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Problem solution · C++

Minimum Difference in Sums After Removal of Elements

Minimum Difference in Sums After Removal of Elements: a C++ solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Sliding window or two pointers
Source
walkccc LeetCode Solutions
Length
36 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Sliding window or two pointers

For Minimum Difference in Sums After Removal of Elements, the implementation maintains a moving interval and updates only the information that enters or leaves the window.

  1. Choose the invariant that makes a window valid or useful.
  2. Advance the right boundary and add the new element.
  3. Move the left boundary only as needed while maintaining the invariant and updating the answer.

Code notes

  • 36 lines of C++ from the credited upstream file 2163.cpp.
  • The implementation visibly relies on sequence storage, work queue.
  • 2 loop blocks detected.

Complexity

Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMinimum Difference in Sums After Removal of Elements · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  long long minimumDifference(vector<int>& nums) {    const int n = nums.size() / 3;    long ans = LONG_MAX;    long leftSum = 0;    long rightSum = 0;    // The left part should be as small as possible.    priority_queue<int> maxHeap;    // The right part should be as big as possible.    priority_queue<int, vector<int>, greater<>> minHeap;    // minLeftSum[i] := the minimum of the sum of n nums in nums[0..i)    vector<long> minLeftSum(nums.size());     for (int i = 0; i < 2 * n; ++i) {      maxHeap.push(nums[i]);      leftSum += nums[i];      if (maxHeap.size() == n + 1)        leftSum -= maxHeap.top(), maxHeap.pop();      if (maxHeap.size() == n)        minLeftSum[i] = leftSum;    }     for (int i = nums.size() - 1; i >= n; --i) {      minHeap.push(nums[i]);      rightSum += nums[i];      if (minHeap.size() == n + 1)        rightSum -= minHeap.top(), minHeap.pop();      if (minHeap.size() == n)        ans = min(ans, minLeftSum[i - 1] - rightSum);    }     return ans;  }};

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