Question

2163. Minimum Difference in Sums After Removal of Elements

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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

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;
}