hi

[drowsy dove]

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[thorn kernel]

import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;

public class frequency{
    
    public static void main(String [] args){
        
        Scanner scan = new Scanner(System.in);
        
        int n = scan.nextInt();
        int arr[] = new int[n];
        int count = 0;
        int freq = 0;
        int max = 0;
        
        for(int i=0; i<n; i++)
            arr[i] = scan.nextInt();
        
        
        for(int i=0; i < (n / 2) +1; i++){
            
            for(int j=i ; j<n ; j++){
                
                if( arr[i] == arr[j])
                    count++;
                
            }
            
            if(count > freq){
                max = arr[i];
                freq = count;
            }
            
            if(count == freq){
                
                if(arr[i] < max)
                    max = arr[i];
                
            }
            
            count = 0 ;
        }
        
        System.out.print(max);
    }
    
}```

how do i solve this
i've been racking my brain over it for atleast half an hour now but can't seem to get it

here's the question

[untold mango]

well ...... ur using an n^2 (n-squared) algorythm
and ur probably recounting the same bird ID's over and over

i mean if u have
1,3,1,1,3,1,2 then for i=0 u count the frequency of id=1 and at i=2 u count the same id again and at i=3 u count the same id AGAIN

....also, if the highest seen ID only appears in the 2nd half of the array u wont ever count it so thats ....... not good, ur lucky the tests all passed

u can count them in O(n) (all of them) instead of O(n^2)

then u just need to find the highest frequency