2163 - Minimum Difference in Sums After Removal of Elements (Hard)

Problem Link

https://leetcode.com/problems/minimum-difference-in-sums-after-removal-of-elements/

Problem Statement

You are given a 0-indexed integer array nums consisting of 3 * n elements.

You are allowed to remove any subsequence of elements of size exactly n from nums. The remaining 2 * n elements will be divided into two equal parts:

• The first n elements belonging to the first part and their sum is sumfirst.
• The next n elements belonging to the second part and their sum is sumsecond.

The difference in sums of the two parts is denoted as sumfirst - sumsecond.

• For example, if sumfirst = 3 and sumsecond = 2, their difference is 1.
• Similarly, if sumfirst = 2 and sumsecond = 3, their difference is -1.

Return the minimum difference possible between the sums of the two parts after the removal of n elements.

Example 1:

Input: nums = [3,1,2]
Output: -1
Explanation: Here, nums has 3 elements, so n = 1. 
Thus we have to remove 1 element from nums and divide the array into two equal parts.
- If we remove nums[0] = 3, the array will be [1,2]. The difference in sums of the two parts will be 1 - 2 = -1.
- If we remove nums[1] = 1, the array will be [3,2]. The difference in sums of the two parts will be 3 - 2 = 1.
- If we remove nums[2] = 2, the array will be [3,1]. The difference in sums of the two parts will be 3 - 1 = 2.
The minimum difference between sums of the two parts is min(-1,1,2) = -1. 

Example 2:

Input: nums = [7,9,5,8,1,3]
Output: 1
Explanation: Here n = 2. So we must remove 2 elements and divide the remaining array into two parts containing two elements each.
If we remove nums[2] = 5 and nums[3] = 8, the resultant array will be [7,9,1,3]. The difference in sums will be (7+9) - (1+3) = 12.
To obtain the minimum difference, we should remove nums[1] = 9 and nums[4] = 1. The resultant array becomes [7,5,8,3]. The difference in sums of the two parts is (7+5) - (8+3) = 1.
It can be shown that it is not possible to obtain a difference smaller than 1.

Constraints:

• nums.length == 3 * n
• 1 <= n <= 10^5
• 1 <= nums[i] <= 10^5

Approach 1: Two Heaps

Given an array of $3 * n$ elements, we need to remove a subsequence of $n$ elements. At the end, we will have $2 * n$ elements. The answer is the $sum_{first} - sum_{second}$. Therefore, we need to make $sum_{first}$as small as possible and $sum_{second}$ as large as possible.

We use two heaps $p$ and $s$ to record the smallest $n$ elements and the largest $n$ elements and $pre$to record the sum of $p$ and $suf$ to record that of $s$.

Now we handle the middle $n$ elements. From left to right, we check if the element $nums[i]$ is smaller than the top element $q$ from $p$. If so, we should take this element instead and ditch the top one. We update $pre' = pre + nums[i] - q$. At the same time we keep the prefix sum $pv[i]$ in$[n, 2 * n)$. Similarly, we do the same thing from right to left to build the suffix sum $sv[i]$ in the same range.

At the end, we can find out the minimum difference by checking $pv[i - 1] - sv[i]$ between the middle $n$ range.

Written by @wingkwong

class Solution {
public:
    long long minimumDifference(vector<int>& nums) {
        int n = (int) nums.size() / 3;
        long long ans = 1e18;
        priority_queue<long long> p;
        priority_queue<long long, vector<long long>, greater<long long>> s;
        vector<long long> pv(3 * n, 1e18), sv(3 * n, -1e18);
        long long pre = 0, suf = 0;
        for (int i = 0; i < n; i++) {
            p.push(nums[i]); s.push(nums[3 * n - 1 - i]);
            pre += nums[i], suf += nums[3 * n - 1 - i];
        }
        pv[n - 1] = pre, sv[2 * n] = suf;
        for (int i = n; i < 2 * n; i++) {
            long long q = p.top();
            if (nums[i] < q) {
                p.pop();
                pre += nums[i] - q;
                p.push(nums[i]);
            }
            pv[i] = pre;
        }
        for (int i = 2 * n - 1; i >= n; i--) {
            long long q = s.top();
            if (nums[i] > q) {
                s.pop();
                suf += nums[i] - q;
                s.push(nums[i]);
            }
            sv[i] = suf;
        }
        for (int i = n; i <= 2 * n; i++) ans = min(ans, pv[i - 1] - sv[i]);
        return ans;
    }
};