删除元素后和的最小差值

/*
 * Problem: Given an integer array nums containing 3 * n elements,
 * you can delete exactly n elements, leaving 2 * n elements.
 * The remaining elements can be divided into two equal parts.
 * Find the minimum absolute difference between the sums of these two parts.
 */
import java.util.*;
import java.util.stream.IntStream;

public class MinDifferenceStream {
    public static int minDifference(int[] nums) {
        int n = nums.length / 3;
        int[] prefixSum = new int[3 * n + 1];
        IntStream.range(0, 3 * n).forEach(i -> prefixSum[i + 1] = prefixSum[i] + nums[i]);

        PriorityQueue<Integer> maxHeap = new PriorityQueue<>(Collections.reverseOrder());
        int[] maxFirst = new int[2 * n + 1];
        IntStream.range(0, 2 * n).forEach(i -> {
            maxHeap.add(nums[i]);
            if (maxHeap.size() > n) maxHeap.poll();
            maxFirst[i + 1] = maxFirst[i] + maxHeap.peek();
        });

        PriorityQueue<Integer> minHeap = new PriorityQueue<>();
        int[] minSecond = new int[2 * n + 1];
        IntStream.range(0, 2 * n).map(i -> 2 * n - i - 1).forEach(i -> {
            minHeap.add(nums[3 * n - 1 - i]);
            if (minHeap.size() > n) minHeap.poll();
            minSecond[i + 1] = minSecond[i] + minHeap.peek();
        });

        return IntStream.range(0, n + 1).map(i -> maxFirst[n + i] - minSecond[i]).min().orElse(Integer.MAX_VALUE);
    }

    public static void main(String[] args) {
        int[] nums = {7, 9, 5, 8, 1, 3};
        System.out.println(minDifference(nums)); // Output: 1
    }
}