LeetCode in Kotlin
2035. Partition Array Into Two Arrays to Minimize Sum Difference
Hard
You are given an integer array nums of 2 * n integers. You need to partition nums into two arrays of length n to minimize the absolute difference of the sums of the arrays. To partition nums, put each element of nums into one of the two arrays.
Return the minimum possible absolute difference.
Example 1:
Input: nums = [3,9,7,3]
Output: 2
Explanation: One optimal partition is: [3,9] and [7,3]. The absolute difference between the sums of the arrays is abs((3 + 9) - (7 + 3)) = 2.
Example 2:
Input: nums = [-36,36]
Output: 72
Explanation: One optimal partition is: [-36] and [36]. The absolute difference between the sums of the arrays is abs((-36) - (36)) = 72.
Example 3:
Input: nums = [2,-1,0,4,-2,-9]
Output: 0
Explanation: One optimal partition is: [2,4,-9] and [-1,0,-2]. The absolute difference between the sums of the arrays is abs((2 + 4 + -9) - (-1 + 0 + -2)) = 0.
Constraints:
1 <= n <= 15nums.length == 2 * n-107 <= nums[i] <= 107
Solution
class Solution {
fun minimumDifference(nums: IntArray): Int {
if (nums.isEmpty()) {
return -1
}
val n = nums.size / 2
var sum = 0
val arr1: MutableList<MutableList<Int>> = ArrayList()
val arr2: MutableList<MutableList<Int>> = ArrayList()
for (i in 0..n) {
arr1.add(ArrayList())
arr2.add(ArrayList())
if (i < n) {
sum += nums[i]
sum += nums[i + n]
}
}
for (state in 0 until (1 shl n)) {
var sum1 = 0
var sum2 = 0
for (i in 0 until n) {
if (state and (1 shl i) == 0) {
continue
}
val a1 = nums[i]
val a2 = nums[i + n]
sum1 += a1
sum2 += a2
}
val numOfEleInSet = Integer.bitCount(state)
arr1[numOfEleInSet].add(sum1)
arr2[numOfEleInSet].add(sum2)
}
for (i in 0..n) {
arr2[i].sort()
}
var min = Int.MAX_VALUE
for (i in 0..n) {
val sums1: List<Int> = arr1[i]
val sums2: List<Int> = arr2[n - i]
for (s1 in sums1) {
var idx = sums2.binarySearch(sum / 2 - s1)
if (idx < 0) {
idx = -(idx + 1)
}
if (idx < sums1.size) {
min = Math.min(
min,
Math.abs(sum - s1 - sums2[idx] - (sums2[idx] + s1))
)
}
if (idx - 1 >= 0) {
min = Math.min(
min,
Math.abs(
sum - s1 - sums2[idx - 1] -
(sums2[idx - 1] + s1)
)
)
}
}
}
return min
}
}