Given an array arr[] containing N strings and an integer K, the task is to find the number of strings which are isograms and at least of length K.
Examples:
Input: arr[] = {“abcd”, “der”, “erty”}, K = 4
Output: 2
Explanation: All given strings are isograms, but only “abcd” and “erty” are of length at least K. Hence count is 2Input: arr[] = {“ag”, “bka”, “lkmn”, “asdfg”}, K = 2
Output: 4
Explanation: All the strings are isograms and each strings is of length >=K. Hence count is 4.
Approach: This problem can be solved by storing frequencies of all the characters or by using a Set data structure. A string is an isogram if no letter in that string appears more than once. Follow the steps below to solve the given problem.
- Traverse the array of strings arr[], and for each string
- Create a frequency map of characters.
- Wherever any character has a frequency greater than 1, or if the length of the string is less than K, skip the current string.
- Otherwise, if no character has a frequency more than 1, and if the length of the string is less than K, increment the count of the answer.
- Print the count stored in the answer when all the strings are traversed.
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){ // To store the frequencies 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 notint allIsograms(vector<string>& arr, int K){ int ans = 0; for (string x : arr) { if (isIsogram(x) && x.length() >= K) { ans++; } } return ans;}// Driver Codeint main(){ vector<string> arr = { "abcd", "der", "erty" }; int K = 4; // Function call and printing the answer cout << allIsograms(arr, K);} |
Java
// Java code for the above approachimport java.io.*;class GFG { // Function to check if a string // is an isogram or not static boolean isIsogram(String s) { // To store the frequencies int[] freq = new int[26]; for (int i = 0; i < s.length(); i++) { char c = s.charAt(i); freq++; if (freq > 1) { return false; } } return true; } // Function to check if array arr contains // all isograms or not static int allIsograms(String[] arr, int K) { int ans = 0; for (String x : arr) { if (isIsogram(x) && x.length() >= K) { ans++; } } return ans; } // Driver Code public static void main(String[] args) { String arr[] = { "abcd", "der", "erty" }; int K = 4; // Function call and printing the answer System.out.println(allIsograms(arr, K)); }}// 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): # To store the frequencies freq = [0] * 26 for c in s: freq[ord(c) - ord("a")] += 1 if (freq[s.index(c)] > 1): return False return True# Function to check if array arr contains# all isograms or notdef allIsograms (arr, K): ans = 0 for x in arr: if isIsogram(x) and len(x) >= K: ans += 1 return ans# Driver Codearr = ["abcd", "der", "erty"]K = 4# Function call and printing the answerprint(allIsograms(arr, K))# This code is contributed by Saurabh jaiswal |
C#
// C# code for the above approachusing System;class GFG{// Function to check if a string// is an isogram or notstatic bool isIsogram(string s){ // To store the frequencies int[] freq = new int[26]; for(int i = 0; i < s.Length; i++) { char c = s[i]; freq++; if (freq > 1) { return false; } } return true;}// Function to check if array arr contains// all isograms or notstatic int allIsograms(string[] arr, int K){ int ans = 0; foreach(string x in arr) { if (isIsogram(x) && x.Length >= K) { ans++; } } return ans;}// Driver Codepublic static void Main(string[] args){ string[] arr = { "abcd", "der", "erty" }; int K = 4; // Function call and printing the answer Console.WriteLine(allIsograms(arr, K));}}// This code is contributed by ukasp |
Javascript
<script> // JavaScript code for the above approach // Function to check if a string // is an isogram or not const isIsogram = (s) => { // To store the frequencies let freq = new Array(26).fill(0); for (let c in s) { freq[s.charCodeAt(c) - "a".charCodeAt(0)]++; if (freq > 1) { return false; } } return true; } // Function to check if array arr contains // all isograms or not const allIsograms = (arr, K) => { let ans = 0; for (let x in arr) { if (isIsogram(x) && arr[x].length >= K) { ans++; } } return ans; } // Driver Code let arr = ["abcd", "der", "erty"]; let K = 4; // Function call and printing the answer document.write(allIsograms(arr, K)); // This code is contributed by rakeshsahni</script> |
Output
2
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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