Sep 02, 2020

# Day 2

When fame was acquired briefly.

## The Question

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


Constraints:

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

## The Process

Note a few ideas that will help you understand this problem and make the optimal time complexity approach easier for you:

• Consider the edge cases. What are the edge cases? What happens when the array is empty, singleton etc. And what about when k is 0? Consider edge cases that you know are unlikely to be included in a general solution, and take care of those cases right away.
• Then, you understand that for any element x in nums, you’re actually looking for its partner. This partner element is at most k hops away from x, and doesn’t differ from x by more than t.
• So, if you happen to have k elements in your hand, and look at a (k + 1)th element, you ask “does this (k + 1)th guy find a partner in the k I’m already holding on to?” If not, well guess what? The oldest element in the k ones you have can no more find its partner so you can get rid of it and let (k + 1)th guy get in.
• For an example, let’s consider the first one given i.e. nums = [1,2,3,1] | k = 3 | t = 0.
• Now, you’re holding on to the first k elements i.e. [1,2,3]. Your supposed loop considers the next element i.e. [..., 1]. Now if this 1 has a partner in [1,2,3], great, we’re done. But if not, then there’s no problem in throwing the first [1...], because well, all the elements you see now are going to be more than k hops away from him. So you end up with [2,3,1].
• That was the basic processing idea.
• Finally, you understand that upon processing each element x, if the largest number smaller than x or the smallest number greater than x, in the k elements you already have, produce something bounded by t, you have an answer.
• Now it’s your turn to research what data structure can hold onto elements such that it can yeild a ceiling/floor relative to some number efficiently (clearly it will have to keep its elements in a sorted manner…)
• (P.S. think a bit about why ceiling/floor and not just max/min)

## The Code

    public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
int len = nums.length;

// return false when given parameters makes two distinct indices impossible
if (len <= 1 || k <= 0) {
return false;
}

/**
* Important, question definition:
*   Absolute diff between nums[i] and nums[j] <= t
*   Absolute diff between indices i and j <= k
**/

TreeSet<Long> set = new TreeSet();
for (int i = 0; i < nums.length; i++) {
Long number = new Long(nums[i]);
Long ceil = set.ceiling(number);
Long floor = set.floor(number);

if (ceil != null && ceil - number <= t) {
return true;
}

if (floor != null && number - floor <= t) {
return true;
}

}