CPMagical CP Library

Problem solution · C++

Maximum Sum of M Non Overlapping Subarrays II

Maximum Sum of M Non Overlapping Subarrays II: 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
95 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 Maximum Sum of M Non Overlapping Subarrays II, 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

  • 95 lines of C++ from the credited upstream file maximum-sum-of-m-non-overlapping-subarrays-ii.cpp.
  • The implementation visibly relies on sequence storage, work queue, cached states.
  • 8 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 codeMaximum Sum of M Non Overlapping Subarrays II · C++C++
Use this to learn the idea, then write your own version.
// Time:  O(nlogr)// Space: O(n) // prefix sum, dp, mono deque, wqs binary search, alien trickclass Solution {public:    long long maximumSum(vector<int>& nums, int m, int l, int r) {        static const auto& NEG_INF = numeric_limits<int64_t>::min();         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;        };         vector<int64_t> prefix(size(nums) + 1);        const auto& best_single = [&]() {            auto result = NEG_INF;            deque<int> dq;            for (int i = 1; i <= size(nums); ++i) {                const auto& j = i - l;                if (j >= 0) {                    while (!empty(dq) && prefix[dq.back()] >= prefix[j]) {                        dq.pop_back();                    }                    dq.emplace_back(j);                }                while (!empty(dq) && dq.front() < i - r) {                    dq.pop_front();                }                if (!empty(dq)) {                    result = max(result, prefix[i] - prefix[dq.front()]);                }            }            return result;        };         auto f = [&](int64_t x) -> pair<int64_t, int> {            const auto& better = [](auto v1, auto c1, auto v2, auto c2)  {                return v1 > v2 || (v1 == v2 && c1 < c2);            };             vector<pair<int64_t, int>> dp(size(nums) + 1, make_pair(0LL, 0));            deque<int> dq;            for (int i = 1; i <= size(nums); ++i) {                const auto& j = i - l;                if (j >= 0) {                    while (!empty(dq) && better(dp[j].first - prefix[j], dp[j].second, dp[dq.back()].first - prefix[dq.back()], dp[dq.back()].second)) {                        dq.pop_back();                    }                    dq.emplace_back(j);                }                while (!empty(dq) && dq.front() < i - r) {                    dq.pop_front();                }                dp[i] = dp[i - 1];                if (!empty(dq)) {                    pair<int64_t, int> new_dp = {                        ((dp[dq.front()].first - prefix[dq.front()]) + prefix[i]) - x,                        dp[dq.front()].second + 1                    };                    if (better(new_dp.first, new_dp.second, dp[i].first, dp[i].second)) {                        dp[i] = move(new_dp);                    }                }            }            return dp.back();        };         for (int i = 0; i < size(nums); ++i) {            prefix[i + 1] = prefix[i] + nums[i];        }        const auto& single = best_single();        auto [dp, cnt] = f(0);        if (cnt == 0) {            return single;        }        if (cnt <= m) {            return dp;        }        const auto& mx = single;        assert(f(mx).second <= m);        const auto& x = binary_search(static_cast<int64_t>(1), mx, [&](const auto& x) {            return f(x).second <= m;        });        return f(x).first + static_cast<int64_t>(m) * x;    }};

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 ↗