LeetCode – Contains Duplicate II (Java)

Category: Algorithms >> Interview May 29, 2014

Given an array of integers and an integer k, return true if and only if there are two distinct indices i and j in the array such that nums[i] = nums[j] and the difference between i and j is at most k.

Java Solution 1 - HashMap

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

Java Solution 2 - HashSet

public boolean containsNearbyDuplicate(int[] nums, int k) {
    if(nums==null || nums.length<2 || k==0)
        return false;
 
    int i=0; 
 
    HashSet<Integer> set = new HashSet<Integer>();
 
    for(int j=0; j<nums.length; j++){
        if(!set.add(nums[j])){
            return true;
        }            
 
        if(set.size()>=k+1){
            set.remove(nums[i++]);
        }
    }
 
    return false;
}

If you want someone to read your code, please put the code inside <pre><code> and </code></pre> tags. For example:

<pre><code> 
String foo = "bar";
</code></pre>

Alik Elzin

A bit simpler solution – just one pointer `i`:
boolean containsDuplicates(int[] nums, int k) { if (nums == null || nums.length <= 1 || k <= 0) return false;
HashSet exists = new HashSet(k); // TODO: take into control load factor. for (int i = 0 ; i k) { exists.remove(nums[i-k]); } }
return false; }
Alik Elzin

Without the HashSet, you’ll traverse k times for each element, in order to compare to the previous k, meaning O(k*n).
With the HashSet, it’s O(n).
Sumedha Prathipati

Hi ,
We can do this without the set , right?!
ricky

I think if you use two pointers from front and end. then you can do this in O(n).
Vasyl Grygoryev

You are right. My code for this solution:
public boolean containsNearbyDuplicate(int[] nums, int k) { HashSet set = new HashSet();
for (int i = 0; i = k) set.remove(nums[i - k]); }
return false; }
peng li

We can maintain a queue and hashset which only maintain current k elements. But this takes O(k) space.
Cole Thatcher

Is there a way to do this O(n) time and O(1) space though? I’m just thinking how would we go about proving that there isn’t?

LeetCode – Contains Duplicate II (Java)

Related posts: