LeetCode-in-Java
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
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
public class Solution {
public int minimumDifference(int[] nums) {
if (nums == null || nums.length == 0) {
return -1;
}
int n = nums.length / 2;
int sum = 0;
List<List<Integer>> arr1 = new ArrayList<>();
List<List<Integer>> arr2 = new ArrayList<>();
for (int i = 0; i <= n; i++) {
arr1.add(new ArrayList<>());
arr2.add(new ArrayList<>());
if (i < n) {
sum += nums[i];
sum += nums[i + n];
}
}
for (int state = 0; state < (1 << n); state++) {
int sum1 = 0;
int sum2 = 0;
for (int i = 0; i < n; i++) {
if ((state & (1 << i)) == 0) {
continue;
}
int a1 = nums[i];
int a2 = nums[i + n];
sum1 += a1;
sum2 += a2;
}
int numOfEleInSet = Integer.bitCount(state);
arr1.get(numOfEleInSet).add(sum1);
arr2.get(numOfEleInSet).add(sum2);
}
for (int i = 0; i <= n; i++) {
Collections.sort(arr2.get(i));
}
int min = Integer.MAX_VALUE;
for (int i = 0; i <= n; i++) {
List<Integer> sums1 = arr1.get(i);
List<Integer> sums2 = arr2.get(n - i);
for (int s1 : sums1) {
int idx = Collections.binarySearch(sums2, sum / 2 - s1);
if (idx < 0) {
idx = -(idx + 1);
}
if (idx < sums1.size()) {
min =
Math.min(
min,
Math.abs((sum - s1 - sums2.get(idx)) - (sums2.get(idx) + s1)));
}
if (idx - 1 >= 0) {
min =
Math.min(
min,
Math.abs(
(sum - s1 - sums2.get(idx - 1))
- (sums2.get(idx - 1) + s1)));
}
}
}
return min;
}
}