I have tried to solve this code coach problem of Deja vu in python. Correct me with my code.

Question

https://www.sololearn.com/coach/54?ref=app

You aren't paying attention and you accidentally type a bunch of random letters on your keyboard. You want to know if you ever typed the same letter twice, or if they are all unique letters.

Task:
If you are given a string of random letters, your task is to evaluate whether any letter is repeated in the string or if you only hit unique keys while you typing.

Input Format:
A string of random letter characters (no numbers or other buttons were pressed).

# code goes from here
strng = input()
count = {}
for i in strng:
if i in count:
count[i] += 1
else:
count[i] = 1
for s in count:
if count[s] == 1:
print ("Unique")
elif count[s] > 1:
print ("Deja Vu")
break

Answer 1

s = input()
c = 0
for l in s:
if s.count(l) > 1:
c += s.count(l)

c -= int(c/2)
if c > 1:
print('Deja Vu')
else:
print('Unique')

this will work

Answer 2

#DIY test
in_str = list(str(input()))
str_len = len(in_str)

set_letters = set(in_str)
set_len = len(set_letters)

if str_len == set_len:
print("Unique")
else:
print("Deja Vu")

Answer 3

TASK:
If you are given a string of random letters, your task is to evaluate whether any letter is repeated in the string or if you only hit unique keys while you typing.

Input Format: /
A string of random letter characters (no numbers or other buttons were pressed).
==MY ANSWER IN JAVA PROGRAMMING===
import java.util.*;
public class Program
{
public static void main(String[] args) {
Scanner ip=new Scanner(System.in);
String name=ip.nextLine();
char[] c=name.toCharArray();
int count=0;
for(int i=0;i<c.length;i++)
{
for(int j=i+1;j<c.length;j++)
{
if(c[i]==c[j])
{
count++;
}
}
}
if(count==0)
System.out.print("Unique");
else
System.out.print("Deja Vu");
}
}

Answer 4

its the line: "for s in count:"

count, i beleive, is a dictionary. It has key and value pairs that need to be seperated.

only_one = True
for key, value in count.items():
if value > 1:
print("DejaVu")
only_one = False
break
if only_one:
print("Unique")

Answer 5

Actually the code is working in this way but the problem is it's giving only the output as 'unique', even where it suppose to give the output as 'Deja vu' that's the problem here with this code.

Answer 6

So it working or its not working?

Answer 7

I will post the problem here

Answer 8

You aren't paying attention and you accidentally type a bunch of random letters on your keyboard. You want to know if you ever typed the same letter twice, or if they are all unique letters.

Task:
If you are given a string of random letters, your task is to evaluate whether any letter is repeated in the string or if you only hit unique keys while you typing.

Input Format:
A string of random letter characters (no numbers or other buttons were pressed).

Output Format:
A string that says 'Deja Vu' if any letter is repeated in the input string, or a string that says 'Unique' if there are no repeats.

Sample Input:
aaaaaaaghhhhjkll

Sample Output:
Deja Vu

Answer 9

That's not needed as the code isn't difficult. I saw and explained your problem in the first post. You created a dictionary, then didn't access it the right way

Answer 10

Try running my code

Answer 11

How to do it then

Answer 12

Dont need to yours wont work, just delete your code from the line I mentioned and put the code I supplied in.

Answer 13

K i will try

Answer 14

Same output, no change

Answer 15

post the code. And not just text, put it in the playground and link it here

Answer 16

Thank you it worked. But, i can't understand the logic can you explain that

Answer 17

only_one = True # keeps track of if there are duplicates

# when you have a dictionary and call <dict>.items() it gives a list of tuples. In each tuple, the first value is the key and the second is the value
for key, value in count.items():
# this check will print a message if there ARE duplicates
if value > 1:
print("DejaVu")
# if duplicates, the duplicate tracker is set to False and then breaks out of loop
only_one = False
break
# if duplicate tracker was not set to False, print the positive message saying it's a unique string
if only_one:
print("Unique")

Answer 18

Only_one is a variable, right?

Answer 19

And it is dealing with Boolean operations

Answer 20

Yes, it's a variable using boolean True/False