Contains Duplicate III

Given an array of integers, find out whether there are two distinct indicesiandjin the array such that theabsolutedifference betweennums[i]andnums[j]is at mosttand theabsolutedifference betweeniandjis at mostk.

Example 1:

Input: nums = [1,2,3,1], k = 3, t = 0
Output: true

Example 2:

Input: nums = [1,0,1,1], k = 1, t = 2
Output: true

Example 3:

Input: nums = [1,5,9,1,5,9], k = 2, t = 3
Output: false

Solution

Solution that would take a lot of time.

public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
    for (int i = 0; i < nums.length; ++i) {
        for (int j = Math.max(i - k, 0); j < i; ++j) {
            if (Math.abs(nums[i] - nums[j]) <= t) return true;
        }
    }
    return false;
}

More optimized solution using binary search tree.

public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
    TreeSet<Integer> set = new TreeSet<>();
    for (int i = 0; i < nums.length; ++i) {
        // Find the successor of current element
        Integer s = set.ceiling(nums[i]);
        if (s != null && s <= nums[i] + t) return true;

        // Find the predecessor of current element
        Integer g = set.floor(nums[i]);
        if (g != null && nums[i] <= g + t) return true;

        set.add(nums[i]);
        if (set.size() > k) {
            set.remove(nums[i - k]);
        }
    }
    return false;
}

Solution using buckets.

public class Solution {
    // Get the ID of the bucket from element value x and bucket width w
    // In Java, `-3 / 5 = 0` and but we need `-3 / 5 = -1`.
    private long getID(long x, long w) {
        return x < 0 ? (x + 1) / w - 1 : x / w;
    }

    public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
        if (t < 0) return false;
        Map<Long, Long> d = new HashMap<>();
        long w = (long)t + 1;
        for (int i = 0; i < nums.length; ++i) {
            long m = getID(nums[i], w);
            // check if bucket m is empty, each bucket may contain at most one element
            if (d.containsKey(m))
                return true;
            // check the neighbor buckets for almost duplicate
            if (d.containsKey(m - 1) && Math.abs(nums[i] - d.get(m - 1)) < w)
                return true;
            if (d.containsKey(m + 1) && Math.abs(nums[i] - d.get(m + 1)) < w)
                return true;
            // now bucket m is empty and no almost duplicate in neighbor buckets
            d.put(m, (long)nums[i]);
            if (i >= k) d.remove(getID(nums[i - k], w));
        }
        return false;
    }
}