LeetCode-Problems/Hard Problems/2163_Minimum_Difference_in_Sums_After_Removal_of_Elements.cpp at main · UdaySharmaGitHub/LeetCode-Problems · GitHub

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
/*
2163. 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
*/
class Solution {
public:
    long long minimumDifference(vector<int>& nums) {
        int n = nums.size();
        int k = n / 3;

        vector<long long> leftMins(n), rightMaxs(n);
        priority_queue<int> maxLeftHeap; // max-heap for smallest k
        priority_queue<int, vector<int>, greater<int>> minRightHeap; // min-heap for largest k

        long long leftSum = 0, rightSum = 0, minDiff = LLONG_MAX;

        // Build leftMins
        for (int i = 0; i < k; ++i) {
            maxLeftHeap.push(nums[i]);
            leftSum += nums[i];
        }
        leftMins[k - 1] = leftSum;

        for (int i = k; i < n - k; ++i) {
            if (!maxLeftHeap.empty() && nums[i] < maxLeftHeap.top()) {
                leftSum += nums[i] - maxLeftHeap.top();
                maxLeftHeap.pop();
                maxLeftHeap.push(nums[i]);
            }
            leftMins[i] = leftSum;
        }

        // Build rightMaxs
        for (int i = n - 1; i >= n - k; --i) {
            minRightHeap.push(nums[i]);
            rightSum += nums[i];
        }
        rightMaxs[n - k] = rightSum;

        for (int i = n - k - 1; i >= k - 1; --i) {
            if (!minRightHeap.empty() && nums[i] > minRightHeap.top()) {
                rightSum += nums[i] - minRightHeap.top();
                minRightHeap.pop();
                minRightHeap.push(nums[i]);
            }
            rightMaxs[i] = rightSum;
        }

        // Compute min difference
        for (int i = k - 1; i < n - k; ++i) {
            minDiff = min(minDiff, leftMins[i] - rightMaxs[i + 1]);
        }

        return minDiff;
    }
};