In this HackerRank Java Dequeue problem in the java programming language, you are given N integers. You need to find the maximum number of unique integers among all the possible contiguous subarrays of size M.

HackerRank Java Dequeue problem solution


HackerRank Java Dequeue problem solution.

import java.util.*;

public class test {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        final Deque<Integer> deque = new ArrayDeque<Integer>();
        final Map<Integer, Integer> map = new HashMap<Integer, Integer>();
        final int n = in.nextInt();
        final int m = in.nextInt();

        int res = 0;
        for (int i = 0; i < n; i++) {
            final int num = in.nextInt();
            deque.addLast(num);
            if (map.containsKey(num)) {
                map.put(num, map.get(num).intValue() + 1);
            } else {
                map.put(num, 1);
            }

            if (deque.size() == m + 1) {
                final int key = deque.removeFirst();
                final int v = map.get(key);
                if (v == 1) {
                    map.remove(key);
                } else {
                    map.put(key, v - 1);
                }
            }

            final int cnt = map.size();
            if (cnt > res) { res = cnt; }
        }
        System.out.println(res);
    }
}


Second solution

    import java.util.*;
    public class test {
        public static void main(String[] args) {
            Scanner in = new Scanner(System.in);
            Deque<Integer> deque = new ArrayDeque<Integer>();
            HashMap<Integer, Integer> freqs = new HashMap<Integer, Integer>();
            int n = in.nextInt();
            int m = in.nextInt();
            int ans = 0, countDistinct = 0;
            
            for (int i = 0; i < n; i++) {
                int num = in.nextInt();
                deque.addLast(num);
                if (freqs.get(num) == null) freqs.put(num,0);
                
                freqs.put(num,freqs.get(num)+1);
                if (freqs.get(num)==1) countDistinct++;
                
                if (deque.size()==m+1){
                    int rem = deque.removeFirst();
                    freqs.put(rem,freqs.get(rem)-1);
                    if (freqs.get(rem) == 0) countDistinct--;
                }
                if (deque.size()==m){
                    if (countDistinct > ans) ans = countDistinct;
                }
               
            }
            System.out.println(ans);
        }
    }


A solution in java8 programming.

    import java.util.*;

public class test {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        Deque<Integer> deque = new ArrayDeque<>();
        HashSet<Integer> set = new HashSet<>();
        
        int n = in.nextInt();
        int m = in.nextInt();
        int max = Integer.MIN_VALUE;

        for (int i = 0; i < n; i++) {
            int input = in.nextInt();
            
            deque.add(input);
            set.add(input);
            
            if (deque.size() == m) {
                if (set.size() > max) max = set.size();
                int first = deque.remove();
                if (!deque.contains(first)) set.remove(first);
            }
        }
        
        System.out.println(max);
    }
}