+ 3
Password validation
I tried to solve this.. I got all the 9 conditions correct but the last 4 were wrong. Somebody help me!! p=input("") #string input count=0 special=['!', '@', '#', '
#x27;, '%', '&', '*'] a=0 d=0 sl=0 for i in range(len(p)): if(p[i].isalpha()): a=a+1 elif(p[i].isdigit()): d=d+1 elif (p[i] in special): sl=sl+1; count=a+d+sl if(count>=11 and sl>=2 and d>=2 and a>=7): print("Strong") else: print("Weak") This is my code in python.☝️☝️ And the below one is in C 👇👇. include <stdlib.h> #include <stdio.h> #include <string.h> main() { char l[1000]; int i,v,a=0,d=0,s=0,count=0; //printf("\n enter passeord:"); gets (l); for(i=0;i<strlen(l);i++) { if(l[i]=='!'|| l[i]=='@'|| l[i]=='#'||l[i]=='#x27;|| l[i]=='%'|| l[i]=='&' ||l[i]=='*') s++; else if((l[i]>='A'&&l[i]<='Z')|| (l[i]>='a'&&l[i]<='z')) a++; else if(isdigit(l[i])) d++; } if(d>=2 && s>=2 && a>=7) printf("Strong"); else printf("Weak"); return 0; }10 Answers
+ 2
Just consider this code :
https://code.sololearn.com/cV57TqZAIuhp/?ref=app
+ 5
This will help you if you like this to analyze
https://code.sololearn.com/cOjvkbJvYPg2/?ref=app
+ 3
# Python code:
a, b, c = input(), 0, 0
for x in a:
if x.isnumeric(): b += 1
if x in "!@#$%&*": c += 1
if b > 1 and c > 1 and len(a) > 6:
print("Strong")
else: print("Weak")
# Hope this helps
+ 2
What is need of condition count>=11 in if block..? May i know about the logic?
edit: nivedha srinivasan
you are counting alphabets into variable 'a' but you need length of input string which also includes special chars and digits.. so take
a = len(p) next only need special chars and digits separately..
in c : a = strlen(l);
hope it helps you..
+ 1
In first, I understand more python than c so my comment will be for the python code.
Firstly you don't need to check if in password given are characters cause this is not a criteria of the challenge.
Secondly there are three cases you should check :
1-The length of password.
2-At least two numbers.
3-At least two special characters.
So in python you may create three different functions to check those points or you just can run the check in a single for loop. For 2 and 3 you may need a variable to store the count of them. For the 1 just use len() function.
For example:
#this is without functions
pass = str(input())
nums = list(range(10))
special_c = [#special characters needed]
n = 0
s = 0
for char in pass:
if char in nums:
n++
elif char in special_c:
s++
#the final check
if len(pass)>=7 and n>=2 and s>=2:
print('Strong')
else:
print('Weak')
You got it ?😀
+ 1
The same for the c code
+ 1
Thank you ,guys.completed 🥳🥳
+ 1
No problem 👍. Ask anytime
+ 1
// C code:
#include <stdio.h>
int main() {
char a[25];
char b[7] = {'!','@','#','#x27;,'&','*','%'};
fgets(a, 25, stdin);
int x=0, y=0, i;
for (i=0;a[i]!='\0';i++) {
for (int k=0;k<10;k++) {
if (k == a[i] - 48) x++;
if (a[i] == b[k * (k < 7)]) y++;
}
}
if (x > 1 && y > 1 && i > 6) {
puts("Strong");
}
else puts("Weak");
}
// Hope this helps
0
password = list(input())
str = " ".join(password) #a string with space b/w the items
import re
symb = re.findall("[!@#\$%&\*]+", str)
num = re.findall(r"[0-9]+", str)
filt = filter((lambda x: x in symb,str), (password))
print ("Strong" if len(symb) >= 2 and len(num) >= 2 and len(password) >=7 else "Weak")