<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 DSU:
    def __init__(self, size) -> None:
        self.parent = [i for i in range(size)]
        self.rank = [1]*size
        
    def find(self, node):
        if self.parent[node] != node:
            self.parent[node] = self.find(self.parent[node])
            
        return self.parent[node]
    
    def union(self, nodeA , nodeB):
        x = self.find(nodeA)
        y = self.find(nodeB)
        
        if x == y:
            return
        
        if self.rank[x] > self.rank[y]:
            self.parent[y] = x
            self.rank[x]+=self.rank[y]
        else:
            self.parent[x] = y
            self.rank[y]+=self.rank[x]

class Solution:
    def lexicographicallySmallestArray(self, nums: List[int], limit: int) -> List[int]:
        n = len(nums)

        pairs = []
        res = [0]*len(nums)
        dsu = DSU(n)
        for i in range(0,len(nums)):
            pairs.append((nums[i] , i))
        pairs = sorted(pairs)

        for i in range(0, len(nums)-1):
            if abs(pairs[i][0] - pairs[i+1][0]) <= limit:
                dsu.union(pairs[i][1],  pairs[i+1][1])

        mp = collections.defaultdict(list)

        for i in range(0,len(nums)):
            j = dsu.find(i)
            mp[j]+=[(nums[i], i)]

        for k , v in mp.items():
            a = [val for val,_ in v]
            b = [idx for _, idx in v]
            a = sorted(a)
            b = sorted(b)

            for i , num in zip(b, a):
                res[i] = num

        return(res)
        


   


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