2163. Minimum Difference in Sums After Removal of Elements

Problem Link

Description

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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

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Python3

class Solution:
    def minimumDifference(self, nums: List[int]) -> int:
        M = len(nums)
        N = M // 3
 
        prefixMin, currMin = [sum(nums[:N])], sum(nums[:N])
        pq = [-x for x in nums[:N]] # maxHeap
        heapify(pq)
 
        for i in range(N, 2 * N):
            x = -heappop(pq)
            currMin -= x
            nxt = min(x, nums[i])
            currMin += nxt
            heappush(pq, -nxt)
            prefixMin.append(currMin)
        
        suffixMax, currMax = [sum(nums[2 * N:])], sum(nums[2 * N:])
        pq = [x for x in nums[2 * N:]] # minHeap
        heapify(pq)
 
        for i in range(2 * N - 1, N - 1, -1):
            x = heappop(pq)
            currMax -= x
            nxt = max(x, nums[i])
            currMax += nxt
            heappush(pq, nxt)
            suffixMax.append(currMax)
        
        suffixMax.reverse()
 
        res = inf
        for a, b in zip(prefixMin, suffixMax):
            res = min(res, a - b)
 
        return res