", "gold", "cat", "puffer", "damsel"] print(count_contains_x(fish)) 0 strlist = ["X", "x"] print(count_contains_x(strlist)) 2 list_of_one_x = ["x"] print(count_contains_x(list_of_one_x)) 1LAB RESTRICTIONS, PLEASE READ: Do not add any imports, the ones that you need will be given to you. You may not use any lists or list methods, or any while loops. Within your loops, you MUST NOT use any break or continue statements. Please also do not use try-except statements, you should be able to anticipate or prevent any errors from happening at all!
- import Prelude hiding (lookup) data Exp = Num Int | Add Exp Exp | Sub Exp Exp | Mul Exp Exp | Div Exp Exp | Var String | Let String Exp Exp | App Exp Exp | Fnc String Exp deriving (Show) type Env = [(String, Val)] data Val = CVal Int | FnVal String Exp Env | Error String deriving (Show) {-Implement a function lookup:: Env -> String -> Val that looks up the value of a variable within a context. This function returns the value of the variable if found and returns an error otherwise. -} lookup :: Env -> String -> Val -- code here-- {- Implement a function eval:: Exp -> Env -> Val to evaluate an expression with an initial context and return a value. Errors occur in the following cases: (a) Left-hand-side of a function application is not a function. (b) Use an undefined variable (c) Arithmetic expression where operand is not a number -} eval :: Env-> Exp -> Val -- code here --111111 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 count_E=s.count('E') 20 count_e=s.count('e') 21 count_res= count_E + count_e 22 print('Count: ', count_res) 23 24 A string containing a text is given. Find the number of words starting with the letter "e". Text example: Ezheviku dlya ezhat Prinesli dva ezha. Ezheviku ele-ele Ezhata vozle eli ş'eli. (7 words) ### s = "Ezheviku dlya ezhat "Prinesli dva ezha." \ "Ezheviku ele-ele" \ "Ezhata vozle eli s'eli."Revorse the vewels def reverse_vowels(text): Given a text string, create and return a new string constructed by finding all its vowels (for simplicity, in this problem vowels are the letters found in the string 'aeiouAEIOU') and reversing their order, while keeping all other characters exactly as they were in their original positions. However, to make the result look prettier, the capitalization of each moved vowel must be the same as that of the vowel that was originally in the target position. For example, reversing the vowels of 'Ilkka' should produce 'Alkki' instead of 'alkkI'. Applying this operation to random English sentences seems to occasionally give them a curious pseudo-Mediterranean vibe. Along with many possible other ways to perform this square dance, one straightforward way to reverse the vowels starts with collecting all vowels of text into a separate list, and initializing the result to an empty string. After that, iterate through all positions of the original text.…
- Revorse the vewels def reverse_vowels(text): Given a text string, create and return a new string constructed by finding all its vowels (for simplicity, in this problem vowels are the letters found in the string 'aeiouAEIOU') and reversing their order, while keeping all other characters exactly as they were in their original positions. However, to make the result look prettier, the capitalization of each moved vowel must be the same as that of the vowel that was originally in the target position. For example, reversing the vowels of 'Ilkka' should produce 'Alkki' instead of 'alkkI'. Applying this operation to random English sentences seems to occasionally give them a curious pseudo-Mediterranean vibe.Along with many possible other ways to perform this square dance, one straightforward way to reverse the vowels starts with collecting all vowels of text into a separate list, and initializing the result to an empty string. After that, iterate through all positions of the original text.…In Python, The function add_char_counts takes three parameters: a string s and two strings of length 1, char1 and char2, and returns the total number of times either char1 or char2 appears in s. For example, the word "scallions" has 2 s's and 1 n, so add_char_counts("scallions", "s", "n") returns 3. Hint: Use a string method to help solve this problem (count). You do not need an accumulator. For example: Test Result print(add_char_counts('Indiana', 'a', 'b')) 2 print(add_char_counts('BAHAMAS', 'A', 'B')) 4 print(add_char_counts('Mississippi', 'i', 'M')) 5 print(add_char_counts("spam spam spam spam", "s", "p")) 8generate a series of 25 random integers between -100 and 100 inclusive. Store the random integers in a list. Count how many of them are zeroes, positive, and negative. Calculate and display the sum and average of the negative integers, the average of the positive integers, and the average of the 25 integers. Sample output: Generate 25 random integers. Count +ve, -ve and zeroes. Calculatte their sum and average -84 97 48 22 -56 -89 -35 -99 86 -28 -38 51 -21 -46 16 83 8 36 79 -42 -90 21 -78 -42 -4 There are 15 negative integers. The sum of these 15 negative integers is equal to -831. The average of these 15 negative integers is equal to -55.4. There are 10 positive integers. The sum of these 10 positive integers is equal to 468. The average of these 15 positive integers is equal to 46.8. The sum of these 25 positive integers is equal to -363. The average of these 25 positive integers is equal to -14.52.