Given an array of integers and an integer k, find out whether there are two distinct indices i and j in the array such that nums[i] = nums[j] and the absolute difference between i and j is at most k.

My original solution using a HashMap:

class Solution {
    public boolean containsNearbyDuplicate(int[] nums, int k) {
        Map<Integer, Integer> map = new HashMap<>();
        for(int i=0;i<nums.length;i++){
            if(map.containsKey(nums[i])){
                if(i-map.get(nums[i])>k) map.put(nums[i],i);
                else return true;
            }else map.put(nums[i],i);
        }
        return false;
    }
}

However, in the discuss, sbdy suggest that using set has better performance.

Source

public boolean containsNearbyDuplicate(int[] nums, int k) {
        Set<Integer> set = new HashSet<Integer>();
        for(int i = 0; i < nums.length; i++){
            if(i > k) set.remove(nums[i-k-1]);
            if(!set.add(nums[i])) return true;
            // cannot add into the set, duplicates found
        }
        return false;
        // no duplicates found
 }

Explanation:

It iterates over the array using a sliding window.
The front of the window is at i, the rear of the window is k steps back.
The elements within that window are maintained using a set.
While adding new element to the set, if add() returns false, it means the element already exists in the set.
At that point, we return true.

set.add() returns a boolean type -> slight performance improvement avoid using set.contains()

So is set preferred over a map? Not actually.. Maybe its the Map.containsKey() method that costs more time.