Make Lexicographically Smallest Array by

lcpy gen -n 2948   # by problem number
lcpy gen -s make_lexicographically_smallest_array_by_swapping_elements   # by problem name

Input: nums = [1,5,3,9,8], limit = 2
Output: [1,3,5,8,9]
Explanation: 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.

Input: nums = [1,7,6,18,2,1], limit = 3
Output: [1,6,7,18,1,2]
Explanation: 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.

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

from itertools import pairwise


class Solution:
    # Time: O(n log n)
    # Space: O(n)
    def lexicographically_smallest_array(self, nums: list[int], limit: int) -> list[int]:
        order = sorted(range(len(nums)), key=lambda i: nums[i])
        result: list[int] = [0] * len(nums)
        group: list[int] = [order[0]]
        for prev, idx in pairwise(order):
            if nums[idx] - nums[prev] > limit:
                self._assign_group(result, group, nums)
                group = []
            group.append(idx)
        self._assign_group(result, group, nums)
        return result

    def _assign_group(self, result: list[int], indices: list[int], nums: list[int]) -> None:
        values = sorted(nums[i] for i in indices)
        for pos, val in zip(sorted(indices), values, strict=True):
            result[pos] = val