2163. Minimum Difference in Sums After Removal of Elements

Description

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 <= 105
• 1 <= nums[i] <= 105

Solutions

Solution 1: Priority Queue (Max and Min Heap) + Prefix and Suffix Sum + Enumeration of Split Points

The problem is essentially equivalent to finding a split point in $nums$, dividing the array into two parts. In the first part, select the smallest $n$ elements, and in the second part, select the largest $n$ elements, so that the difference between the sums of the two parts is minimized.

We can use a max heap to maintain the smallest $n$ elements in the prefix, and a min heap to maintain the largest $n$ elements in the suffix. We define $pre[i]$ as the sum of the smallest $n$ elements among the first $i$ elements of the array $nums$, and $suf[i]$ as the sum of the largest $n$ elements from the $i$-th element to the last element of the array. During the process of maintaining the max and min heaps, update the values of $pre[i]$ and $suf[i]$.

Finally, we enumerate the split points in the range of $i \in [n, 2n]$, calculate the value of $pre[i] - suf[i + 1]$, and take the minimum value.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.

Python3JavaC++GoTypeScript

class Solution:
    def minimumDifference(self, nums: List[int]) -> int:
        m = len(nums)
        n = m // 3

        s = 0
        pre = [0] * (m + 1)
        q1 = []
        for i, x in enumerate(nums[: n * 2], 1):
            s += x
            heappush(q1, -x)
            if len(q1) > n:
                s -= -heappop(q1)
            pre[i] = s

        s = 0
        suf = [0] * (m + 1)
        q2 = []
        for i in range(m, n, -1):
            x = nums[i - 1]
            s += x
            heappush(q2, x)
            if len(q2) > n:
                s -= heappop(q2)
            suf[i] = s

        return min(pre[i] - suf[i + 1] for i in range(n, n * 2 + 1))

class Solution {
    public long minimumDifference(int[] nums) {
        int m = nums.length;
        int n = m / 3;
        long s = 0;
        long[] pre = new long[m + 1];
        PriorityQueue<Integer> pq = new PriorityQueue<>((a, b) -> b - a);
        for (int i = 1; i <= n * 2; ++i) {
            int x = nums[i - 1];
            s += x;
            pq.offer(x);
            if (pq.size() > n) {
                s -= pq.poll();
            }
            pre[i] = s;
        }
        s = 0;
        long[] suf = new long[m + 1];
        pq = new PriorityQueue<>();
        for (int i = m; i > n; --i) {
            int x = nums[i - 1];
            s += x;
            pq.offer(x);
            if (pq.size() > n) {
                s -= pq.poll();
            }
            suf[i] = s;
        }
        long ans = 1L << 60;
        for (int i = n; i <= n * 2; ++i) {
            ans = Math.min(ans, pre[i] - suf[i + 1]);
        }
        return ans;
    }
}

class Solution {
public:
    long long minimumDifference(vector<int>& nums) {
        int m = nums.size();
        int n = m / 3;

        using ll = long long;
        ll s = 0;
        ll pre[m + 1];
        priority_queue<int> q1;
        for (int i = 1; i <= n * 2; ++i) {
            int x = nums[i - 1];
            s += x;
            q1.push(x);
            if (q1.size() > n) {
                s -= q1.top();
                q1.pop();
            }
            pre[i] = s;
        }
        s = 0;
        ll suf[m + 1];
        priority_queue<int, vector<int>, greater<int>> q2;
        for (int i = m; i > n; --i) {
            int x = nums[i - 1];
            s += x;
            q2.push(x);
            if (q2.size() > n) {
                s -= q2.top();
                q2.pop();
            }
            suf[i] = s;
        }
        ll ans = 1e18;
        for (int i = n; i <= n * 2; ++i) {
            ans = min(ans, pre[i] - suf[i + 1]);
        }
        return ans;
    }
};

func minimumDifference(nums []int) int64 {
    m := len(nums)
    n := m / 3
    s := 0
    pre := make([]int, m+1)
    q1 := hp{}
    for i := 1; i <= n*2; i++ {
        x := nums[i-1]
        s += x
        heap.Push(&q1, -x)
        if q1.Len() > n {
            s -= -heap.Pop(&q1).(int)
        }
        pre[i] = s
    }
    s = 0
    suf := make([]int, m+1)
    q2 := hp{}
    for i := m; i > n; i-- {
        x := nums[i-1]
        s += x
        heap.Push(&q2, x)
        if q2.Len() > n {
            s -= heap.Pop(&q2).(int)
        }
        suf[i] = s
    }
    ans := int64(1e18)
    for i := n; i <= n*2; i++ {
        ans = min(ans, int64(pre[i]-suf[i+1]))
    }
    return ans
}

type hp struct{ sort.IntSlice }

func (h hp) Less(i, j int) bool { return h.IntSlice[i] < h.IntSlice[j] }
func (h *hp) Push(v any)        { h.IntSlice = append(h.IntSlice, v.(int)) }
func (h *hp) Pop() any {
    a := h.IntSlice
    v := a[len(a)-1]
    h.IntSlice = a[:len(a)-1]
    return v
}

function minimumDifference(nums: number[]): number {
    const m = nums.length;
    const n = Math.floor(m / 3);
    let s = 0;
    const pre: number[] = Array(m + 1);
    const q1 = new MaxPriorityQueue();
    for (let i = 1; i <= n * 2; ++i) {
        const x = nums[i - 1];
        s += x;
        q1.enqueue(x, x);
        if (q1.size() > n) {
            s -= q1.dequeue().element;
        }
        pre[i] = s;
    }
    s = 0;
    const suf: number[] = Array(m + 1);
    const q2 = new MinPriorityQueue();
    for (let i = m; i > n; --i) {
        const x = nums[i - 1];
        s += x;
        q2.enqueue(x, x);
        if (q2.size() > n) {
            s -= q2.dequeue().element;
        }
        suf[i] = s;
    }
    let ans = Number.MAX_SAFE_INTEGER;
    for (let i = n; i <= n * 2; ++i) {
        ans = Math.min(ans, pre[i] - suf[i + 1]);
    }
    return ans;
}

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