Answer 1

https://code.sololearn.com/c9UQlZRStECI/?ref=app

Answer 2

name=input("Enter your name ")
print(name[-1]+name[:len(name)-1])

U need some changes in this program after that u can make . Store your first letter in variable and append it in last

Answer 3

https://code.sololearn.com/cpflVlg8E3sT/?ref=app

Answer 4

try yourself
hint
use split, loop, append
i= i[1:] + i[0]

Answer 5

This is the code 👇👇

https://code.sololearn.com/c4nvuZGCB6uT/?ref=app

Answer 6

You can use .replace() string function.

Answer 7

Hey мʀυηαlι ʀαנρυт here is the code

https://code.sololearn.com/cBARd2rIE8aX/?ref=app

although I am a beginner with python but so my code might not be much efficient but still I tried to answer your question to best of my knowledge...

Answer 8

мʀυηαlι ʀαנρυт Kindly post your attempt.

Answer 9

𝐊𝐢𝐢𝐛𝐨 𝐆𝐡𝐚𝐲𝐚𝐥

Yeah its look like coad coach
And it would be same as pig latin challenge
There u have given an input , then u have to put the first the first letter of the the word to last and add 'ay' to it

https://code.sololearn.com/cKz15Eg6S788/?ref=app

Answer 10

Suppose,
s='string'
s[-1]='z'----replaces the last letter with 'z'

Answer 11

easiest way to do this is :

string=input ()

for i in str:
i.replace(‘i[-1]’,’ the object with which you want to replace ’)

leave the last parameter blank if you just want to delete that

Answer 12

Use syntax str

Answer 13

var letter = prompt("Enter letters...");

letter = letter.toLowerCase();

if (letter == "a" || letter =="i" || letter =="e"|| letter =="u");
console.log("Vowel");

else
console.log("consonant");

why not work else . please help me

Answer 14

For example {the first }5

Answer 15

Indexing

Answer 16

words = str(input("Input Sentence:")).split()
for word in words:
print(word[1:] + word[0])

Answer 17

Edit:
new_word = word[1:-1] + word[0]

example:

https://code.sololearn.com/c65bf5Ebfnk9/#py

Answer 18

t=input()
print (t[0:-1])

Answer 19

I wish i wouldn't be late.
It's like Pig Latin.
Be DRY.

https://code.sololearn.com/cjPG0fwFdZKs/?ref=app

Answer 20

Hi