[LeetCode] 219. Contains Duplicate II

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.

 Thought process:
Iterate through the array. Use a hash map to record number -> index. If duplicate numbers are found, check if the difference between their indices is <= k. If not, update the mapped index to the larger one.

 Solution 1:
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public 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]) && i - map.get(nums[i]) <= k) {
                return true;
            }
            map.put(nums[i], i);
        }
        
        return false;
    }
}

Time complexity: O(n).

 A cleaner way to solve this problem is to use a set and maintain a window of size k, i. e. if the set's size exceeds k, remove the earliest item.

 Solution 2:
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public class Solution {
    public boolean containsNearbyDuplicate(int[] nums, int k) {
        Set<Integer> set = new HashSet<>();
        
        for (int i = 0; i < nums.length; i++) {
            if (set.contains(nums[i])) {
                return true;
            }
            set.add(nums[i]);
            if (set.size() > k) {
                set.remove(nums[i - k]);
            }
        }
        
        return false;
    }
}

Time complexity: O(n).
Location: San Jose, CA, USA

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