Longest subsequence where each character appears at least k times

Here given code implementation process.

import java.util.HashMap;
/*
    Java program for
    Longest subsequence where each character appears at least k times
*/
public class Subsequence
{
	public void kCharacterSubsequence(String text, int k)
	{
		// Get the length of given text 
		int n = text.length();
		if (n == 0 || k < 0)
		{
			return;
		}
		HashMap < Character, Integer > record = 
          new HashMap < Character, Integer > ();
		String result = "";
		for (int i = 0; i < n; ++i)
		{
			if (record.containsKey(text.charAt(i)))
			{
				// Increase frequency
				record.put(text.charAt(i), 
                           record.get(text.charAt(i)) + 1);
			}
			else
			{
				// Add new key value 
				record.put(text.charAt(i), 1);
			}
		}
      	// Add element which is occurring at least K times
		for (int i = 0; i < n; ++i)
		{
			if (record.get(text.charAt(i)) >= k)
			{   
				result = result + text.charAt(i);
			}
		}
		if (result.length() == 0)
		{
			System.out.print("None");
		}
		else
		{
			System.out.println(result);
		}
	}
	public static void main(String[] args)
	{
		Subsequence task = new Subsequence();
		String text = "abcpcdxyabzdaced";
		int k = 2;
		// Example A
		// str = abccdabdaced
		// k = 2 occurrence at least 2 times
		// [
		//      p : 1
		//      a : 3    3 >= k
		//      b : 2    2 >= k
		//      c : 3    3 >= k
		//      d : 3    3 >= k
		//      e : 1
		//      x : 1
		//      y : 1
		//      z : 1
		// ]
		// Output : abccdabdacd
		task.kCharacterSubsequence(text, k);
		k = 3;
		// Example B
		// str = abccdabdaced
		// k = 3 occurrence at least 3 times
		// [
		//      p : 1
		//      a : 3    3 >= k
		//      b : 2 
		//      c : 3    3 >= k
		//      d : 3    3 >= k
		//      e : 1
		//      x : 1
		//      y : 1
		//      z : 1
		// ]
		// Output : accdadacd
		task.kCharacterSubsequence(text, k);
	}
}

Output

abccdabdacd
accdadacd
// Include header file
#include <iostream>
#include <string>
#include <unordered_map>

using namespace std;
/*
    C++ program for
    Longest subsequence where each character 
    appears at least k times
*/
class Subsequence
{
	public: void kCharacterSubsequence(string text, int k)
	{
		// Get the length of given text 
		int n = text.length();
		if (n == 0 || k < 0)
		{
			return;
		}
		unordered_map < char, int > record;
		string result = "";
		for (int i = 0; i < n; ++i)
		{
			if (record.find(text[i]) != record.end())
			{
				// Increase frequency
				record[text[i]] = record[text[i]] + 1;
			}
			else
			{
				// Add new key value 
				record[text[i]] = 1;
			}
		}
		// Add element which is occurring at least K times
		for (int i = 0; i < n; ++i)
		{
			if (record[text[i]] >= k)
			{
				result = result  +  text[i];
			}
		}
		if (result.length() == 0)
		{
			cout << "None";
		}
		else
		{
			cout << result << endl;
		}
	}
};
int main()
{
	Subsequence *task = new Subsequence();
	string text = "abcpcdxyabzdaced";
	int k = 2;
	// Example A
	// str = abccdabdaced
	// k = 2 occurrence at least 2 times
	// [
	//      p : 1
	//      a : 3    3 >= k
	//      b : 2    2 >= k
	//      c : 3    3 >= k
	//      d : 3    3 >= k
	//      e : 1
	//      x : 1
	//      y : 1
	//      z : 1
	// ]
	// Output : abccdabdacd
	task->kCharacterSubsequence(text, k);
	k = 3;
	// Example B
	// str = abccdabdaced
	// k = 3 occurrence at least 3 times
	// [
	//      p : 1
	//      a : 3    3 >= k
	//      b : 2 
	//      c : 3    3 >= k
	//      d : 3    3 >= k
	//      e : 1
	//      x : 1
	//      y : 1
	//      z : 1
	// ]
	// Output : accdadacd
	task->kCharacterSubsequence(text, k);
	return 0;
}

Output

abccdabdacd
accdadacd
// Include namespace system
using System;
using System.Collections.Generic;
/*
    Csharp program for
    Longest subsequence where each character appears at least k times
*/
public class Subsequence
{
	public void kCharacterSubsequence(String text, int k)
	{
		// Get the length of given text 
		int n = text.Length;
		if (n == 0 || k < 0)
		{
			return;
		}
		Dictionary < char, int > record = 
          new Dictionary < char, int > ();
		String result = "";
		for (int i = 0; i < n; ++i)
		{
			if (record.ContainsKey(text[i]))
			{
				// Increase frequency
				record[text[i]] = record[text[i]] + 1;
			}
			else
			{
				// Add new key value 
				record.Add(text[i], 1);
			}
		}
		// Add element which is occurring at least K times
		for (int i = 0; i < n; ++i)
		{
			if (record[text[i]] >= k)
			{
				result = result + text[i];
			}
		}
		if (result.Length == 0)
		{
			Console.Write("None");
		}
		else
		{
			Console.WriteLine(result);
		}
	}
	public static void Main(String[] args)
	{
		Subsequence task = new Subsequence();
		String text = "abcpcdxyabzdaced";
		int k = 2;
		// Example A
		// str = abccdabdaced
		// k = 2 occurrence at least 2 times
		// [
		//      p : 1
		//      a : 3    3 >= k
		//      b : 2    2 >= k
		//      c : 3    3 >= k
		//      d : 3    3 >= k
		//      e : 1
		//      x : 1
		//      y : 1
		//      z : 1
		// ]
		// Output : abccdabdacd
		task.kCharacterSubsequence(text, k);
		k = 3;
		// Example B
		// str = abccdabdaced
		// k = 3 occurrence at least 3 times
		// [
		//      p : 1
		//      a : 3    3 >= k
		//      b : 2 
		//      c : 3    3 >= k
		//      d : 3    3 >= k
		//      e : 1
		//      x : 1
		//      y : 1
		//      z : 1
		// ]
		// Output : accdadacd
		task.kCharacterSubsequence(text, k);
	}
}

Output

abccdabdacd
accdadacd
package main
import "fmt"
/*
    Go program for
    Longest subsequence where each character appears at 
    least k times
*/

func kCharacterSubsequence(text string, k int) {
	// Get the length of given text 
	var n int = len(text)
	if n == 0 || k < 0 {
		return
	}
	var record = make(map[byte] int)
	var result string = ""
	for i := 0 ; i < n ; i++ {
		if _, found := record[text[i]] ; found {
			// Increase frequency
			record[text[i]] = record[text[i]] + 1
		} else {
			// Add new key value 
			record[text[i]] = 1
		}
	}
	// Add element which is occurring at least K times
	for i := 0 ; i < n ; i++ {
		if record[text[i]] >= k {
			result = result + string(text[i])
		}
	}
	if len(result) == 0 {
		fmt.Print("None")
	} else {
		fmt.Println(result)
	}
}
func main() {

	var text string = "abcpcdxyabzdaced"
	var k int = 2
	// Example A
	// str = abccdabdaced
	// k = 2 occurrence at least 2 times
	// [
	//      p : 1
	//      a : 3    3 >= k
	//      b : 2    2 >= k
	//      c : 3    3 >= k
	//      d : 3    3 >= k
	//      e : 1
	//      x : 1
	//      y : 1
	//      z : 1
	// ]
	// Output : abccdabdacd
	kCharacterSubsequence(text, k)
	k = 3
	// Example B
	// str = abccdabdaced
	// k = 3 occurrence at least 3 times
	// [
	//      p : 1
	//      a : 3    3 >= k
	//      b : 2 
	//      c : 3    3 >= k
	//      d : 3    3 >= k
	//      e : 1
	//      x : 1
	//      y : 1
	//      z : 1
	// ]
	// Output : accdadacd
	kCharacterSubsequence(text, k)
}

Output

abccdabdacd
accdadacd
<?php
/*
    Php program for
    Longest subsequence where each character appears at least k times
*/
class Subsequence
{
	public	function kCharacterSubsequence($text, $k)
	{
		// Get the length of given text 
		$n = strlen($text);
		if ($n == 0 || $k < 0)
		{
			return;
		}
		$record = array();
		$result = "";
		for ($i = 0; $i < $n; ++$i)
		{
			if (array_key_exists($text[$i], $record))
			{
				// Increase frequency
				$record[$text[$i]] = $record[$text[$i]] + 1;
			}
			else
			{
				// Add new key value 
				$record[$text[$i]] = 1;
			}
		}
		// Add element which is occurring at least K times
		for ($i = 0; $i < $n; ++$i)
		{
			if ($record[$text[$i]] >= $k)
			{
				$result = $result.strval($text[$i]);
			}
		}
		if (strlen($result) == 0)
		{
			echo("None");
		}
		else
		{
			echo($result.
				"\n");
		}
	}
}

function main()
{
	$task = new Subsequence();
	$text = "abcpcdxyabzdaced";
	$k = 2;
	// Example A
	// str = abccdabdaced
	// k = 2 occurrence at least 2 times
	// [
	//      p : 1
	//      a : 3    3 >= k
	//      b : 2    2 >= k
	//      c : 3    3 >= k
	//      d : 3    3 >= k
	//      e : 1
	//      x : 1
	//      y : 1
	//      z : 1
	// ]
	// Output : abccdabdacd
	$task->kCharacterSubsequence($text, $k);
	$k = 3;
	// Example B
	// str = abccdabdaced
	// k = 3 occurrence at least 3 times
	// [
	//      p : 1
	//      a : 3    3 >= k
	//      b : 2 
	//      c : 3    3 >= k
	//      d : 3    3 >= k
	//      e : 1
	//      x : 1
	//      y : 1
	//      z : 1
	// ]
	// Output : accdadacd
	$task->kCharacterSubsequence($text, $k);
}
main();

Output

abccdabdacd
accdadacd
/*
    Node JS program for
    Longest subsequence where each character appears at least k times
*/
class Subsequence
{
	kCharacterSubsequence(text, k)
	{
		// Get the length of given text 
		var n = text.length;
		if (n == 0 || k < 0)
		{
			return;
		}
		var record = new Map();
		var result = "";
		for (var i = 0; i < n; ++i)
		{
			if (record.has(text.charAt(i)))
			{
				// Increase frequency
				record.set(text.charAt(i), 
                           record.get(text.charAt(i)) + 1);
			}
			else
			{
				// Add new key value 
				record.set(text.charAt(i), 1);
			}
		}
		// Add element which is occurring at least K times
		for (var i = 0; i < n; ++i)
		{
			if (record.get(text.charAt(i)) >= k)
			{
				result = result + text.charAt(i);
			}
		}
		if (result.length == 0)
		{
			process.stdout.write("None");
		}
		else
		{
			console.log(result);
		}
	}
}

function main()
{
	var task = new Subsequence();
	var text = "abcpcdxyabzdaced";
	var k = 2;
	// Example A
	// str = abccdabdaced
	// k = 2 occurrence at least 2 times
	// [
	//      p : 1
	//      a : 3    3 >= k
	//      b : 2    2 >= k
	//      c : 3    3 >= k
	//      d : 3    3 >= k
	//      e : 1
	//      x : 1
	//      y : 1
	//      z : 1
	// ]
	// Output : abccdabdacd
	task.kCharacterSubsequence(text, k);
	k = 3;
	// Example B
	// str = abccdabdaced
	// k = 3 occurrence at least 3 times
	// [
	//      p : 1
	//      a : 3    3 >= k
	//      b : 2 
	//      c : 3    3 >= k
	//      d : 3    3 >= k
	//      e : 1
	//      x : 1
	//      y : 1
	//      z : 1
	// ]
	// Output : accdadacd
	task.kCharacterSubsequence(text, k);
}
main();

Output

abccdabdacd
accdadacd
#    Python 3 program for
#    Longest subsequence where each character appears at least k times
class Subsequence :
	def kCharacterSubsequence(self, text, k) :
		#  Get the length of given text 
		n = len(text)
		if (n == 0 or k < 0) :
			return
		
		record = dict()
		result = ""
		i = 0
		while (i < n) :
			if ((text[i] in record.keys())) :
				#  Increase frequency
				record[text[i]] = record.get(text[i]) + 1
			else :
				#  Add new key value 
				record[text[i]] = 1
			
			i += 1
		
		i = 0
		#  Add element which is occurring at least K times
		while (i < n) :
			if (record.get(text[i]) >= k) :
				result = result + str(text[i])
			
			i += 1
		
		if (len(result) == 0) :
			print("None", end = "")
		else :
			print(result)
		
	

def main() :
	task = Subsequence()
	text = "abcpcdxyabzdaced"
	k = 2
	#  Example A
	#  str = abccdabdaced
	#  k = 2 occurrence at least 2 times
	#  [
	#       p : 1
	#       a : 3    3 >= k
	#       b : 2    2 >= k
	#       c : 3    3 >= k
	#       d : 3    3 >= k
	#       e : 1
	#       x : 1
	#       y : 1
	#       z : 1
	#  ]
	#  Output : abccdabdacd
	task.kCharacterSubsequence(text, k)
	k = 3
	#  Example B
	#  str = abccdabdaced
	#  k = 3 occurrence at least 3 times
	#  [
	#       p : 1
	#       a : 3    3 >= k
	#       b : 2 
	#       c : 3    3 >= k
	#       d : 3    3 >= k
	#       e : 1
	#       x : 1
	#       y : 1
	#       z : 1
	#  ]
	#  Output : accdadacd
	task.kCharacterSubsequence(text, k)

if __name__ == "__main__": main()

Output

abccdabdacd
accdadacd
#    Ruby program for
#    Longest subsequence where each character appears at least k times
class Subsequence 
	def kCharacterSubsequence(text, k) 
		#  Get the length of given text 
		n = text.length
		if (n == 0 || k < 0) 
			return
		end

		record = Hash.new()
		result = ""
		i = 0
		while (i < n) 
			if (record.key?(text[i])) 
				#  Increase frequency
				record[text[i]] = record[text[i]] + 1
			else
 
				#  Add new key value 
				record[text[i]] = 1
			end

			i += 1
		end

		i = 0
		#  Add element which is occurring at least K times
		while (i < n) 
			if (record[text[i]] >= k) 
				result = result + text[i].to_s
			end

			i += 1
		end

		if (result.length == 0) 
			print("None")
		else
 
			print(result, "\n")
		end

	end

end

def main() 
	task = Subsequence.new()
	text = "abcpcdxyabzdaced"
	k = 2
	#  Example A
	#  str = abccdabdaced
	#  k = 2 occurrence at least 2 times
	#  [
	#       p : 1
	#       a : 3    3 >= k
	#       b : 2    2 >= k
	#       c : 3    3 >= k
	#       d : 3    3 >= k
	#       e : 1
	#       x : 1
	#       y : 1
	#       z : 1
	#  ]
	#  Output : abccdabdacd
	task.kCharacterSubsequence(text, k)
	k = 3
	#  Example B
	#  str = abccdabdaced
	#  k = 3 occurrence at least 3 times
	#  [
	#       p : 1
	#       a : 3    3 >= k
	#       b : 2 
	#       c : 3    3 >= k
	#       d : 3    3 >= k
	#       e : 1
	#       x : 1
	#       y : 1
	#       z : 1
	#  ]
	#  Output : accdadacd
	task.kCharacterSubsequence(text, k)
end

main()

Output

abccdabdacd
accdadacd
import scala.collection.mutable._;
/*
    Scala program for
    Longest subsequence where each character appears at least k times
*/
class Subsequence()
{
	def kCharacterSubsequence(text: String, k: Int): Unit = {
		// Get the length of given text 
		var n: Int = text.length();
		if (n == 0 || k < 0)
		{
			return;
		}
		var record = new HashMap[Character, Int]();
		var result: String = "";
		var i: Int = 0;
		while (i < n)
		{
			if (record.contains(text.charAt(i)))
			{
				// Increase frequency
				record.addOne(text.charAt(i), 
                              record.get(text.charAt(i)).get + 1);
			}
			else
			{
				// Add new key value 
				record.addOne(text.charAt(i), 1);
			}
			i += 1;
		}
		i = 0;
		// Add element which is occurring at least K times
		while (i < n)
		{
			if (record.get(text.charAt(i)).get >= k)
			{
				result = result + text.charAt(i).toString();
			}
			i += 1;
		}
		if (result.length() == 0)
		{
			print("None");
		}
		else
		{
			println(result);
		}
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: Subsequence = new Subsequence();
		var text: String = "abcpcdxyabzdaced";
		var k: Int = 2;
		// Example A
		// str = abccdabdaced
		// k = 2 occurrence at least 2 times
		// [
		//      p : 1
		//      a : 3    3 >= k
		//      b : 2    2 >= k
		//      c : 3    3 >= k
		//      d : 3    3 >= k
		//      e : 1
		//      x : 1
		//      y : 1
		//      z : 1
		// ]
		// Output : abccdabdacd
		task.kCharacterSubsequence(text, k);
		k = 3;
		// Example B
		// str = abccdabdaced
		// k = 3 occurrence at least 3 times
		// [
		//      p : 1
		//      a : 3    3 >= k
		//      b : 2 
		//      c : 3    3 >= k
		//      d : 3    3 >= k
		//      e : 1
		//      x : 1
		//      y : 1
		//      z : 1
		// ]
		// Output : accdadacd
		task.kCharacterSubsequence(text, k);
	}
}

Output

abccdabdacd
accdadacd
import Foundation;
/*
    Swift 4 program for
    Longest subsequence where each character appears at least k times
*/
class Subsequence
{
	func kCharacterSubsequence(_ data: String, _ k: Int)
	{
		// Get the length of given text 
		let n: Int = data.count;
		if (n == 0 || k < 0)
		{
			return;
		}
      	let text = Array(data);
		var record = [Character : Int]();
		var result: String = "";
		var i: Int = 0;
		while (i < n)
		{
			if (record.keys.contains(text[i]))
			{
				// Increase frequency
				record[text[i]] = record[text[i]]! + 1;
			}
			else
			{
				// Add new key value 
				record[text[i]] = 1;
			}
			i += 1;
		}
		i = 0;
		// Add element which is occurring at least K times
		while (i < n)
		{
			if (record[text[i]]! >= k)
			{
				result = result + String(text[i]);
			}
			i += 1;
		}
		if (result.count == 0)
		{
			print("None", terminator: "");
		}
		else
		{
			print(result);
		}
	}
}
func main()
{
	let task: Subsequence = Subsequence();
	let text: String = "abcpcdxyabzdaced";
	var k: Int = 2;
	// Example A
	// str = abccdabdaced
	// k = 2 occurrence at least 2 times
	// [
	//      p : 1
	//      a : 3    3 >= k
	//      b : 2    2 >= k
	//      c : 3    3 >= k
	//      d : 3    3 >= k
	//      e : 1
	//      x : 1
	//      y : 1
	//      z : 1
	// ]
	// Output : abccdabdacd
	task.kCharacterSubsequence(text, k);
	k = 3;
	// Example B
	// str = abccdabdaced
	// k = 3 occurrence at least 3 times
	// [
	//      p : 1
	//      a : 3    3 >= k
	//      b : 2 
	//      c : 3    3 >= k
	//      d : 3    3 >= k
	//      e : 1
	//      x : 1
	//      y : 1
	//      z : 1
	// ]
	// Output : accdadacd
	task.kCharacterSubsequence(text, k);
}
main();

Output

abccdabdacd
accdadacd
/*
    Kotlin program for
    Longest subsequence where each character appears at least k times
*/
class Subsequence
{
	fun kCharacterSubsequence(text: String, k: Int): Unit
	{
		// Get the length of given text 
		val n: Int = text.length;
		if (n == 0 || k < 0)
		{
			return;
		}
		var record = HashMap < Char, Int > ();
		var result: String = "";
		var i: Int = 0;
		while (i < n)
		{
			if (record.containsKey(text.get(i)))
			{
				// Increase frequency
				record.put(text.get(i), 
                           record.getValue(text.get(i)) + 1);
			}
			else
			{
				// Add new key value 
				record.put(text.get(i), 1);
			}
			i += 1;
		}
		i = 0;
		// Add element which is occurring at least K times
		while (i < n)
		{
			if (record.getValue(text.get(i)) >= k)
			{
				result = result + text.get(i).toString();
			}
			i += 1;
		}
		if (result.length == 0)
		{
			print("None");
		}
		else
		{
			println(result);
		}
	}
}
fun main(args: Array < String > ): Unit
{
	val task: Subsequence = Subsequence();
	val text: String = "abcpcdxyabzdaced";
	var k: Int = 2;
	// Example A
	// str = abccdabdaced
	// k = 2 occurrence at least 2 times
	// [
	//      p : 1
	//      a : 3    3 >= k
	//      b : 2    2 >= k
	//      c : 3    3 >= k
	//      d : 3    3 >= k
	//      e : 1
	//      x : 1
	//      y : 1
	//      z : 1
	// ]
	// Output : abccdabdacd
	task.kCharacterSubsequence(text, k);
	k = 3;
	// Example B
	// str = abccdabdaced
	// k = 3 occurrence at least 3 times
	// [
	//      p : 1
	//      a : 3    3 >= k
	//      b : 2 
	//      c : 3    3 >= k
	//      d : 3    3 >= k
	//      e : 1
	//      x : 1
	//      y : 1
	//      z : 1
	// ]
	// Output : accdadacd
	task.kCharacterSubsequence(text, k);
}

Output

abccdabdacd
accdadacd


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