LeetCode in Kotlin
1330. Reverse Subarray To Maximize Array Value
Hard
You are given an integer array nums. The value of this array is defined as the sum of |nums[i] - nums[i + 1]| for all 0 <= i < nums.length - 1.
You are allowed to select any subarray of the given array and reverse it. You can perform this operation only once.
Find maximum possible value of the final array.
Example 1:
Input: nums = [2,3,1,5,4]
Output: 10
Explanation: By reversing the subarray [3,1,5] the array becomes [2,5,1,3,4] whose value is 10.
Example 2:
Input: nums = [2,4,9,24,2,1,10]
Output: 68
Constraints:
1 <= nums.length <= 3 * 104-105 <= nums[i] <= 105
Solution
class Solution {
private fun getAbsoluteDifference(a: Int, b: Int): Int {
return Math.abs(a - b)
}
fun maxValueAfterReverse(nums: IntArray): Int {
val n = nums.size
var result = 0
for (i in 0 until n - 1) {
result += getAbsoluteDifference(nums[i], nums[i + 1])
}
var minLine = Int.MIN_VALUE
var maxLine = Int.MAX_VALUE
for (i in 0 until n - 1) {
minLine = Math.max(minLine, Math.min(nums[i], nums[i + 1]))
maxLine = Math.min(maxLine, Math.max(nums[i], nums[i + 1]))
}
var diff = Math.max(0, (minLine - maxLine) * 2)
for (i in 1 until n - 1) {
diff = Math.max(
diff,
getAbsoluteDifference(nums[0], nums[i + 1]) -
getAbsoluteDifference(nums[i], nums[i + 1]),
)
}
for (i in 0 until n - 1) {
diff = Math.max(
diff,
getAbsoluteDifference(nums[n - 1], nums[i]) -
getAbsoluteDifference(nums[i + 1], nums[i]),
)
}
return result + diff
}
}