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 2
Input: 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++
#include <bits/stdc++.h>
using namespace std;
bool isDuogram(string s, int K)
{
vector<int> freq(26, 0);
for (char c : s) {
freq++;
}
for (int i = 0; i < 26; i++)
if (freq[i] > K)
return false;
return true;
}
int countDuograms(vector<string>& arr, int K)
{
int ans = 0;
for (string x : arr) {
if (isDuogram(x, K)) {
ans++;
}
}
return ans;
}
int main()
{
vector<string> arr = { "abab", "derdee", "erre" };
int K = 2;
cout << countDuograms(arr, K);
}
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Java
class GFG{
static boolean isDuogram(String s, int K)
{
int []freq = new int[26];
for (char c : s.toCharArray()) {
freq++;
}
for (int i = 0; i < 26; i++)
if (freq[i] > K)
return false;
return true;
}
static int countDuograms(String[] arr, int K)
{
int ans = 0;
for (String x : arr) {
if (isDuogram(x, K)) {
ans++;
}
}
return ans;
}
public static void main(String[] args)
{
String []arr = { "abab", "derdee", "erre" };
int K = 2;
System.out.print(countDuograms(arr, K));
}
}
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Python3
def isDuogram (s, K) :
freq = [0] * 26
for c in s:
freq[ord(c) - ord("a")] += 1
for i in range(26):
if (freq[i] > K):
return False
return True
def countDuograms (arr, K) :
ans = 0
for x in arr:
if (isDuogram(x, K)):
ans += 1
return ans
arr = ["abab", "derdee", "erre"]
K = 2
print(countDuograms(arr, K))
|
C#
using System;
class GFG
{
static bool isDuogram(String s, int K)
{
int[] freq = new int[26];
foreach (char c in s.ToCharArray())
{
freq[(int)c - (int)'a']++;
}
for (int i = 0; i < 26; i++)
if (freq[i] > K)
return false;
return true;
}
static int countDuograms(String[] arr, int K)
{
int ans = 0;
foreach (String x in arr)
{
if (isDuogram(x, K))
{
ans++;
}
}
return ans;
}
public static void Main()
{
String[] arr = { "abab", "derdee", "erre" };
int K = 2;
Console.Write(countDuograms(arr, K));
}
}
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Javascript
<script>
const isDuogram = (s, K) => {
let freq = new Array(26).fill(0);
for (let c in s) {
freq[s.charCodeAt(c) - "a".charCodeAt(0)]++;
}
for (let i = 0; i < 26; i++)
if (freq[i] > K)
return false;
return true;
}
const countDuograms = (arr, K) => {
let ans = 0;
for (let x in arr) {
if (isDuogram(arr[x], K)) {
ans++;
}
}
return ans;
}
let arr = ["abab", "derdee", "erre"];
let K = 2;
document.write(countDuograms(arr, K));
</script>
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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)