<p>You are given a <strong>0-indexed</strong> array of <strong>positive</strong> integers <code>nums</code> and a <strong>positive</strong> integer <code>limit</code>.</p>

<p>In one operation, you can choose any two indices <code>i</code> and <code>j</code> and swap <code>nums[i]</code> and <code>nums[j]</code> <strong>if</strong> <code>|nums[i] - nums[j]| &lt;= limit</code>.</p>

<p>Return <em>the <strong>lexicographically smallest array</strong> that can be obtained by performing the operation any number of times</em>.</p>

<p>An array <code>a</code> is lexicographically smaller than an array <code>b</code> if in the first position where <code>a</code> and <code>b</code> differ, array <code>a</code> has an element that is less than the corresponding element in <code>b</code>. For example, the array <code>[2,10,3]</code> is lexicographically smaller than the array <code>[10,2,3]</code> because they differ at index <code>0</code> and <code>2 &lt; 10</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,5,3,9,8], limit = 2
<strong>Output:</strong> [1,3,5,8,9]
<strong>Explanation:</strong> Apply the operation 2 times:
- Swap nums[1] with nums[2]. The array becomes [1,3,5,9,8]
- Swap nums[3] with nums[4]. The array becomes [1,3,5,8,9]
We cannot obtain a lexicographically smaller array by applying any more operations.
Note that it may be possible to get the same result by doing different operations.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,7,6,18,2,1], limit = 3
<strong>Output:</strong> [1,6,7,18,1,2]
<strong>Explanation:</strong> Apply the operation 3 times:
- Swap nums[1] with nums[2]. The array becomes [1,6,7,18,2,1]
- Swap nums[0] with nums[4]. The array becomes [2,6,7,18,1,1]
- Swap nums[0] with nums[5]. The array becomes [1,6,7,18,1,2]
We cannot obtain a lexicographically smaller array by applying any more operations.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,7,28,19,10], limit = 3
<strong>Output:</strong> [1,7,28,19,10]
<strong>Explanation:</strong> [1,7,28,19,10] is the lexicographically smallest array we can obtain because we cannot apply the operation on any two indices.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= limit &lt;= 10<sup>9</sup></code></li>
</ul>

class Node:
    def __init__(self,index,val):
        self.index = index
        self.val = val

class Neighborhood:
    def __init__(self,node,limit):
        self.limit = limit
        self.upper = node.val + limit
        self.group = [node]
        

    def join(self,node):
        if node.val <= self.upper:
            self.group.append(node)
            self.upper = max(self.upper, node.val + self.limit)
            return True
        return False

class Solution:
    def lexicographicallySmallestArray(self, nums: List[int], limit: int) -> List[int]:
        # union-find w/ index being node_num and val being value
        # sort the connections as best as possible
        node_map = []
        for i in range(len(nums)):
            node_map.append(Node(i,nums[i]))
        # sort indices to make construction of neighborhoods linear
        nums_index = [(nums[i],i) for i in range(len(nums))]
        nums_index.sort()

        res = [0 for i in range(len(nums))]
        first_node = node_map[nums_index[0][1]]
        neighborhoods = [Neighborhood(first_node,limit)]

        for i in range(1,len(nums_index)):
            curr_node = node_map[nums_index[i][1]]
            if not neighborhoods[-1].join(curr_node):
                neighborhoods.append(Neighborhood(curr_node,limit))

        for nbh in neighborhoods:
            local_nodes = nbh.group
            local_nodes.sort(key=lambda x : x.val)
            idx_hist = [node.index for node in local_nodes]
            idx_hist.sort()
            for i in range(len(idx_hist)):
                res[idx_hist[i]] = local_nodes[i].val

        return res
        
        
63712