# 0220. Contains Duplicate III

<https://leetcode.com/problems/contains-duplicate-iii>

## Description

Given an integer array `nums` and two integers `k` and `t`, return `true` if there are **two distinct indices** `i` and `j` in the array such that `abs(nums[i] - nums[j]) <= t` and `abs(i - j) <= k`.

**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
```

**Constraints:**

* `0 <= nums.length <= 2 * 104`
* `-231 <= nums[i] <= 231 - 1`
* `0 <= k <= 104`
* `0 <= t <= 231 - 1`

## ac1: TreeSet

`TreeSet` is Black-Red Tree, self balance tree, sorted.

<https://leetcode.com/problems/contains-duplicate-iii/discuss/61655/Java-O(N-lg-K)-solution>

```java
aclass Solution {
    public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
        TreeSet<Long> set = new TreeSet<Long>();

        for (int i = 0; i < nums.length; i++) {
            long val = (long) nums[i];

            Long smaller = set.floor(val);
            if (smaller != null && val - smaller <= t) return true;

            Long larger = set.ceiling(val);
            if (larger != null && larger - val <= t) return true;

            set.add(val);

            if (set.size() > k) set.remove((long)nums[i-k]);
        }

        return false;
    }
}
// O(nlogk)
```

## ac2: Bucket

<https://leetcode.com/problems/contains-duplicate-iii/discuss/61645/AC-O(N)-solution-in-Java-using-buckets-with-explanation>

```java
class Solution {
    public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
        if (t < 0) return false;

        Map<Long, Long> buckets = new HashMap<Long, Long>();
        long size = (long)t + 1;

        for (int i = 0; i < nums.length; i++) {
            long ival = (long) nums[i];
            long label = getId(ival, size);

            if (buckets.containsKey(label)) return true;

            if (buckets.containsKey(label - 1) && Math.abs(buckets.get(label-1) - ival) <= t) return true;

            if (buckets.containsKey(label + 1) && Math.abs(buckets.get(label+1) - ival) <= t) return true;

            buckets.put(label, ival);

            if (i >= k) {
                buckets.remove(getId((long)nums[i-k], size));
            }

        }

        return false;
    }

    private Long getId(long num, long size) {
        return num < 0 ?  num / size - 1: num / size;
        // in Java, -3 / 5 == 0, but we need -3 / 5 == -1, so we minus 1.
    }
}

// O(n)
// Why "Math.abs(buckets.get(label-1) - ival) <= t)"? Because 1 bucket can only has one element, if it has more than 1, "buckets.containsKey(label)" will return true earlier.
```
