2948. Make Lexicographically Smallest Array by Swapping Elements

Problem Link

Description

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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 <= 105
• 1 <= nums[i] <= 109
• 1 <= limit <= 109

Solution

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Python3

class Solution:
    def lexicographicallySmallestArray(self, nums: List[int], limit: int) -> List[int]:
        N = len(nums)
        A = sorted(range(N), key = lambda i : nums[i])
        i = 0
        res = [None] * N
        
        while i < N:
            curr = [A[i]]
            j = i + 1
            
            while j < N and nums[A[j]] - nums[A[j - 1]] <= limit:
                curr.append(A[j])
                j += 1
 
            indices = sorted(curr)
            V = sorted(curr, key = lambda i: nums[i])
            for index, v in zip(indices, V):
                res[index] = nums[v]
            
            i = j
        
        return res