Given an array arr containing N strings, the task is to check if all strings are isogram or not. If they are, print Yes, otherwise No.
An Isogram is a word in which no letter occurs more than once.
Examples:
Input: arr[] = {“abcd”, “derg”, “erty”}
Output: YesInput: arr[] = {“agka”, “lkmn”}
Output: No
Approach: A string is an isogram if no letter in that string appears more than once. Now to solve this problem,
- Traverse the array arr, and for each string
- Create a frequency map of characters.
- Wherever any character has a frequency greater than 1, print No and return.
- Otherwise, after the whole array is traversed, print Yes.
Below is the implementation of the above approach:
C++
// C++ code for the above approach#include <bits/stdc++.h>using namespace std;// Function to check if a string// is an isogram or notbool isIsogram(string s){ vector<int> freq(26, 0); for (char c : s) { freq++; if (freq > 1) { return false; } } return true;}// Function to check if array arr contains// all isograms or notbool allIsograms(vector<string>& arr){ for (string x : arr) { if (!isIsogram(x)) { return false; } } return true;}// Driver Codeint main(){ vector<string> arr = { "abcd", "derg", "erty" }; if (allIsograms(arr)) { cout << "Yes"; return 0; } cout << "No";} |
Java
// Java program for the above approachimport java.io.*;class GFG { // Function to check if a string // is an isogram or not static boolean isIsogram(String s) { int freq[] = new int[26]; char S[] = s.toCharArray(); for (char c : S) { freq++; if (freq > 1) { return false; } } return true; } // Function to check if array arr contains // all isograms or not static boolean allIsograms(String arr[]) { for (String x : arr) { if (isIsogram(x) == false) { return false; } } return true; } // Driver Code public static void main(String[] args) { String arr[] = { "abcd", "derg", "erty" }; if (allIsograms(arr) == true) { System.out.println("Yes"); } else { System.out.println("No"); } }} // This code is contributed by Potta Lokesh |
Python3
# Python code for the above approach# Function to check if a string# is an isogram or notdef isIsogram (s): 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 check if array arr contains# all isograms or notdef allIsograms (arr): for x in arr: if (not isIsogram(x)): return False return True# Driver Codearr = ["abcd", "derg", "erty"]if (allIsograms(arr)): print("Yes")else: print("No")# This code is contributed by Saurabh Jaiswal |
C#
// C# program for the above approachusing System;public class GFG { // Function to check if a string // is an isogram or not static bool isIsogram(String s) { int []freq = new int[26]; char []S = s.ToCharArray(); foreach (char c in S) { freq++; if (freq > 1) { return false; } } return true; } // Function to check if array arr contains // all isograms or not static bool allIsograms(String []arr) { foreach (String x in arr) { if (isIsogram(x) == false) { return false; } } return true; } // Driver Code public static void Main(String[] args) { String []arr = { "abcd", "derg", "erty" }; if (allIsograms(arr) == true) { Console.WriteLine("Yes"); } else { Console.WriteLine("No"); } }}// This code is contributed by 29AjayKumar |
Javascript
<script> // JavaScript code for the above approach // Function to check if a string // is an isogram or not const isIsogram = (s) => { let freq = new Array(26).fill(0); for (let c in s) { freq[s.charCodeAt(c) - 'a'.charCodeAt(0)]++; if (freq[s.charCodeAt(c) - 'a'.charCodeAt(0)] > 1) { return false; } } return true; } // Function to check if array arr contains // all isograms or not const allIsograms = (arr) => { for (let x in arr) { if (!isIsogram(arr[x])) { return false; } } return true; } // Driver Code let arr = ["abcd", "derg", "erty"]; if (allIsograms(arr)) document.write("Yes"); else document.write("No"); // This code is contributed by rakeshsahni</script> |
Output
Yes
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)
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