Count of Isogram strings in given Array of Strings

Given an array arr[] containing N strings, the task is to find the count of strings which are isograms. A string is an isogram if no letter in that string appears more than once.

Examples:

Input: arr[] = {"abcd", "derg", "erty"}
Output: 3
Explanation: All given strings are isograms. In all the strings no character 
is present more than once. Hence count is 3

Input: arr[] = {"agka", "lkmn"}
Output: 1
Explanation: Only string "lkmn" is isogram. In the string "agka" 
the character 'a' is present twice. Hence count is 1.

Approach: Greedy approach can be used for solving this problem. Traverse each string in the given string array and check if that is isogram or not. To do that follow the steps mentioned below:

• Traverse the array of string and follow the below steps for each string:
• Create a frequency map of characters.
• Wherever any character has a frequency greater than 1, skip the current string and move to the next one.
• If no character has frequency more than 1, increment the count of answer by 1.
• Return the count stored in answer when all the strings are traversed.

Below is the implementation of the above approach:

#include <bits/stdc++.h>
using namespace std;

// Function to check
// if a string is an isogram
bool isIsogram(string s)
{
    // Loop to check
    // if string is isogram or not
    vector<int> freq(26, 0);
    for (char c : s) {
        freq[c - 'a']++;
        if (freq[c - 'a'] > 1) {
            return false;
        }
    }
    return true;
}

// Function to count the number of isograms
int countIsograms(vector<string>& arr)
{
    int ans = 0;

    // Loop to iterate the string array
    for (string x : arr) {
        if (isIsogram(x)) {
            ans++;
        }
    }
    return ans;
}

// Driver Code
int main()
{
    vector<string> arr = { "abcd", "derg", "erty" };

    // Count of isograms in string array arr[]
    cout << countIsograms(arr) << endl;
    return 0;
}

// Java program for the above approach
import java.util.ArrayList;

class GFG {

    // Function to check
    // if a String is an isogram
    static boolean isIsogram(String s) {

        // Loop to check
        // if String is isogram or not
        int[] freq = new int[26];
        for (int i = 0; i < 26; i++) {
            freq[i] = 0;
        }

        for (char c : s.toCharArray()) {
            freq[c - 'a']++;
            if (freq[c - 'a'] > 1) {
                return false;
            }
        }
        return true;
    }

    // Function to count the number of isograms
    static int countIsograms(ArrayList<String> arr) {
        int ans = 0;

        // Loop to iterate the String array
        for (String x : arr) {
            if (isIsogram(x)) {
                ans++;
            }
        }
        return ans;
    }

    // Driver Code
    public static void main(String args[]) {
        ArrayList<String> arr = new ArrayList<String>();

        arr.add("abcd");
        arr.add("derg");
        arr.add("erty");

        // Count of isograms in String array arr[]
        System.out.println(countIsograms(arr));
    }
}

// This code is contributed by gfgking

# Function to check
# if a string is an isogram
def isIsogram(s):

    # Loop to check
    # if string is isogram or not
    freq = [0]*(26)
    for c in s:
        freq[ord(c) - ord('a')] += 1
        if (freq[ord(c) - ord('a')] > 1):
            return False

    return True

# Function to count the number of isograms
def countIsograms(arr):
    ans = 0

    # Loop to iterate the string array
    for x in arr:
        if (isIsogram(x)):
            ans += 1

    return ans

# Driver Code
if __name__ == "__main__":

    arr = ["abcd", "derg", "erty"]

    # Count of isograms in string array arr[]
    print(countIsograms(arr))

    # This code is contributed by ukasp.

// C# program for the above approach
using System;
using System.Collections;

class GFG
{
    
// Function to check
// if a string is an isogram
static bool isIsogram(string s)
{
  
    // Loop to check
    // if string is isogram or not
    int []freq = new int[26];
    for(int i = 0; i < 26; i++) {
        freq[i] = 0;
    }
    
    foreach (char c in s) {
        freq[c - 'a']++;
        if (freq[c - 'a'] > 1) {
            return false;
        }
    }
    return true;
}

// Function to count the number of isograms
static int countIsograms(ArrayList arr)
{
    int ans = 0;

    // Loop to iterate the string array
    foreach (string x in arr) {
        if (isIsogram(x)) {
            ans++;
        }
    }
    return ans;
}

// Driver Code
public static void Main()
{
    ArrayList arr = new ArrayList();
    
    arr.Add("abcd");
    arr.Add("derg");
    arr.Add("erty");

    // Count of isograms in string array arr[]
    Console.WriteLine(countIsograms(arr));
}
}

// This code is contributed by Samim Hossain Mondal.

<script>

    // Function to check
    // if a string is an isogram
    const isIsogram = (s) => {
    
        // Loop to check
        // if string is isogram or not
        let freq = new Array(26).fill(0);
        for (let c in s) {
            freq[s.charCodeAt(c) - "0".charCodeAt(0)]++;
            if (freq[c] > 1) {
                return false;
            }
        }
        return true;
    }

    // Function to count the number of isograms
    const countIsograms = (arr) => {
        let ans = 0;

        // Loop to iterate the string array
        for (let x in arr) {
            if (isIsogram(x)) {
                ans++;
            }
        }
        return ans;
    }

    // Driver Code
    let arr = ["abcd", "derg", "erty"];

    // Count of isograms in string array arr[]
    document.write(countIsograms(arr));

    // This code is contributed by rakeshsahni

</script>

3

Time Complexity: O(N*M), where N is the size of the array and M is the size of the longest string
Auxiliary Space: O(1)