leetcode - 2948. Make Lexicographically Smallest Array by Swapping Elements

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Description

You are given a 0-indexed array of positive integers nums and a positive integer limit.

In one operation, you can choose any two indices i and j and swap nums[i] and nums[j] if |nums[i] - nums[j]| <= limit.

Return the lexicographically smallest array that can be obtained by performing the operation any number of times.

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

Example 1:

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.

Example 2:

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.

Example 3:

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.

Constraints:

1 <= nums.length <= 10^5
1 <= nums[i] <= 10^9
1 <= limit <= 10^9

Solution

If a could swap with b, b could swap with c, then a could swap with c. So use a sliding window to union all sets, and then start with the smallest number, find its final location.

Time complexity: $o (n)$
Space complexity: $o (n)$

Code

class UnionFindSet:def __init__(self, nums: list) -> None:self.parent = [i for i in nums]self.height = [1] * len(nums)def find(self, x: int) -> int:if self.parent[x] != x:self.parent[x] = self.find(self.parent[x])return self.parent[x]def union(self, x: int, y: int) -> None:x, y = self.find(x), self.find(y)if x == y:return if self.height[x] < self.height[y]:self.parent[x] = yelse:self.parent[y] = xif self.height[x] == self.height[y]:self.height[x] += 1def same(self, x: int, y: int) -> bool:return self.find(x) == self.find(y)class Solution:def lexicographicallySmallestArray(self, nums: List[int], limit: int) -> List[int]:sorted_num = sorted([(index, item) for index, item in enumerate(nums)], key=lambda x: x[1])sorted_num_val = sorted(nums)union_find = UnionFindSet(list(range(len(nums))))# sliding window to union all setsleft = 0for right in range(len(sorted_num)):while sorted_num[right][1] - sorted_num[left][1] > limit:left += 1union_find.union(sorted_num[right][0], sorted_num[left][0])swap_map = {}# key: parent index, value: children indexs (indexs that could be swapped)for i in range(len(nums)):parent_i = union_find.find(i)if parent_i not in swap_map:swap_map[parent_i] = []heapq.heappush(swap_map[parent_i], i)# start from the smallest numberres = [0] * len(nums)for i in range(len(sorted_num)):index_in_num = sorted_num[i][0]candidates = swap_map[union_find.find(index_in_num)]target_index = heapq.heappop(candidates)res[target_index] = sorted_num[i][1]return res