Today I have done leetcode's July LeetCoding Challenge with cpp.

July LeetCoding Challenge 18

Description

Minimum Difference in Sums After Removal of Elements

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

Solution

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

    vector<long long> result(n + 1);
    priority_queue<int> pqF(nums.begin(), nums.begin() + n);
    long long sum = accumulate(nums.begin(), nums.begin() + n, 0LL);
    for(int i = 0; i <= n; ++i) {
      result[i] = sum;
      if(pqF.top() <= nums[n + i]) continue;
      sum -= pqF.top();
      sum += nums[n + i];
      pqF.pop();
      pqF.push(nums[n + i]);
    }

    priority_queue<int, vector<int>, greater<int>> pqS(nums.begin() + 2 * n, nums.end());
    sum = accumulate(nums.begin() + 2 * n, nums.end(), 0LL);
    for(int i = n; i >= 0; --i) {
      result[i] -= sum;
      if(pqS.top() >= nums[n + i - 1]) continue;
      sum -= pqS.top();
      sum += nums[n + i - 1];
      pqS.pop();
      pqS.push(nums[n + i - 1]);
    }

    return *min_element(result.begin(), result.end());
  }
};

// Accepted
// 110/110 cases passed (70 ms)
// Your runtime beats 99.23 % of cpp submissions
// Your memory usage beats 99.48 % of cpp submissions (131.9 MB)