Let A1n be an array with n numbers An element x in A1n is sa

Let A[1..n] be an array with n numbers. An element x in A[1..n] is said to be cool if it appears more than n 2 times in A[1..n]. For instance, 5 is cool in [5, 1, 5, 2, 5, 2, 5], but in [1, 4, 2, 6, 5, 4, 4, 4], no element is cool. Design a O(n) time algorithm that decides whether or not there is a cool element in A[1..n]. If there is no cool element, the algorithm must return “No”. If there is a cool element, the algorithm must return the value of the cool element. Explain why your algorithm is correct and why it does take O(n) time.

Solution

package com.app.test;

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class CheggAlgorithm {

int[ ] numbers = new int[n];
   public CheggAlgorithm() {

   }

   public void coolElement(int coolElementArray[])
   {
   int count=0,coolelement,first,curr_count=0,freq_num=0;
   for(int j=0;j<coolElementArray.length;j++){
       first = coolElementArray[j];
       for(int k=1;k<coolElementArray.length;k++){
           if(first==coolElementArray[k] && first!=freq_num){
           curr_count++;
       }
           if(count<curr_count)
           {
               count=curr_count;
               curr_count =1;
               freq_num = first;
           }
   }
   }
   if(count>coolElementArray.length/2){
       System.out.println(\"The Cool Element is \"+freq_num);
   }
   else
   {
       System.out.println(\"There is no any Cool Element\");
   }
   }
   public static void main(String[] args) {
       CheggAlgorithm algo = new CheggAlgorithm();
       Scanner scan = new Scanner(System.in);
       System.out.println(\"Enter the values do yo want to enter the array? \");
int n = scan.nextInt();
System.out.println(n);
for(int loopindex=0;loopindex<n;loopindex++)
{
   algo.numbers[loopindex] = scan.nextInt();
}
algo.coolElement(algo.numbers);
   }

}