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: Priority Queue

• Time complexity: O(nlogn)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {number[]} nums
 * @return {number}
 */
const minimumDifference = function (nums) {
  const n = nums.length / 3;
  const leftHeap = new MaxPriorityQueue();
  const rightHeap = new MinPriorityQueue();
  const minLeftSum = Array.from({ length: n * 3 }, () => null);
  let leftSum = 0;
  let rightSum = 0;
  let result = Number.MAX_SAFE_INTEGER;

  for (let index = 0; index < 2 * n; index++) {
    const num = nums[index];

    leftSum += num;
    leftHeap.enqueue(num);

    if (leftHeap.size() === n + 1) {
      const maxNum = leftHeap.dequeue();

      leftSum -= maxNum;
    }

    if (leftHeap.size() === n) {
      minLeftSum[index] = leftSum;
    }
  }

  for (let index = n * 3 - 1; index >= n; index--) {
    const num = nums[index];

    rightSum += num;
    rightHeap.enqueue(num);

    if (rightHeap.size() === n + 1) {
      const minNum = rightHeap.dequeue();

      rightSum -= minNum;
    }

    if (rightHeap.size() === n) {
      const difference = minLeftSum[index - 1] - rightSum;

      result = Math.min(difference, result);
    }
  }

  return result;
};