Check if a String is Isogram or Not

Problem

Given a string S of lowercase alphabets, check if it is isogram or not. An Isogram is a string in which no letter occurs more than once.

Examples:

Example 1:

Input:
S = machine
Output: 1
Explanation: machine is an isogram as no letter has appeared twice. Hence we print 1.

Example 2:

Input:
S = geeks
Output: 0
Explanation: geeks is not an isogram as 'e' appears twice. Hence we print 0.

Your task:

This is a function problem. You only need to complete the function isIsogram() that takes a string as a parameter and returns either true or false.

• Expected Time Complexity: $O(N)$
• Expected Auxiliary Space: $O$(Number of distinct characters)

Note: N = |S|

Constraints:

• $1 <= |s| <= 10^3$

Solution

Python

def isIsogram(self,s):
    length = len(s)
    mapHash = [0] * 26
    for i in range(length):
        mapHash[ord(s[i]) - ord('a')] += 1
        if (mapHash[ord(s[i]) - ord('a')] > 1):
            return False
    return True

Java

static boolean isIsogram(String data){
    int length = data.length();
    int[] mapHash = new int[26]; 
    for (int i = 0; i < length; i++) {
        char currentChar = data.charAt(i);
        if (Character.isLetter(currentChar)) {
            currentChar = Character.toLowerCase(currentChar); 
            mapHash[currentChar - 'a']++;
            if (mapHash[currentChar - 'a'] > 1) {
                return false;
            }
        }
    }
    return true;
}

C++

bool isIsogram(string s) {
    int length = s.length();
    vector<int> mapHash(26, 0); 
    for (int i = 0; i < length; i++) {
        char currentChar = s[i];
        if (isalpha(currentChar)) {
            currentChar = tolower(currentChar); 
            mapHash[currentChar - 'a']++;
            if (mapHash[currentChar - 'a'] > 1) {
                return false;
            }
        }
    }
    return true;
}

C

int isIsogram(const char* s) {
    int mapHash[26] = {0}; 
    int length = 0;
    const char* ptr = s;
    while (*ptr != '\0') {
        length++;
        ptr++;
    }
    for (int i = 0; i < length; i++) {
        char currentChar = s[i];
        if (isalpha(currentChar)) {
            currentChar = tolower(currentChar);
            mapHash[currentChar - 'a']++;
            if (mapHash[currentChar - 'a'] > 1) {
                return 0; 
            }
        }
    }
    return 1; 
}

• Time Complexity: $O(N)$
• Auxiliary Space: $O$(Number of distinct characters)