LeetCode in Kotlin
3357. Minimize the Maximum Adjacent Element Difference
Hard
You are given an array of integers nums. Some values in nums are missing and are denoted by -1.
You can choose a pair of positive integers (x, y) exactly once and replace each missing element with either x or y.
You need to minimize the maximum absolute difference between adjacent elements of nums after replacements.
Return the minimum possible difference.
Example 1:
Input: nums = [1,2,-1,10,8]
Output: 4
Explanation:
By choosing the pair as (6, 7), nums can be changed to [1, 2, 6, 10, 8].
The absolute differences between adjacent elements are:
|1 - 2| == 1|2 - 6| == 4|6 - 10| == 4|10 - 8| == 2
Example 2:
Input: nums = [-1,-1,-1]
Output: 0
Explanation:
By choosing the pair as (4, 4), nums can be changed to [4, 4, 4].
Example 3:
Input: nums = [-1,10,-1,8]
Output: 1
Explanation:
By choosing the pair as (11, 9), nums can be changed to [11, 10, 9, 8].
Constraints:
2 <= nums.length <= 105nums[i]is either -1 or in the range[1, 109].
Solution
import kotlin.math.abs
import kotlin.math.max
import kotlin.math.min
class Solution {
fun minDifference(nums: IntArray): Int {
val n = nums.size
var maxAdj = 0
var mina = Int.Companion.MAX_VALUE
var maxb = Int.Companion.MIN_VALUE
for (i in 0..<n - 1) {
val a = nums[i]
val b = nums[i + 1]
if (a > 0 && b > 0) {
maxAdj = max(maxAdj, abs((a - b)))
} else if (a > 0 || b > 0) {
mina = min(mina, max(a, b))
maxb = max(maxb, max(a, b))
}
}
var res = 0
for (i in 0..<n) {
if ((i > 0 && nums[i - 1] == -1) || nums[i] > 0) {
continue
}
var j = i
while (j < n && nums[j] == -1) {
j++
}
var a = Int.Companion.MAX_VALUE
var b = Int.Companion.MIN_VALUE
if (i > 0) {
a = min(a, nums[i - 1])
b = max(b, nums[i - 1])
}
if (j < n) {
a = min(a, nums[j])
b = max(b, nums[j])
}
if (a <= b) {
if (j - i == 1) {
res = max(res, min((maxb - a), (b - mina)))
} else {
res = max(
res,
min(
maxb - a,
min(b - mina, (maxb - mina + 2) / 3 * 2),
),
)
}
}
}
return max(maxAdj, (res + 1) / 2)
}
}