[LeetCode] 220. Contains Duplicate III

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

 Thought process:
Iterate through the array. Use a binary search tree (TreeSet in Java) to keep track of the most recent k numbers. For each number, get the ceiling (smallest number that's greater than num) and floor (greatest number that's less than num) of it, and check if either's difference with num is at most t. 
If the tree's size exceeds k, remove the least recent number.
Note: there may be integer overflow when subtracting a negative number from a positive number.

 Solution 1:
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public class Solution {
    public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
        TreeSet<Integer> set = new TreeSet<>();
        
        for (int i = 0; i < nums.length; i++) {
            Integer ceiling = set.ceiling(nums[i]);
            Integer floor = set.floor(nums[i]);
            if ((ceiling != null && (long) ceiling - (long) nums[i] <= t) || (floor != null && (long) nums[i] - (long) floor <= t)) {
                return true;
            }
            set.add(nums[i]);
            if (set.size() > k) {
                set.remove(nums[i - k]);
            }
        }
        
        return false;
    }
}

Time complexity:
for loop takes O(n). Adding and removing numbers takes O(logk). The overall time complexity is O(nlogk).

 There is also a O(n) solution:
  1. Divide numbers into buckets of size t + 1. For example, numbers between 0 and t are in bucket 0, and numbers between t + 1 and 2t + 1 are in bucket 1.
  2. Iterate through the array. Use a hash map to store bucket -> number mappings. Maintain a window of size k, i.e. there are at most k mappings in the map.
  3. This gives me three cases as I iterate:
    1. Map contains the same bucket: return true, because I know the numbers satisfy both requirements.
    2. Neighboring buckets contain a number whose difference with the current number <= t: return true.
    3. If neither of 1 and 2 happens, put the mapping of current bucket -> number into the map. If current index is larger than k, evict the mapping of the bucket of array[index - k]. It is safe to remove this mapping because its index will never be within k difference with a future number.
Note: there may be three cases of integer overflow:
  1. Bucket size: if t = Integer.MAX_VALUE.
  2. Neighboring buckets: if t = 0, then size of bucket = 1. If there is Integer.MAX_VALUE in the array, its neighbor bucket will overflow.
  3. Difference between numbers.

Solution 2:
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public class Solution {
    public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
        if (t < 0) {
            return false;
        }
        
        long size = t + 1;
        Map<Long, Long> map = new HashMap<>();
        
        for (int i = 0; i < nums.length; i++) {
            long bucket = getBucket(nums[i], size);
            if (map.containsKey(bucket)) {
                return true;
            }
            if (map.containsKey(bucket - 1) && (long) nums[i] - map.get(bucket - 1) <= t) {
                return true;
            }
            if (map.containsKey(bucket + 1) && map.get(bucket + 1) - (long) nums[i] <= t) {
                return true;
            }
            map.put(bucket, (long) nums[i]);
            if (i >= k) {
                map.remove(getBucket(nums[i - k], size));
            }
        }
        
        return false;
    }
    
    private long getBucket(int number, long size) {
        return number < 0 ? number / size - 1 : number / size;
    }
}

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

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