Check if a string is Isogram or not using Hash-Map

Check if a string is Isogram or not using sorting:

To solve the problem follow the below idea:

In this, the count of characters of the string is stored in the hashmap, and wherever it is found to be greater than 1 for any char, return false else return true at the end

Follow the given steps to solve the problem:

Declare a hashmap to store the count of the characters of the string
Traverse the string
- Increase the count of the current character in the hashmap
- If the count of the current character is greater than one then return false
Return true

Below is the implementation of the above approach:

C++

// CPP code to check string is isogram or not
#include <bits/stdc++.h>
 
using namespace std;
 
// function to check isogram
bool check_isogram(string str)
{
 
    int length = str.length();
    int mapHash[26] = { 0 };
 
    // loop to store count of chars and check if it is
    // greater than 1
    for (int i = 0; i < length; i++) {
        mapHash[str[i] - 'a']++;
 
        // if count > 1, return false
        if (mapHash[str[i] - 'a'] > 1) {
            return false;
        }
    }
 
    return true;
}
 
// Driver code
int main()
{
    string str = "geeks";
    string str2 = "computer";
 
    // checking str as isogram
    if (check_isogram(str)) {
        cout << "True" << endl;
    }
    else {
        cout << "False" << endl;
    }
 
    // checking str2 as isogram
    if (check_isogram(str2)) {
        cout << "True" << endl;
    }
    else {
        cout << "False" << endl;
    }
 
    return 0;
}

Java

// Java code to check string is isogram or not
import java.io.*;
 
class GFG {
 
    // function to check isogram
    static boolean check_isogram(String str)
    {
 
        int length = str.length();
        int mapHash[] = new int[26];
 
        // loop to store count of chars and
        // check if it is greater than 1
        for (int i = 0; i < length; i++) {
            mapHash[str.charAt(i) - 'a']++;
 
            // if count > 1, return false
            if (mapHash[str.charAt(i) - 'a'] > 1) {
                return false;
            }
        }
 
        return true;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String str = "geeks";
        String str2 = "computer";
 
        // checking str as isogram
        if (check_isogram(str))
            System.out.println("True");
        else
            System.out.println("False");
 
        // checking str2 as isogram
        if (check_isogram(str2))
            System.out.println("True");
        else
            System.out.println("False");
    }
}
 
// This code contributed by Rajput-Ji

Python3

# Python3 code to check string is isogram or not
 
# function to check isogram
 
 
def check_isogram(string):
 
    length = len(string)
    mapHash = [0] * 26
 
    # loop to store count of chars
    # and check if it is greater than 1
    for i in range(length):
 
        mapHash[ord(string[i]) - ord('a')] += 1
 
        # if count > 1, return false
        if (mapHash[ord(string[i]) - ord('a')] > 1):
 
            return False
 
    return True
 
 
# Driver code
if __name__ == "__main__":
 
    string = "geeks"
    string2 = "computer"
 
    # checking str as isogram
    if (check_isogram(string)):
        print("True")
    else:
        print("False")
 
    # checking str2 as isogram
    if (check_isogram(string2)):
        print("True")
 
    else:
        print("False")
 
# This code is contributed by AnkitRai01

C#

// C# code to check string is isogram or not
using System;
public class GFG {
 
    // function to check isogram
    static bool check_isogram(String str)
    {
 
        int length = str.Length;
        int[] mapHash = new int[26];
 
        // loop to store count of chars and
        // check if it is greater than 1
        for (int i = 0; i < length; i++) {
            mapHash[str[i] - 'a']++;
 
            // if count > 1, return false
            if (mapHash[str[i] - 'a'] > 1) {
                return false;
            }
        }
 
        return true;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        String str = "geeks";
        String str2 = "computer";
 
        // checking str as isogram
        if (check_isogram(str))
            Console.WriteLine("True");
        else
            Console.WriteLine("False");
 
        // checking str2 as isogram
        if (check_isogram(str2))
            Console.WriteLine("True");
        else
            Console.WriteLine("False");
    }
}
 
// This code has been contributed by 29AjayKumar

Javascript

<script>
    // Javascript code to check string is isogram or not 
     
    // function to check isogram 
    function check_isogram(str) 
    { 
 
        let length = str.length; 
        let mapHash = new Array(26);
        mapHash.fill(0);
 
        // loop to store count of chars and 
        // check if it is greater than 1 
        for (let i = 0; i < length; i++) 
        { 
            mapHash[str[i].charCodeAt() - 'a'.charCodeAt()]++; 
 
            // if count > 1, return false 
            if (mapHash[str[i].charCodeAt() - 'a'.charCodeAt()] > 1) 
            { 
                return false; 
            } 
        } 
 
        return true; 
    } 
     
    let str = "geeks"; 
    let str2 = "computer"; 
    
    // checking str as isogram 
    if (check_isogram(str)) 
        document.write("True" + "</br>");
    else
        document.write("False" + "</br>");
    
    // checking str2 as isogram 
    if (check_isogram(str2)) 
          document.write("True" + "</br>");
    else
        document.write("False" + "</br>");
     
    // This code is contributed by divyeshrabadiya07.
</script>

Output

False
True

Time Complexity: O(N)
Auxiliary Space: O(1), as an array of fixed size 26 is used.

Thanks Sahil Bansal for suggesting the above method.
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Check if a string is Isogram or not

Given a word or phrase, check if it is an isogram or not. An Isogram is a word in which no letter occurs more than once

Examples:

Input: Machine
Output: True
Explanation: “Machine” does not have any character repeating, it is an Isogram

Input : Geek
Output : False
Explanation: “Geek” has ‘e’ as repeating character, it is not an Isogram

Tags:

#DSA #Hash #Strings #Hash #Strings