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  1. Prepare
  2. Algorithms
  3. Strings
  4. Pangrams
  5. Discussions

Pangrams

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1950 Discussions

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  • ilovemylittlecr1
     + 0 comments

    Fastest Solution: 

    #include <iostream>
using namespace std;

bool is_pangram(const string& str) {
    int mask = 0; 
    for(const char c : str) {
        if(islower(c)) mask |= 1 << (c - 'a'); 
        else if(isupper(c)) mask |= 1 << (c - 'A'); 
    }

    return mask == (1 << 26) - 1; 
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    string str; 
    getline(cin, str); 

    cout << (is_pangram(str)? "pangram" : "not pangram") << endl; 
    return 0; 
}

  • dsivasuriya
     + 0 comments

    def pangrams(s):

    lo = s.lower()
count = []
sl = ["a","b","c","d","e","f","g","h","i","j","k","l",
"m","n","o","p","q","r","s","t","u","v","w","x","y","z"]

for i in lo:
    if i in sl:
        count.append(i)

n = set(count)
print(n)

if len(n) == 26:
    return('pangram')
else:
    return('not pangram')

  • yashdeora98294
     + 0 comments

    Here is problem solution in python, java, c++, c and javascript programming - https://programmingoneonone.com/hackerrank-pangrams-problem-solution.html

  • mrashidcit
     + 0 comments

    My solution in kotlin covering all the test cases

    fun pangrams(s: String): String {
    
    
    val target = "abcdefghijklmnopqrstuvwxyz"
    val input = s.lowercase()
    
    for(ch in target) {
        if(!input.contains(ch)) 
            return "not pangram"
    }
    
    return "pangram"

}

  • SVVBathini_w
     + 0 comments

    For Python3 Platform

    I wrote the code from scratch just to get more practice

    def pangrams(s):
    s = set(s.lower().replace(" ", ""))
    
    if(len(s) == 26):
        return "pangram"
    else:
        return "not pangram"

s = input()

result = pangrams(s)

print(result)