<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 Solution:
    def lexicographicallySmallestArray(self, nums: List[int], limit: int) -> List[int]:
        """
        identify all of such group of values where difference between consecutive values is less than or equal to limit.

        For example, in the input array nums = [3, 6, 8, 10, 2, 1] and limit = 1, we can identify 3, 2, 1 as one such group.

If we sort these values while keeping them in their identified group, we get the ideal position of the values in that group.
        """
        # create a list of array items with their indexes
        item_ix_list = [(nums[i], i) for i in range(len(nums))]
        # sort the list based on item value
        # having the list sorted guarantees us to find chunks in order, which satisfies the limit
        item_ix_sorted = sorted(item_ix_list, key=lambda x: x[0])

        # previous value to compare with
        prev = item_ix_sorted[0][0]
        # result arr
        res = [0] * len(nums)
        # indices list to keep track of chunk of indices
        chunk_indices = []
        # first item of chunk
        first = 0

        for num, ix in item_ix_sorted:
            if num - prev > limit:
                # if the condition is not satisfied, we are at a new chunk
                # loop over sorted chunk indices, and add the corresponding
                # items to the result arr on this index
                for i in sorted(chunk_indices):
                    res[i] = item_ix_sorted[first][0]
                    # increment value of first pointer
                    first += 1
                
                # reset chunk indices for next chunk
                chunk_indices = []
            
            # meets the condition add the items to chunk
            chunk_indices.append(ix)
            prev = num
        
        # run the loop again for the last chunk
        for i in sorted(chunk_indices):
            res[i] = item_ix_sorted[first][0]
            # increment value of first pointer
            first += 1
        
        return res


        






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