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Code String

Find the second most frequent character in a given string

Here given code implementation process.

/*
    C program for
    Find the second most frequent character in a given string
*/
#include <stdio.h>
#include <string.h>

void secondMostFrequent(char *text)
{
	int n = strlen(text);
	if (n == 0)
	{
		return;
	}
	int first = 0;
	int second = -1;
	int record[256];
	// Set initial count of character record
	for (int i = 0; i < 256; ++i)
	{
		record[i] = 0;
	}
	// Count frequency of character element
	for (int i = 0; i < n; ++i)
	{
		record[text[i]] = record[text[i]] + 1;
	}
	// Find second most occurring character
	for (int i = 0; i < n; ++i)
	{
		if (record[text[i]] > record[text[first]])
		{
			second = first;
			first = i;
		}
		else if (record[text[i]] < record[text[first]])
		{
			if (second == -1 || record[text[second]] < record[text[i]])
			{
				second = i;
			}
		}
	}
	printf("\n Given Text : %s", text);
	if (second == -1)
	{
		printf("\n Second most frequent element not exist");
	}
	else
	{
		printf("\n Second most frequent element is : %c", text[second]);
	}
}
int main(int argc, char
	const *argv[])
{
	// Given Sting text
	char *text1 = "xyoxoxx";
	char *text2 = "abaacdccec";
	char *text3 = "xxyyzz";
	/*
	    Test A
	    ---------
	    text   = "xyoxoxx"
	    "xxxx" = 4
	    "y"    = 1
	    "oo"   = 2
	    -------------------------------
	    o = 2 which is second largest
	*/
	secondMostFrequent(text1);
	/*
	    Test B 
	    ---------
	    text   = "abaacdccec"
	    "aaa"  = 3
	    "b"    = 1
	    "cccc" = 4
	    "d"    = 1
	    -------------------------------
	    a = 3 which is second largest
	*/
	secondMostFrequent(text2);
	/*
	    Test C
	    ---------
	    "xx" = 2
	    "yy" = 2
	    "zz" = 2
	    -------------------------------
	    No second most frequent element
	*/
	secondMostFrequent(text3);
	return 0;
}

Output

 Given Text : xyoxoxx
 Second most frequent element is : o
 Given Text : abaacdccec
 Second most frequent element is : a
 Given Text : xxyyzz
 Second most frequent element not exist
/*
    Java program for
    Find the second most frequent character in a given string
*/
public class Occurrence
{
    public void secondMostFrequent(String text)
    {
        int n = text.length();
        if (n == 0)
        {
            return;
        }
        int first = 0;
        int second = -1;
        int[] record = new int[256];
        // Set initial count of character record
        for (int i = 0; i < 256; ++i)
        {
            record[i] = 0;
        }
        // Count frequency of character element
        for (int i = 0; i < n; ++i)
        {
            record[text.charAt(i)] = record[text.charAt(i)] + 1;
        }
        // Find second most occurring character
        for (int i = 0; i < n; ++i)
        {
            if (record[text.charAt(i)] > record[text.charAt(first)])
            {
                second = first;
                first = i;
            }
            else if (record[text.charAt(i)] < record[text.charAt(first)])
            {
                if (second == -1 || 
                    record[text.charAt(second)] < record[text.charAt(i)])
                {
                    second = i;
                }
            }
        }
        System.out.print("\n Given Text : " + text );

        if (second == -1)
        {
            System.out.print("\n Second most frequent element not exist");
        }
        else
        {
            System.out.print("\n Second most frequent element is : " + 
                             text.charAt(second) );
        }
    }
    public static void main(String[] args)
    {
        Occurrence task = new Occurrence();

        // Given Sting text
        String text1 = "xyoxoxx";
        String text2 = "abaacdccec";
        String text3 = "xxyyzz";
        /*
            Test A
            ---------
            text   = "xyoxoxx"
            "xxxx" = 4
            "y"    = 1
            "oo"   = 2
            -------------------------------
            o = 2 which is second largest
        */
        task.secondMostFrequent(text1);
        /*
            Test B 
            ---------
            text   = "abaacdccec"
            "aaa"  = 3
            "b"    = 1
            "cccc" = 4
            "d"    = 1
            -------------------------------
            a = 3 which is second largest
        */
        task.secondMostFrequent(text2);
        /*
            Test C
            ---------
            "xx" = 2
            "yy" = 2
            "zz" = 2
            -------------------------------
            No second most frequent element
        */
        task.secondMostFrequent(text3);

    }
}

Output

 Given Text : xyoxoxx
 Second most frequent element is : o
 Given Text : abaacdccec
 Second most frequent element is : a
 Given Text : xxyyzz
 Second most frequent element not exist
// Include header file
#include <iostream>
#include <string>

using namespace std;
/*
    C++ program for
    Find the second most frequent character in a given string
*/
class Occurrence
{
	public: void secondMostFrequent(string text)
	{
		int n = text.length();
		if (n == 0)
		{
			return;
		}
		int first = 0;
		int second = -1;
		int record[256];
		// Set initial count of character record
		for (int i = 0; i < 256; ++i)
		{
			record[i] = 0;
		}
		// Count frequency of character element
		for (int i = 0; i < n; ++i)
		{
			record[text[i]] = record[text[i]] + 1;
		}
		// Find second most occurring character
		for (int i = 0; i < n; ++i)
		{
			if (record[text[i]] > record[text[first]])
			{
				second = first;
				first = i;
			}
			else if (record[text[i]] < record[text[first]])
			{
				if (second == -1 || record[text[second]] < record[text[i]])
				{
					second = i;
				}
			}
		}
		cout << "\n Given Text : " << text;
		if (second == -1)
		{
			cout << "\n Second most frequent element not exist";
		}
		else
		{
			cout << "\n Second most frequent element is : " << text[second];
		}
	}
};
int main()
{
	Occurrence *task = new Occurrence();
	// Given Sting text
	string text1 = "xyoxoxx";
	string text2 = "abaacdccec";
	string text3 = "xxyyzz";
	/*
	            Test A
	            ---------
	            text   = "xyoxoxx"
	            "xxxx" = 4
	            "y"    = 1
	            "oo"   = 2
	            -------------------------------
	            o = 2 which is second largest
	        */
	task->secondMostFrequent(text1);
	/*
	            Test B 
	            ---------
	            text   = "abaacdccec"
	            "aaa"  = 3
	            "b"    = 1
	            "cccc" = 4
	            "d"    = 1
	            -------------------------------
	            a = 3 which is second largest
	        */
	task->secondMostFrequent(text2);
	/*
	            Test C
	            ---------
	            "xx" = 2
	            "yy" = 2
	            "zz" = 2
	            -------------------------------
	            No second most frequent element
	        */
	task->secondMostFrequent(text3);
	return 0;
}

Output

 Given Text : xyoxoxx
 Second most frequent element is : o
 Given Text : abaacdccec
 Second most frequent element is : a
 Given Text : xxyyzz
 Second most frequent element not exist
// Include namespace system
using System;
/*
    Csharp program for
    Find the second most frequent character in a given string
*/
public class Occurrence
{
	public void secondMostFrequent(String text)
	{
		int n = text.Length;
		if (n == 0)
		{
			return;
		}
		int first = 0;
		int second = -1;
		int[] record = new int[256];
		// Set initial count of character record
		for (int i = 0; i < 256; ++i)
		{
			record[i] = 0;
		}
		// Count frequency of character element
		for (int i = 0; i < n; ++i)
		{
			record[text[i]] = record[text[i]] + 1;
		}
		// Find second most occurring character
		for (int i = 0; i < n; ++i)
		{
			if (record[text[i]] > record[text[first]])
			{
				second = first;
				first = i;
			}
			else if (record[text[i]] < record[text[first]])
			{
				if (second == -1 || 
                    record[text[second]] < record[text[i]])
				{
					second = i;
				}
			}
		}
		Console.Write("\n Given Text : " + text);
		if (second == -1)
		{
			Console.Write("\n Second most frequent element not exist");
		}
		else
		{
			Console.Write("\n Second most frequent element is : " + text[second]);
		}
	}
	public static void Main(String[] args)
	{
		Occurrence task = new Occurrence();
		// Given Sting text
		String text1 = "xyoxoxx";
		String text2 = "abaacdccec";
		String text3 = "xxyyzz";
		/*
		            Test A
		            ---------
		            text   = "xyoxoxx"
		            "xxxx" = 4
		            "y"    = 1
		            "oo"   = 2
		            -------------------------------
		            o = 2 which is second largest
		        */
		task.secondMostFrequent(text1);
		/*
		            Test B 
		            ---------
		            text   = "abaacdccec"
		            "aaa"  = 3
		            "b"    = 1
		            "cccc" = 4
		            "d"    = 1
		            -------------------------------
		            a = 3 which is second largest
		        */
		task.secondMostFrequent(text2);
		/*
		            Test C
		            ---------
		            "xx" = 2
		            "yy" = 2
		            "zz" = 2
		            -------------------------------
		            No second most frequent element
		        */
		task.secondMostFrequent(text3);
	}
}

Output

 Given Text : xyoxoxx
 Second most frequent element is : o
 Given Text : abaacdccec
 Second most frequent element is : a
 Given Text : xxyyzz
 Second most frequent element not exist
package main
import "fmt"
/*
    Go program for
    Find the second most frequent character in a given string
*/

func secondMostFrequent(text string) {
	var n int = len(text)
	if n == 0 {
		return
	}
	var first int = 0
	var second int = -1
	var record = make([]int ,256)
	// Count frequency of character element
	for i := 0 ; i < n ; i++ {
		record[text[i]] = record[text[i]] + 1
	}
	// Find second most occurring character
	for i := 0 ; i < n ; i++ {
		if record[text[i]] > record[text[first]] {
			second = first
			first = i
		} else if record[text[i]] < record[text[first]] {
			if second == -1 || record[text[second]] < record[text[i]] {
				second = i
			}
		}
	}
	fmt.Print("\n Given Text : ", text)
	if second == -1 {
		fmt.Print("\n Second most frequent element not exist")
	} else {
		fmt.Print("\n Second most frequent element is : ", string(text[second]))
	}
}
func main() {

	// Given Sting text
	var text1 string = "xyoxoxx"
	var text2 string = "abaacdccec"
	var text3 string = "xxyyzz"
	/*
	    Test A
	    ---------
	    text   = "xyoxoxx"
	    "xxxx" = 4
	    "y"    = 1
	    "oo"   = 2
	    -------------------------------
	    o = 2 which is second largest
	*/
	secondMostFrequent(text1)
	/*
	    Test B 
	    ---------
	    text   = "abaacdccec"
	    "aaa"  = 3
	    "b"    = 1
	    "cccc" = 4
	    "d"    = 1
	    -------------------------------
	    a = 3 which is second largest
	*/
	secondMostFrequent(text2)
	/*
	    Test C
	    ---------
	    "xx" = 2
	    "yy" = 2
	    "zz" = 2
	    -------------------------------
	    No second most frequent element
	*/
	secondMostFrequent(text3)
}

Output

 Given Text : xyoxoxx
 Second most frequent element is : o
 Given Text : abaacdccec
 Second most frequent element is : a
 Given Text : xxyyzz
 Second most frequent element not exist
<?php
/*
    Php program for
    Find the second most frequent character in a given string
*/
class Occurrence
{
    public  function secondMostFrequent($text)
    {
        $n = strlen($text);
        if ($n == 0)
        {
            return;
        }
        $first = 0;
        $second = -1;
        $record = array_fill(0, 256, 0);
        // Count frequency of character element
        for ($i = 0; $i < $n; ++$i)
        {
            $record[ord($text[$i])] = $record[ord($text[$i])] + 1;
        }
        // Find second most occurring character
        for ($i = 0; $i < $n; ++$i)
        {
            if ($record[ord($text[$i])] > $record[ord($text[$first])])
            {
                $second = $first;
                $first = $i;
            }
            else if ($record[ord($text[$i])] < $record[ord($text[$first])])
            {
                if ($second == -1 || 
                    $record[ord($text[$second])] < $record[ord($text[$i])])
                {
                    $second = $i;
                }
            }
        }
        echo("\n Given Text : ".$text);
        if ($second == -1)
        {
            echo("\n Second most frequent element not exist");
        }
        else
        {
            echo("\n Second most frequent element is : ".$text[$second]);
        }
    }
}

function main()
{
    $task = new Occurrence();
    // Given Sting text
    $text1 = "xyoxoxx";
    $text2 = "abaacdccec";
    $text3 = "xxyyzz";
    /*
        Test A
        ---------
        text   = "xyoxoxx"
        "xxxx" = 4
        "y"    = 1
        "oo"   = 2
        -------------------------------
        o = 2 which is second largest
    */
    $task->secondMostFrequent($text1);
    /*
        Test B 
        ---------
        text   = "abaacdccec"
        "aaa"  = 3
        "b"    = 1
        "cccc" = 4
        "d"    = 1
        -------------------------------
        a = 3 which is second largest
    */
    $task->secondMostFrequent($text2);
    /*
        Test C
        ---------
        "xx" = 2
        "yy" = 2
        "zz" = 2
        -------------------------------
        No second most frequent element
    */
    $task->secondMostFrequent($text3);
}
main();

Output

 Given Text : xyoxoxx
 Second most frequent element is : o
 Given Text : abaacdccec
 Second most frequent element is : a
 Given Text : xxyyzz
 Second most frequent element not exist
/*
    Node JS program for
    Find the second most frequent character in a given string
*/
class Occurrence
{
	secondMostFrequent(text)
	{
		var n = text.length;
		if (n == 0)
		{
			return;
		}
		var first = 0;
		var second = -1;
		var record = Array(256).fill(0);
		// Count frequency of character element
		for (var i = 0; i < n; ++i)
		{
			record[text.charCodeAt(i)] = record[text.charCodeAt(i)] + 1;
		}
		// Find second most occurring character
		for (var i = 0; i < n; ++i)
		{
			if (record[text.charCodeAt(i)] > record[text.charCodeAt(first)])
			{
				second = first;
				first = i;
			}
			else if (record[text.charCodeAt(i)] < record[text.charCodeAt(first)])
			{
				if (second == -1 || 
                    record[text.charCodeAt(second)] < record[text.charCodeAt(i)])
				{
					second = i;
				}
			}
		}
		process.stdout.write("\n Given Text : " + text);
		if (second == -1)
		{
			process.stdout.write("\n Second most frequent element not exist");
		}
		else
		{
			process.stdout.write("\n Second most frequent element is : " +
                                 text.charAt(second));
		}
	}
}

function main()
{
	var task = new Occurrence();
	// Given Sting text
	var text1 = "xyoxoxx";
	var text2 = "abaacdccec";
	var text3 = "xxyyzz";
	/*
	    Test A
	    ---------
	    text   = "xyoxoxx"
	    "xxxx" = 4
	    "y"    = 1
	    "oo"   = 2
	    -------------------------------
	    o = 2 which is second largest
	*/
	task.secondMostFrequent(text1);
	/*
	    Test B 
	    ---------
	    text   = "abaacdccec"
	    "aaa"  = 3
	    "b"    = 1
	    "cccc" = 4
	    "d"    = 1
	    -------------------------------
	    a = 3 which is second largest
	*/
	task.secondMostFrequent(text2);
	/*
	    Test C
	    ---------
	    "xx" = 2
	    "yy" = 2
	    "zz" = 2
	    -------------------------------
	    No second most frequent element
	*/
	task.secondMostFrequent(text3);
}
main();

Output

 Given Text : xyoxoxx
 Second most frequent element is : o
 Given Text : abaacdccec
 Second most frequent element is : a
 Given Text : xxyyzz
 Second most frequent element not exist
#    Python 3 program for
#    Find the second most frequent character in a given string
class Occurrence :
	def secondMostFrequent(self, text) :
		n = len(text)
		if (n == 0) :
			return
		
		first = 0
		second = -1
		record = [0] * (256)
		i = 0
		#  Count frequency of character element
		while (i < n) :
			record[ord(text[i])] = record[ord(text[i])] + 1
			i += 1
		
		i = 0
		#  Find second most occurring character
		while (i < n) :
			if (record[ord(text[i])] > record[ord(text[first])]) :
				second = first
				first = i
			elif (record[ord(text[i])] < record[ord(text[first])]) :
				if (second == -1 or 
                    record[ord(text[second])] < record[ord(text[i])]) :
					second = i
				
			
			i += 1
		
		print("\n Given Text : ", text, end = "")
		if (second == -1) :
			print("\n Second most frequent element not exist", end = "")
		else :
			print("\n Second most frequent element is : ", text[second], end = "")
		
	

def main() :
	task = Occurrence()
	#  Given Sting text
	text1 = "xyoxoxx"
	text2 = "abaacdccec"
	text3 = "xxyyzz"
	#    Test A
	#    ---------
	#    text   = "xyoxoxx"
	#    "xxxx" = 4
	#    "y"    = 1
	#    "oo"   = 2
	#    -------------------------------
	#    o = 2 which is second largest
	task.secondMostFrequent(text1)
	#    Test B 
	#    ---------
	#    text   = "abaacdccec"
	#    "aaa"  = 3
	#    "b"    = 1
	#    "cccc" = 4
	#    "d"    = 1
	#    -------------------------------
	#    a = 3 which is second largest
	task.secondMostFrequent(text2)
	#    Test C
	#    ---------
	#    "xx" = 2
	#    "yy" = 2
	#    "zz" = 2
	#    -------------------------------
	#    No second most frequent element
	task.secondMostFrequent(text3)

if __name__ == "__main__": main()

Output

 Given Text :  xyoxoxx
 Second most frequent element is :  o
 Given Text :  abaacdccec
 Second most frequent element is :  a
 Given Text :  xxyyzz
 Second most frequent element not exist
#    Ruby program for
#    Find the second most frequent character in a given string
class Occurrence 
	def secondMostFrequent(text) 
		n = text.length
		if (n == 0) 
			return
		end

		first = 0
		second = -1
		record = Array.new(256) {0}
		i = 0
		#  Count frequency of character element
		while (i < n) 
			record[text[i].ord] = record[text[i].ord] + 1
			i += 1
		end

		i = 0
		#  Find second most occurring character
		while (i < n) 
			if (record[text[i].ord] > record[text[first].ord]) 
				second = first
				first = i
			elsif (record[text[i].ord] < record[text[first].ord]) 
				if (second == -1 || 
                    record[text[second].ord] < record[text[i].ord]) 
					second = i
				end

			end

			i += 1
		end

		print("\n Given Text : ", text)
		if (second == -1) 
			print("\n Second most frequent element not exist")
		else
 
			print("\n Second most frequent element is : ", text[second])
		end

	end

end

def main() 
	task = Occurrence.new()
	#  Given Sting text
	text1 = "xyoxoxx"
	text2 = "abaacdccec"
	text3 = "xxyyzz"
	#    Test A
	#    ---------
	#    text   = "xyoxoxx"
	#    "xxxx" = 4
	#    "y"    = 1
	#    "oo"   = 2
	#    -------------------------------
	#    o = 2 which is second largest
	task.secondMostFrequent(text1)
	#    Test B 
	#    ---------
	#    text   = "abaacdccec"
	#    "aaa"  = 3
	#    "b"    = 1
	#    "cccc" = 4
	#    "d"    = 1
	#    -------------------------------
	#    a = 3 which is second largest
	task.secondMostFrequent(text2)
	#    Test C
	#    ---------
	#    "xx" = 2
	#    "yy" = 2
	#    "zz" = 2
	#    -------------------------------
	#    No second most frequent element
	task.secondMostFrequent(text3)
end

main()

Output

 Given Text : xyoxoxx
 Second most frequent element is : o
 Given Text : abaacdccec
 Second most frequent element is : a
 Given Text : xxyyzz
 Second most frequent element not exist
import scala.collection.mutable._;
/*
    Scala program for
    Find the second most frequent character in a given string
*/
class Occurrence()
{
	def secondMostFrequent(text: String): Unit = {
		var n: Int = text.length();
		if (n == 0)
		{
			return;
		}
		var first: Int = 0;
		var second: Int = -1;
		var record: Array[Int] = Array.fill[Int](256)(0);
		var i: Int = 0;
		// Count frequency of character element
		while (i < n)
		{
			record(text.charAt(i)) = record(text.charAt(i)) + 1;
			i += 1;
		}
		i = 0;
		// Find second most occurring character
		while (i < n)
		{
			if (record(text.charAt(i)) > record(text.charAt(first)))
			{
				second = first;
				first = i;
			}
			else if (record(text.charAt(i)) < record(text.charAt(first)))
			{
				if (second == -1 || 
                    record(text.charAt(second)) < record(text.charAt(i)))
				{
					second = i;
				}
			}
			i += 1;
		}
		print("\n Given Text : " + text);
		if (second == -1)
		{
			print("\n Second most frequent element not exist");
		}
		else
		{
			print("\n Second most frequent element is : " + text.charAt(second));
		}
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: Occurrence = new Occurrence();
		// Given Sting text
		var text1: String = "xyoxoxx";
		var text2: String = "abaacdccec";
		var text3: String = "xxyyzz";
		/*
		    Test A
		    ---------
		    text   = "xyoxoxx"
		    "xxxx" = 4
		    "y"    = 1
		    "oo"   = 2
		    -------------------------------
		    o = 2 which is second largest
		*/
		task.secondMostFrequent(text1);
		/*
		    Test B 
		    ---------
		    text   = "abaacdccec"
		    "aaa"  = 3
		    "b"    = 1
		    "cccc" = 4
		    "d"    = 1
		    -------------------------------
		    a = 3 which is second largest
		*/
		task.secondMostFrequent(text2);
		/*
		    Test C
		    ---------
		    "xx" = 2
		    "yy" = 2
		    "zz" = 2
		    -------------------------------
		    No second most frequent element
		*/
		task.secondMostFrequent(text3);
	}
}

Output

 Given Text : xyoxoxx
 Second most frequent element is : o
 Given Text : abaacdccec
 Second most frequent element is : a
 Given Text : xxyyzz
 Second most frequent element not exist
import Foundation;
/*
    Swift 4 program for
    Find the second most frequent character in a given string
*/
class Occurrence
{
	func secondMostFrequent(_ data: String)
	{
		let text = Array(data);
		let n: Int = text.count;
		if (n == 0)
		{
			return;
		}
		var first: Int = 0;
		var second: Int = -1;
		var record: [Int] = Array(repeating: 0, count: 256);
		var i: Int = 0;
		var index: Int = 0;
		// Count frequency of character element
		while (i < n)
		{
			index = Int(UnicodeScalar(String(text[i]))!.value);
			record[index] += 1;
			i += 1;
		}
		i = 0;
		// Find second most occurring character
		while (i < n)
		{
			index = Int(UnicodeScalar(String(text[i]))!.value);
			let f = Int(UnicodeScalar(String(text[first]))!.value);
	
			if (record[index] > record[f])
			{
				second = first;
				first = i;
			}
			else if (record[index] < record[f])
			{
				if (second == -1 ||
                    record[Int(UnicodeScalar(String(text[second]))!.value)] 
                    < record[index])
				{
					second = i;
				}
			}
			i += 1;
		}
		print("\n Given Text : ", data, terminator: "");
		if (second == -1)
		{
			print("\n Second most frequent element not exist", 
                  terminator: "");
		}
		else
		{
			print("\n Second most frequent element is : ", 
                  text[second], terminator: "");
		}
	}
}
func main()
{
	let task: Occurrence = Occurrence();
	// Given Sting text
	let text1: String = "xyoxoxx";
	let text2: String = "abaacdccec";
	let text3: String = "xxyyzz";
	/*
	    Test A
	    ---------
	    text   = "xyoxoxx"
	    "xxxx" = 4
	    "y"    = 1
	    "oo"   = 2
	    -------------------------------
	    o = 2 which is second largest
	*/
	task.secondMostFrequent(text1);
	/*
	    Test B 
	    ---------
	    text   = "abaacdccec"
	    "aaa"  = 3
	    "b"    = 1
	    "cccc" = 4
	    "d"    = 1
	    -------------------------------
	    a = 3 which is second largest
	*/
	task.secondMostFrequent(text2);
	/*
	    Test C
	    ---------
	    "xx" = 2
	    "yy" = 2
	    "zz" = 2
	    -------------------------------
	    No second most frequent element
	*/
	task.secondMostFrequent(text3);
}
main();

Output

 Given Text :  xyoxoxx
 Second most frequent element is :  o
 Given Text :  abaacdccec
 Second most frequent element is :  a
 Given Text :  xxyyzz
 Second most frequent element not exist
/*
    Kotlin program for
    Find the second most frequent character in a given string
*/
class Occurrence
{
	fun secondMostFrequent(text: String): Unit
	{
		val n: Int = text.length;
		if (n == 0)
		{
			return;
		}
		var first: Int = 0;
		var second: Int = -1;
		val record: Array < Int > = Array(256)
		{
			0
		};
		var i: Int = 0;
		// Count frequency of character element
		while (i < n)
		{
			record[text.get(i).toInt()] = record[text.get(i).toInt()] + 1;
			i += 1;
		}
		i = 0;
		// Find second most occurring character
		while (i < n)
		{
			if (record[text.get(i).toInt()] > record[text.get(first).toInt()])
			{
				second = first;
				first = i;
			}
			else if (record[text.get(i).toInt()] < 
                     record[text.get(first).toInt()])
			{
				if (second == -1 || 
                    record[text.get(second).toInt()] <
                    record[text.get(i).toInt()])
				{
					second = i;
				}
			}
			i += 1;
		}
		print("\n Given Text : " + text);
		if (second == -1)
		{
			print("\n Second most frequent element not exist");
		}
		else
		{
			print("\n Second most frequent element is : " + text.get(second));
		}
	}
}
fun main(args: Array < String > ): Unit
{
	val task: Occurrence = Occurrence();
	// Given Sting text
	val text1: String = "xyoxoxx";
	val text2: String = "abaacdccec";
	val text3: String = "xxyyzz";
	/*
	    Test A
	    ---------
	    text   = "xyoxoxx"
	    "xxxx" = 4
	    "y"    = 1
	    "oo"   = 2
	    -------------------------------
	    o = 2 which is second largest
	*/
	task.secondMostFrequent(text1);
	/*
	    Test B 
	    ---------
	    text   = "abaacdccec"
	    "aaa"  = 3
	    "b"    = 1
	    "cccc" = 4
	    "d"    = 1
	    -------------------------------
	    a = 3 which is second largest
	*/
	task.secondMostFrequent(text2);
	/*
	    Test C
	    ---------
	    "xx" = 2
	    "yy" = 2
	    "zz" = 2
	    -------------------------------
	    No second most frequent element
	*/
	task.secondMostFrequent(text3);
}

Output

 Given Text : xyoxoxx
 Second most frequent element is : o
 Given Text : abaacdccec
 Second most frequent element is : a
 Given Text : xxyyzz
 Second most frequent element not exist

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