Yep. But that's not the case, as the problem description clearly states.

Well then all you would need to do is .replace() any of your special chars with nothing. See my edited solution, I removed all spaces.

from collections import Counter
print("pangram" if len(Counter(input().lower().replace(' ',''))) == 26 else "not pangram")

The perfect solution is here, that will work also in case of using other character in string(hyphens,commas,etc.)

x="qwertyuiopasdfghjklzxcvbnm"
x=set(x)
if(set(input().lower()).intersection(x)==x):
    print("pangram")
else:
    print("not pangram")

for i in range(26): if (chr(65+i) not in s) and (chr(97+i) not in s) : print "not pangram" exit() print "pangram"

create array storing characters a-z & A-Z compare with string char if match store 1 in another array check finally ur sum is 26 or not

string s; int sum =0; vector a(26,0); vector b(26); vector c(26);

for(int i=0;i<26;i++){ b[i]='a'+i;}

for(int i=0;i<26;i++){ c[i]='A'+i;}

        getline(cin,s);
for(int i=0;i<s.length();i++){
    for(int j=0;j<26;j++){
        if(s[i]==b[j] || s[i]==c[j]){
            a[j]=1;
        }
    }
}


 for(int i : a){
    sum=sum+i;
}
if(sum==26) cout << "pangram";
else cout << "not pangram";

A simpler way to complete this problem would be to convert all the upper case letters to lower case letters. Here is my solution.

char s[1000];   fgets(s,1000,stdin);
for(int i=0; s[i]!='\0';i++)
    s[i]=tolower(s[i]); //converted to lower case
int counter=0;
 for(int a=97;a<=122;a++)
     { for(int i=0; s[i]!='\0';i++)
         if((int)s[i]==a) {counter++; break;} //compares the                                                       //ASCII value  

     }
if(counter==26) cout<<"pangram";
else cout<<"not pangram";

you could also do it with only one loop here it is is java

public static void main(String[] args){
    Scanner in=new Scanner(System.in);
    String str=in.nextLine();
    str=str.toLowerCase();
    boolean isPanagram=true;
    if(str.length()<26){
        isPanagram=false;
    }else {
        boolean[] allLetters = new boolean[26];
        for (int i = 0; i < allLetters.length; i++) {
            allLetters[i] = false;
        }
        for (int i = 0; i < str.length(); i++) {
            if((int)str.charAt(i)!=32) {
                allLetters[(int) str.charAt(i) - 97] = true;
            }
        }
        for (int i = 0; i < allLetters.length; i++) {
            if (!allLetters[i]) {
                isPanagram = false;
            }
        }
    }
    if(isPanagram){
        System.out.println("pangram");
    }else{
        System.out.println("not pangram");
    }
}

Much simpler version.

for(int i = 0; i < s.length(); i++) { 
    s.at(i) = tolower(s.at(i)); 
} 
sort(s.begin(), s.end()); 
s.erase(unique(s.begin(), s.end()), s.end()); 
s.erase(s.begin());

if(s.length() == 26) 
    cout << "pangram" << endl; 
else 
    cout << "not pangram" << endl;

fgets(s , sizeof(s) , stdin);
int c =65,check=0;
while(c <= 90 && c>=65)
    {
    for(int i=0 ; i<strlen(s) ; i++)
        {
        if((int)s[i]==c || (int)s[i]==(c+32))
        {

            check++;

            break;
        }

    }
    c++;
}

if(check == 26)
    cout<<"pangram";
else
    cout<<"not pangram";

include

using namespace std; int main() { string s;
getline(cin,s,'\n'); int sl,i,ele,a[27]={0},sum=0; sl=s.length(); for(i=0;i91) a[int(s[i])-97]=1; } for(i=0;i<27;i++) sum+=a[i]; if(sum==26) cout<<"pangram"; else cout<<"not pangram"; return 0; }

we can directly compare the alphabets with their ASCII code values

int main() { string s; getline(cin,s);

int a_=96,A_=64;
int totalCount=0;
int alpCount=0;
int size=s.size();

for(int j=0;j<26;j++) { alpCount=0; a_++; A_++;

  for(int i=0;i<size;i++)
   {
     if(s[i]==a_|| s[i]==A_)
         {
           alpCount++;

         }
    }
  if(alpCount>=1)
      {
      totalCount++;


      }

}
if(totalCount==26)
    cout<<"pangram";
else
    cout<<"not pangram";
return 0;

}

int main() { string s;

// New ---
getline(cin, s);

bool isAPangram = true;
string alphabet = "abcdefghijklmnopqrstuvwxyz";

// New ---
 isAPangram = std::all_of(alphabet.begin(), alphabet.end(), 
                         [s](char i){                             
                             char upperCase = toupper(i);
                             bool hasLetter = count(s.begin(), s.end(),i)>0;
                             hasLetter |= count(s.begin(), s.end(),upperCase)>0;
                             //cout <<i<<hasLetter;
                             return hasLetter;});

if (isAPangram)
    {
    cout<<"pangram";        
}
else{
    cout<<"not pangram";
}
return 0;

} ​

(C++) This is technically this is O(n*m) where n is the number of words and m is the size of the word. This would mean n*m is the number of charcters in the line, so this is O(n). You also don't need to do weird stuff to toss out the whitespace like you would if you saved a string with getline.

    // a is index 0, b is index 1, ... etc
    // we use this to check if it was used in the line
    bool alphabet[26] = {false};
    
    string word;
    while(cin >> word) { // let cin ignore whitespace
        for(int i = 0; i < word.length(); i++){
            // this will give the an index between 0 and 25
            int position = tolower(word[i]) - 'a';
            alphabet[position] = true;
        }
    }
    
    for(int i = 0; i < 26; i++) {
        // if there was a letter not included
        if(!alphabet[i]) {
            cout << "not pangram" << endl;
            return 0;
        }
    }
	
	cout << "pangram" << endl;    
    return 0;

Here is a solution with less loops

set<int> set{};
string s{};

while (cin >> s) {
    for (auto c : s) {
        set.insert(tolower(c));
    }
}

if (set.size() == 26) {
    cout << "pangram";
} else {
    cout << "not pangram";
}

the problem statement says it is only alphabets and space.

you can use sets i suppose. The way I understand what you are trying to do is to store every unique instance of the letter in a set right?
Personally, though sets might seem easier for this problem, i would suggest using sets when you have large sets of values and you want that strict time limit of log N. For this problem though, i wanted a simple algorithm because that makes error finding that much easier. Happy coding :)

scanf("%s",str); This statement will not be able to scan the entire string. if(!(s[i]==1 && c[i]==1)) you need to use '||'

I'm getting something similar my code works except for big strings. I'm using getline. do i have to use scanf ?

then how are we supposed to counter that? we cannot use scanf() because that will stop scanning at a whitespace right?

why is it empty ? where am i doing it wrong ?

You are right. But now it's too hard to read and trace.

Thank you very much for your help. I have understood that clearly

My apologies.. true. I think I misplaced my comment.

Probably the remove, the string ' ' doesn't exist, you'd need to do s1.remove(' '.toCharArray()[0]) because it's a set of Character so you need to remove the char ' ' not the string ' '