Medium
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
import kotlin.math.abs
class Solution {
fun lexicographicallySmallestArray(nums: IntArray, limit: Int): IntArray {
val n = nums.size
val nodes = Array(n) { i -> Node(i, nums[i]) }
nodes.sortWith { a: Node, b: Node ->
Integer.signum(
a.value - b.value,
)
}
var group = 1
nodes[0].group = group
for (i in 1 until n) {
if (abs(nodes[i].value - nodes[i - 1].value) <= limit) {
nodes[i].group = group
} else {
nodes[i].group = ++group
}
}
val groupBase = IntArray(group + 1)
for (i in n - 1 downTo 0) {
groupBase[nodes[i].group] = i
}
val groupIndex = IntArray(n)
for (node in nodes) {
groupIndex[node.id] = node.group
}
val ans = IntArray(n)
for (i in 0 until n) {
val index = groupBase[groupIndex[i]]
ans[i] = nodes[index].value
groupBase[groupIndex[i]]++
}
return ans
}
private class Node(var id: Int, var value: Int) {
var group: Int = 0
}
}