# 220.Contains-Duplicate-III

## 220. Contains Duplicate III

## 题目地址

[https://leetcode.com/problems/contains-duplicate-iii/](https://leetcode.com/problems/contains-duplicate-iii/)

## 题目描述

```
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.

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
```

## 代码

### Approach #1 TreeSet - ceiling & floor

```java
class Solution {
  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;
  }
}
```

### Approach #2 Buckets

```java
class Solution {
  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);
      if (d.containsKey(m))        return true;
      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;
      }
      d.put(m, (long)nums[i]);
    }
    return false;
  }

   private long getID(long x, long w) {
      return x < 0 ? (x + 1) / w - 1 : x / w;
   }

}
```
