Skip to content
Related Articles

Related Articles

Improve Article
Check whether a given string is Heterogram or not
  • Last Updated : 08 Jun, 2021

Given a string S. The task is to check whether a the given string is Heterogram or not. A heterogram is a word, phrase, or sentence in which no letter of the alphabet occurs more than once. 

Examples: 

Input : S = "the big dwarf only jumps"
Output : Yes
Each alphabet in the string S is occurred
only once.

Input : S = "geeksforgeeks" 
Output : No
Since alphabet 'g', 'e', 'k', 's' occurred
more than once.

The idea is to make a hash array of size 26, initialised to 0. Traverse each alphabet of the given string and mark 1 in the corresponding hash array position if that alphabet is encounter first time, else return false.

Below is the implementation of this approach: 

C++




// C++ Program to check whether the given string is Heterogram or not.
#include<bits/stdc++.h>
using namespace std;
 
bool isHeterogram(char s[], int n)
{
    int hash[26] = { 0 };
     
    // traversing the string.
    for (int i = 0; i < n; i++)
    {
        // ignore the space
        if (s[i] != ' ')
        {
            // if already encountered
            if (hash[s[i] - 'a'] == 0)
                hash[s[i] - 'a'] = 1;
                 
            // else return false.
            else
                return false;
        }
    }
     
    return true;
}
 
// Driven Program
int main()
{
    char s[] = "the big dwarf only jumps";
    int n = strlen(s);
    (isHeterogram(s, n))?(cout << "YES"):(cout << "NO");
    return 0;
}

Java




// Java Program to check whether the
// given string is Heterogram or not.
class GFG {
         
    static boolean isHeterogram(String s, int n)
    {
        int hash[] = new int[26];
         
        // traversing the string.
        for (int i = 0; i < n; i++)
        {
            // ignore the space
            if (s.charAt(i) != ' ')
            {
                // if already encountered
                if (hash[s.charAt(i) - 'a'] == 0)
                    hash[s.charAt(i) - 'a'] = 1;
                     
                // else return false.
                else
                    return false;
            }
        }
         
        return true;
    }
     
// Driver code
public static void main (String[] args)
{
    String s = "the big dwarf only jumps";
    int n = s.length();
     
    if(isHeterogram(s, n))
        System.out.print("YES");
    else
        System.out.print("NO");
}
}
 
// This code is contributed by Anant Agarwal.

Python3




# Python3 code to check
# whether the given
# string is Heterogram
# or not.
 
def isHeterogram(s, n):
    hash = [0] * 26
     
    # traversing the
    # string.
    for i in range(n):
         
        # ignore the space
        if s[i] != ' ':
             
            # if already
            # encountered
            if hash[ord(s[i]) - ord('a')] == 0:
                hash[ord(s[i]) - ord('a')] = 1
             
            # else return false.
            else:
                return False
     
    return True
 
# Driven Code
s = "the big dwarf only jumps"
n = len(s)
 
print("YES" if isHeterogram(s, n) else "NO")
 
# This code is contributed by "Sharad_Bhardwaj".

C#




// C# Program to check whether the
// given string is Heterogram or not.
using System;
 
class GFG {
         
    static bool isHeterogram(string s, int n)
    {
        int []hash = new int[26];
         
        // traversing the string.
        for (int i = 0; i < n; i++)
        {
            // ignore the space
            if (s[i] != ' ')
            {
                // if already encountered
                if (hash[s[i] - 'a'] == 0)
                    hash[s[i] - 'a'] = 1;
                     
                // else return false.
                else
                    return false;
            }
        }
         
        return true;
    }
     
    // Driver code
    public static void Main ()
    {
        string s = "the big dwarf only jumps";
        int n = s.Length;
         
        if(isHeterogram(s, n))
            Console.WriteLine("YES");
        else
            Console.WriteLine("NO");
    }
}
 
// This code is contributed by Vt_m.

PHP




<?php
// PHP Program to check
// whether the given string
// is Heterogram or not.
 
function isHeterogram($s, $n)
{
    $hash = array();
    for($i = 0; $i < 26; $i++)
        $hash[$i] = 0;
     
    // traversing the string.
    for ($i = 0; $i < $n; $i++)
    {
        // ignore the space
        if ($s[$i] != ' ')
        {
            // if already encountered
            if ($hash[ord($s[$i]) -
                      ord('a')] == 0)
                $hash[ord($s[$i]) -
                      ord('a')] = 1;
                 
            // else return false.
            else
                return false;
        }
    }    
    return true;
}
 
// Driven Code
$s = "the big dwarf only jumps";
$n = strlen($s);
if (isHeterogram($s, $n))
    echo ("YES");
else
    echo ("NO");
     
// This code is contributed by
// Manish Shaw(manishshaw1)
?>

Javascript




<script>
 
// Javascript program to check whether
// the given string is Heterogram or not.
function isHeterogram(s, n)
{
    var hash = Array(26).fill(0);
     
    // Traversing the string.
    for(var i = 0; i < n; i++)
    {
         
        // Ignore the space
        if (s[i] != ' ')
        {
             
            // If already encountered
            if (hash[s[i].charCodeAt(0) -
                      'a'.charCodeAt(0)] == 0)
                hash[s[i].charCodeAt(0) -
                      'a'.charCodeAt(0)] = 1;
                 
            // Else return false.
            else
                return false;
        }
    }
    return true;
}
 
// Driver code
var s = "the big dwarf only jumps";
var n = s.length;
 
(isHeterogram(s, n)) ? (document.write("YES")) :
                       (document.write("NO"));
                        
// This code is contributed by rutvik_56
 
</script>

Output: 



YES

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer DSA Live Classes 




My Personal Notes arrow_drop_up
Recommended Articles
Page :