CPMagical CP Library

Problem solution · C++

Minimum Partition Score

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

Technique
Dynamic programming
Source
Kamyu LeetCode Solutions
Length
108 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

Dynamic programming

For Minimum Partition Score, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

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

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

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 · 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;    }}; // Time:  O(n * k)// Space: O(n)// prefix sum, dp, convex hull trickclass Solution2 {public:    long long minPartitionScore(vector<int>& nums, int k) {        static const int64_t INF = numeric_limits<int64_t>::max();         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);        for (int i = 0; i < size(nums); ++i) {            prefix[i + 1] = prefix[i] + nums[i];        }        vector<int64_t> dp(size(nums) + 1, INF);        dp[0] = 0;        for (int j = 0; j < k; ++j) {            vector<int64_t> new_dp(size(nums) + 1, INF);            deque<pair<int64_t, int64_t>> hull;            for (int i = j; i < static_cast<int>(size(nums)); ++i) {                if (dp[i] != INF) {                    const auto& x = prefix[i];                    const auto& line = pair(-x, dp[i] + (x * x - x) / 2);                    while (size(hull) >= 2 && !check(hull[size(hull) - 2], hull[size(hull) - 1], line)) {                        hull.pop_back();                    }                    hull.emplace_back(line);                }                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();                }                new_dp[i + 1] = get<0>(hull[0]) * x + get<1>(hull[0]) + (x * x + x) / 2;                 }            dp = move(new_dp);        }        return dp.back();    }};

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗