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Problem solution · C++

Maximum Subarray Sum After at Most K Swaps

Maximum Subarray Sum After at Most K Swaps: 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
64 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 Subarray Sum After at Most K Swaps, 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

  • 64 lines of C++ from the credited upstream file maximum-subarray-sum-after-at-most-k-swaps.cpp.
  • The implementation visibly relies on sequence storage, work queue, cached states.
  • 5 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 Subarray Sum After at Most K Swaps · C++C++
Use this to learn the idea, then write your own version.
// Time:  O(n^2 * logk)// Space: O(n) // prefix sum, heap, dpclass Solution {public:    long long maxSum(vector<int>& nums, int k) {        const auto& update = [&](auto& heap, int64_t total, int x) {            if (k == 0) {                return total;            }            heap.emplace(x);            total += x;            if (size(heap) == k + 1) {                total -= heap.top(); heap.pop();            }            return total;        };         const auto& cnt = accumulate(cbegin(nums), cend(nums), 0, [](const auto& accu, const auto& x) {            return accu + (x < 0 ? 1 : 0);        });        if (cnt == size(nums)) {            return ranges::max(nums);        }        if (cnt <= k) {  // optional to boost up performance            return accumulate(cbegin(nums), cend(nums), 0LL, [](const auto& accu, const auto& x) {                return accu + (x >= 0 ? x : 0);            });        }        vector<int64_t> prefix(size(nums) + 1);        for (int i = 0; i < size(nums); ++i) {            prefix[i + 1] = prefix[i] + nums[i];        }        int64_t result = 0;        vector<int64_t> dp(size(nums));        for (int i = 0; i < size(nums); ++i) {            priority_queue<int, vector<int>, greater<int>> max_heap;            int64_t total1 = 0;            for (int j = i; j < size(nums); ++j) {                if (nums[j] < 0) {                    total1 = update(max_heap, total1, -nums[j]);                }                dp[j] = -total1;            }            priority_queue<int, vector<int>, greater<int>> min_heap;            int64_t total2 = 0;            for (int j = 0; j < i; ++j) {                if (nums[j] >= 0) {                    total2 = update(min_heap, total2, nums[j]);                }            }            for (int j = size(nums) - 1; j >= i; --j) {                result = max(result, (prefix[j + 1] - prefix[i]) - dp[j] + total2);                if (nums[j] >= 0) {                    total2 = update(min_heap, total2, nums[j]);                    result = max(result, total2);                }            }        }        return result;    }};

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