HackerRank Pangrams Problem Solution

In this post, we will solve HackerRank Pangrams Problem Solution.

A pangram is a string that contains every letter of the alphabet. Given a sentence determine whether it is a pangram in the English alphabet. Ignore case. Return either pangram or not pangram as appropriate.
Example
s = ‘The quick brown fox jumps over the lazy dog’
The string contains all letters in the English alphabet, so return pangram.

Function Description

Complete the function pangrams in the editor below. It should return the string pangram if the input string is a pangram. Otherwise, it should return not pangram.

pangrams has the following parameter(s):

string s: a string to test

Returns

string: either pangram or not pangram

Input Format

A single line with string s.

Sample Input

Sample Input 0

We promptly judged antique ivory buckles for the next prize

Sample Output 0

pangram

Sample Explanation 0

All of the letters of the alphabet are present in the string.

Sample Input 1

We promptly judged antique ivory buckles for the prize

Sample Output 1

not pangram

Sample Explanation 0

The string lacks an x.

Pangrams C Solution

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>

#define S_MAX 1000
#define N_ALPHABET 26

int main(void) {
	int i, j, k, nbr_letters;
    int c;
	char s[S_MAX+1];
    
    char alpha [][N_ALPHABET+1] = {
        "abcdefghijklmnopqrstuvwxyz",
        "ABCDEFGHIJKLMNOPQRSTUVWXYZ",
        ""
    };
	int n[N_ALPHABET] = {0};
	
    nbr_letters = 0;
	fgets(s, S_MAX+1, stdin);
	for(i=0; s[i] != '\0'; i++) {
        for(j=0; alpha[j][0] != '\0'; j++) {
            for (k=0; alpha[j][k] != '\0'; k++) {
                if (s[i] == alpha[j][k] && n[k] == 0) {
                    nbr_letters++;
                    n[k]++;
                }
            }
        }
    }
	if (nbr_letters == N_ALPHABET)
	    puts("pangram");
	else
	    puts("not pangram");
	    
	return EXIT_SUCCESS;
}

Pangrams C++ Solution

#include <iostream>
#include <cctype>
#include <bitset>
using namespace std;


int main() {
    char c;
    static bitset<26> b{false};
    while (cin >> c) {
        c = tolower(c);
        if (isalpha(c)) {
            b[c] = true;
        }
    }
    
    if (b.count() != 26) {
        cout << "not ";
    }
    cout << "pangram" << endl;
    return 0;
}

Pangrams C Sharp Solution

using System;
using System.Collections.Generic;
using System.IO;
using System.Collections;
class Solution {
    static void Main(String[] args) {
        string sentence = Console.ReadLine();
        sentence = sentence.ToLower().Replace(" ", "");
        
        Hashtable hash = new Hashtable();
        
        for(int i = 0; i < sentence.Length; i++)
        {
            try{hash.Add(sentence[i], 1);}
            catch(Exception e){}
        }
        
        if(hash.Count != 26)
            {
            Console.WriteLine("not pangram");
        }
        else
            {
            Console.WriteLine("pangram");
}
    }
}

Pangrams Java Solution

import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;

class Result {

    /*
     * Complete the 'pangrams' function below.
     *
     * The function is expected to return a STRING.
     * The function accepts STRING s as parameter.
     */

    public static String pangrams(String s) {
    // Write your code here
    if(s.length() < 26){
        return "not pangram";
    }else{
        List<Character> result = new ArrayList<Character>();
        for (char ch: s.toLowerCase().toCharArray()) {
            if(!result.contains(ch) && ch != ' '){
                result.add(ch);
            }
        }
        if(result.size() < 26){
            return "not pangram";
        }else{
            return "pangram";
        }
    }

    }

}

public class Solution {
    public static void main(String[] args) throws IOException {
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));

        String s = bufferedReader.readLine();

        String result = Result.pangrams(s);

        bufferedWriter.write(result);
        bufferedWriter.newLine();

        bufferedReader.close();
        bufferedWriter.close();
    }
}

Pangrams JavaScript Solution

function processData(input) {
    //Enter your code here
    D = {} 
    chars = input.toLowerCase().split(''); 
    chars = chars.filter(function(c) {return 'a' <= c && c <= 'z'; });

    for (i in chars) { 
        D[chars[i]] = 1; 
    }
    if (Object.keys(D).length == 26) { 
        console.log("pangram");
    }
    else { 
        console.log("not pangram"); 
    }
}

process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
    _input += input;
});

process.stdin.on("end", function () {
   processData(_input);
});

Pangrams Python Solution

import string

s = input()
s = s.lower()
isPangram = True

for letter in string.ascii_lowercase:
    isPangram = isPangram and letter in s

if isPangram:
    print('pangram')
else:
    print('not pangram')