package com.interview.array;

import java.util.Comparator;
import java.util.HashMap;
import java.util.Map;
import java.util.PriorityQueue;

/**
 * http://www.careercup.com/question?id=6026101998485504
 * This answer two questions.
 * Group elements in size m such that in every group only unique elements are possible.
 * It also answers question where rearrange array such that same elements are exactly m
 * distance from each other
 */
class Pair{
    int num;
    int count;
    Pair(int num,int count){
        this.count = count;
        this.num = num;
    }
}

class Comparators implements Comparator<Pair>{

    @Override
    public int compare(Pair o1, Pair o2) {
        if(o1.count <= o2.count){
            return 1;
        }else{
            return -1;
        }
    }
    
}

public class GroupElementsInSizeM {

    public boolean group(int input[],int m){
        Map<Integer,Integer> count = new HashMap<Integer,Integer>();
        for(Integer i : input){
            int c = 1;
            if(count.containsKey(i)){
                c = count.get(i);
                c++;
            }
            count.put(i, c);
        }
        
        PriorityQueue<Pair> maxHeap = new PriorityQueue<Pair>(count.size(),new Comparators());
        for(Integer s : count.keySet()){
            int c = count.get(s);
            //if any count is greater than len/m then this arrangement is not possible
            if(c > Math.ceil(input.length*1.0/m)){
                return false;
            }
            maxHeap.offer(new Pair(s,c));
        }
        int current = 0;
        int start = current;
        while(maxHeap.size() > 0){
            Pair p = maxHeap.poll();
            int i =0;
            while(i < p.count){
                input[start] = p.num;
                start = start + m;
                if(start >= input.length){
                    current++;
                    start = current;
                }
                i++;
            }
        }
        return true;
    }
    
    public static void main(String args[]){
        int input[] = {2,1,5,1,3,5,3,3,4};
        int input1[] = {1,2,3,8,8,8,7,8};
        GroupElementsInSizeM gps = new GroupElementsInSizeM();
        boolean r = gps.group(input1, 3);
        System.out.println(r);
        for(int i=0; i < input1.length; i++){
            System.out.println(input1[i]);
        }
    }
}