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Problem solution · C++

Minimum Partition Score II

Minimum Partition Score II: a C++ solution using binary search. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Binary search
Source
Kamyu LeetCode Solutions
Length
67 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

Binary search

For Minimum Partition Score II, the implementation exploits a monotonic condition to discard half of the remaining search space after every check.

  1. Identify the ordered answer range or sorted search domain.
  2. Write a predicate whose truth changes only once.
  3. Move the appropriate boundary after each midpoint check and return the final feasible position.

Code notes

  • 67 lines of C++ from the credited upstream file minimum-partition-score-ii.cpp.
  • The implementation visibly relies on sequence storage, work queue.
  • 6 loop blocks detected.

Complexity

Multiply the logarithmic number of midpoint checks by the cost of one predicate evaluation.

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

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMinimum Partition Score II · C++C++
Use this to learn the idea, then write your own version.
// Time:  O(n * log(n * r)) = O(nlogn + nlogr), r = max(nums)// Space: O(n) // prefix sum, dp, convex hull trick, wqs binary search, alien trickclass Solution {public:    long long minPartitionScore(vector<int>& nums, int k) {        const auto& binary_search = [](auto left, auto right, const auto& check) {            while (left <= right) {                const auto& mid = left + (right - left) / 2;                if (check(mid)) {                    right = mid - 1;                } else {                    left = mid + 1;                }            }            return left;        };         const auto& check = [](const auto& l1, const auto& l2, const auto& l3) {            return (get<1>(l2) - get<1>(l1)) * (get<0>(l2) - get<0>(l3)) < (get<1>(l3) - get<1>(l2)) * (get<0>(l1) - get<0>(l2));        };         vector<int64_t> prefix(size(nums) + 1);        const auto& max_lambda = [&]() {            int64_t mx = 0;            const auto& total = prefix.back() * (prefix.back() + 1) / 2;            for (int i = 1; i < size(nums); ++i) {                const auto& c1 = prefix[i], &c2 = prefix.back() - prefix[i];                mx = max(mx, total - (c1 * (c1 + 1) / 2 + c2 * (c2 + 1) / 2));            }            return mx;        };         const auto& f = [&](auto l) {            int64_t dp = 0;            int cnt = 0;            deque<tuple<int64_t, int64_t, int>> hull = {{0, 0, 0}};            for (int i = 0; i < size(nums); ++i) {                const auto& x = prefix[i + 1];                while (size(hull) >= 2 && get<0>(hull[0]) * x + get<1>(hull[0]) > get<0>(hull[1]) * x + get<1>(hull[1])) {                    hull.pop_front();                }                dp = (get<0>(hull[0]) * x + get<1>(hull[0])) + (x * x + x) / 2 + l;                cnt = get<2>(hull[0]) + 1;                const auto& line = tuple(-x, dp + (x * x - x) / 2, cnt);                while (size(hull) >= 2 && !check(hull[hull.size() - 2], hull[hull.size() - 1], line)) {                    hull.pop_back();                }                hull.emplace_back(line);            }            return pair(dp, cnt);        };         for (int i = 0; i < size(nums); ++i) {            prefix[i + 1] = prefix[i] + nums[i];        }        const auto& mx = max_lambda();        assert(f(mx).second == 1);        const auto& l = binary_search(static_cast<int64_t>(0), mx, [&](const auto& l) {            return f(l).second <= k;        });        return f(l).first - k * l;    }};

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