<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:
    # O (n log n): reference https://leetcode.com/problems/make-lexicographically-smallest-array-by-swapping-elements/
    def lexicographicallySmallestArray(self, nums: List[int], limit: int) -> List[int]:
        # sort the array
        n = len(nums)
        arr = sorted([(nums[i], i) for i in range(n)])
        loops = defaultdict(list)  # store the indices
        loop_num = 0
        result = [0] * n
        for i in range(n): # go through the sorted list                
            if i>0 and arr[i][0] - arr[i-1][0] > limit:
                loop_num += 1
            loops[loop_num].append(arr[i][1]) # store the indices
        for loop_num in loops.keys():
            index = loops[loop_num]
            sorted_index = sorted(index)
            for j in range(len(index)):
                result[sorted_index[j]] = nums[index[j]]
        return result

    # O(n^2) solution
    def on2_lexicographicallySmallestArray(self, nums: List[int], limit: int) -> List[int]:
        ans = nums.copy()
        while True:
            # basically find the best in each swap 
            for i in range(len(nums)):
                for j in range(i+1, len(nums)):
                    if nums[i] > nums[j] and nums[i]-nums[j]<=limit:
                        # print(nums, nums[i], nums[j], i, j)
                        nums[i], nums[j] = nums[j], nums[i]
                        if self.lex_smaller(nums, ans):
                            ans = nums.copy()
                        # swap back to original
                        nums[i], nums[j] = nums[j], nums[i]
            if ans == nums:
                break
            nums = ans.copy()
        return ans
    
    def lex_smaller(self, nums, ans):
        for i in range(len(nums)):
            if nums[i] < ans[i]:
                return True
        return False
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