Duplicate within K Distance in an Array

Given an integer array arr[] and an integer k, determine whether there exist two indices i and j such that arr[i] == arr[j] and |i - j| ≤ k. If such a pair exists, return 'Yes', otherwise return 'No'.

Examples:

Input: k = 3, arr[] = [1, 2, 3, 4, 1, 2, 3, 4]
Output: No
Explanation: Each element in the given array arr[] appears twice and the distance between every element and its duplicate is 4.

Input: k = 3, arr[] = [1, 2, 3, 1, 4, 5]
Output: Yes
Explanation: 1 is present at index 0 and 3.

Input: k = 3, arr[] = [1, 2, 3, 4, 5]
Output: No
Explanation: There is no duplicate element in arr[].

[Naive Approach] - O(n * k) Time and O(1) Space

The idea is to run two loops. The outer loop picks every index i as a starting index, and the inner loop compares all elements which are within k distance of i, i.e. i + k.

Below is given the implementation:

Yes

Time Complexity: O(n * k), for each element of the array arr[], we are iterating up to next k elements.
Auxiliary Space: O(1)

[Expected Approach] - Using HashSet - O(n) Time and O(k) Space

The idea is to use HashSet to store elements of the array arr[] and check if there is any duplicate present within a k distance. Also remove elements that are present at more than k distance from the current element. Following is a detailed algorithm.

1. Create an empty HashSet. 
2. Traverse all elements from left to right. Let the current element be 'arr[i]' 
   • If the current element 'arr[i]' is present in a HashSet, then return true. 
   • Else add arr[i] to hash and remove arr[i-k] from hash if i >= k

Below is given the implementation:

Yes

Time Complexity: O(n), as we are iterating through elements only once.
Auxiliary Space: O(k), to store the k elements in HashSet.