Given an array arr[] containing N strings and an integer K, the task is to find the count of strings with the frequency of each character at most K
Examples:
Input: arr[] = { “abab”, “derdee”, “erre” }, K = 2
Output: 2
Explanation: Strings with character frequency at most 2 are “abab”, “erre”. Hence count is 2Input: arr[] = {“ag”, “ka”, “nanana”}, K = 3
Output: 1
Approach: The idea is to traverse the array of strings, and for each string create a frequency map of characters. Wherever any character has a frequency greater than K, skip the current string. Otherwise, if no character has a frequency more than K, increment the count of answer. Print the count stored in the answer when all the strings are traversed. Follow the steps below to solve the problem:
- Define a function isDuogram(string s, int K) and perform the following tasks:
- Initialize the vector freq[26] with values 0.
- Traverse over the string s using the variable c and perform the following tasks:
- Increase the count of freq by 1.
- Iterate over the range [0. 26) using the variable i and perform the following tasks:
- If freq[i] is greater than K then return false.
- After performing the above steps, return true as the answer.
- Initialize the variable ans as 0 to store the answer.
- Traverse over the array arr[] using the variable x and perform the following tasks:
- Call the function isDuogram(x, K) and if the function returns true then increase the count of ans by 1.
- After performing the above steps, print the value of ans as the answer.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to check if a string// is an duogram or notbool isDuogram(string s, int K){ // Array to store the frequency // of characters vector<int> freq(26, 0); // Count the frequency for (char c : s) { freq++; } // Check if the frequency is greater // than K or not for (int i = 0; i < 26; i++) if (freq[i] > K) return false; return true;}// Function to check if array arr contains// all duograms or notint countDuograms(vector<string>& arr, int K){ // Store the answer int ans = 0; // Traverse the strings for (string x : arr) { // Check the condition if (isDuogram(x, K)) { ans++; } } return ans;}// Driver Codeint main(){ vector<string> arr = { "abab", "derdee", "erre" }; int K = 2; cout << countDuograms(arr, K);} |
Java
// Java program for the above approachclass GFG{// Function to check if a String// is an duogram or notstatic boolean isDuogram(String s, int K){ // Array to store the frequency // of characters int []freq = new int[26]; // Count the frequency for (char c : s.toCharArray()) { freq++; } // Check if the frequency is greater // than K or not for (int i = 0; i < 26; i++) if (freq[i] > K) return false; return true;}// Function to check if array arr contains// all duograms or notstatic int countDuograms(String[] arr, int K){ // Store the answer int ans = 0; // Traverse the Strings for (String x : arr) { // Check the condition if (isDuogram(x, K)) { ans++; } } return ans;}// Driver Codepublic static void main(String[] args){ String []arr = { "abab", "derdee", "erre" }; int K = 2; System.out.print(countDuograms(arr, K));}}// This code is contributed by 29AjayKumar |
Python3
# Python program for the above approach# Function to check if a string# is an duogram or notdef isDuogram (s, K) : # Array to store the frequency # of characters freq = [0] * 26 # Count the frequency for c in s: freq[ord(c) - ord("a")] += 1 # Check if the frequency is greater # than K or not for i in range(26): if (freq[i] > K): return False return True# Function to check if array arr contains# all duograms or notdef countDuograms (arr, K) : # Store the answer ans = 0 # Traverse the strings for x in arr: # Check the condition if (isDuogram(x, K)): ans += 1 return ans# Driver Codearr = ["abab", "derdee", "erre"]K = 2print(countDuograms(arr, K))# This code is contributed by Saurabh Jaiswal |
C#
// C# program for the above approachusing System;class GFG{ // Function to check if a String // is an duogram or not static bool isDuogram(String s, int K) { // Array to store the frequency // of characters int[] freq = new int[26]; // Count the frequency foreach (char c in s.ToCharArray()) { freq[(int)c - (int)'a']++; } // Check if the frequency is greater // than K or not for (int i = 0; i < 26; i++) if (freq[i] > K) return false; return true; } // Function to check if array arr contains // all duograms or not static int countDuograms(String[] arr, int K) { // Store the answer int ans = 0; // Traverse the Strings foreach (String x in arr) { // Check the condition if (isDuogram(x, K)) { ans++; } } return ans; } // Driver Code public static void Main() { String[] arr = { "abab", "derdee", "erre" }; int K = 2; Console.Write(countDuograms(arr, K)); }}// This code is contributed by gfgking |
Javascript
<script> // JavaScript program for the above approach // Function to check if a string // is an duogram or not const isDuogram = (s, K) => { // Array to store the frequency // of characters let freq = new Array(26).fill(0); // Count the frequency for (let c in s) { freq[s.charCodeAt(c) - "a".charCodeAt(0)]++; } // Check if the frequency is greater // than K or not for (let i = 0; i < 26; i++) if (freq[i] > K) return false; return true; } // Function to check if array arr contains // all duograms or not const countDuograms = (arr, K) => { // Store the answer let ans = 0; // Traverse the strings for (let x in arr) { // Check the condition if (isDuogram(arr[x], K)) { ans++; } } return ans; } // Driver Code let arr = ["abab", "derdee", "erre"]; let K = 2; document.write(countDuograms(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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