java - How to check if elements in an array are sorted according to number of occurences? -

i want check if elements in array sorted according number of occurrences in. assuming grouped if elements same. int [] y = {7, 7, 7, 8, 8, 5, 5, 5, 5} should return false while int [] y = {5, 5, 5, 5, 7, 7, 7, 8, 8} should return true. there 4 5s, 3 7s , 2 8s, should arranged way.

this have far:

import java.io.*; import java.util.*;   public class testing2 {    public static void main(string[] args){     int [] y = {7, 7, 7, 8, 8, 5, 5, 5, 5};     system.out.println(issorted(y));  // should return false    }    public static boolean issorted(int[] y){      hashmap<integer,integer> hash = new hashmap<integer,integer>();      for(int i=0; i<y.length; i++){       if(hash.containskey(y[i])){         hash.put(y[i], hash.get(y[i]) + 1);       } else {         hash.put(y[i], 1);       }     }     system.out.println(hash);     return true;  // code not complete yet   } } 

it print:

{5=4, 7=3, 8=2} true 

so have number of occurrences each element. do next?

you can solve in single pass of array; here hints:

while iterating through array, keep track of current number (e.g. 5, 7 or 8) , quantity. when finish next group of numbers, compare previous number/quantity. then, return true or false appropriately.