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Problem solution · C++

Maximum Total Subarray Value II

Maximum Total Subarray Value II: a C++ solution using segment tree or range structure. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Segment tree or range structure
Source
Kamyu LeetCode Solutions
Length
219 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

Segment tree or range structure

For Maximum Total Subarray Value II, the implementation stores interval information in a range-query data structure so updates and queries avoid rescanning the full input.

  1. Choose the aggregate stored for each interval or prefix.
  2. Build or initialize the structure from the input.
  3. Apply updates and combine the affected nodes to answer each query.

Code notes

  • 219 lines of C++ from the credited upstream file maximum-total-subarray-value-ii.cpp.
  • The implementation visibly relies on sequence storage, hash lookup, work queue.
  • 15 loop blocks detected.

Complexity

Count the build once, then multiply the logarithmic update or query path by the number of operations.

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 Total Subarray Value II · C++C++
Use this to learn the idea, then write your own version.
// Time:  O((n + k) * logn)// Space: O(n + k) // heap, sort, two pointersclass Solution {private:    struct TupleHash {        template <typename... T>        std::size_t operator()(const std::tuple<T...>& t) const {            return apply([](const auto&... args) {                std::size_t seed = 0;                ((seed ^= std::hash<std::decay_t<decltype(args)>>{}(args) +                         0x9e3779b9 + (seed << 6) + (seed >> 2)), ...);                return seed;            }, t);        }    }; public:    long long maxTotalValue(vector<int>& nums, int k) {        vector<int> idxs(size(nums));        iota(begin(idxs), end(idxs), 0);        sort(begin(idxs), end(idxs), [&](const auto& a, const auto& b) {            return nums[a] != nums[b] ? nums[a] < nums[b] : a < b;        });        const auto& nxt = [&](int left, int right, int i, int j) {            for (; !(left <= idxs[i] && idxs[i] <= right); ++i);            for (; !(left <= idxs[j] && idxs[j] <= right); --j);            return tuple(i, j);        };         unordered_map<tuple<int, int>, tuple<int, int>, TupleHash> lookup;        lookup[{0, size(idxs) - 1}] = {0, size(idxs) - 1};        using P = pair<int, pair<int, int>>;        priority_queue<P> max_heap;        max_heap.push({nums[idxs[size(idxs) - 1]] - nums[idxs[0]], {0, size(idxs) - 1}});        int64_t result = 0;        while (k) {            const auto [v, l_r] = max_heap.top();  max_heap.pop();            const auto& [l, r] = l_r;            const auto& [i, j] = lookup[{l, r}];            const auto& nl = min(idxs[i], idxs[j]);            const auto& nr = max(idxs[i], idxs[j]);            const int64_t c = min((nl - l + 1) * (r - nr + 1), k);            k -= c;            result += c * v;            if (nl + 1 <= r && !lookup.count({nl + 1, r})) {                const auto& [ni, nj] = nxt(nl + 1, r, i, j);                lookup[{nl + 1, r}] = {ni, nj};                max_heap.push({nums[idxs[nj]] - nums[idxs[ni]], {nl + 1, r}});            }            if (l <= nr - 1 && !lookup.count({l, nr - 1})) {                const auto& [ni, nj] = nxt(l, nr - 1, i, j);                lookup[{l, nr - 1}] = {ni, nj};                max_heap.push({nums[idxs[nj]] - nums[idxs[ni]], {l, nr - 1}});            }        }        return result;    }}; // Time:  O((n + k) * logn)// Space: O(nlogn)// heap, rmq, sparse tableclass Solution2 {public:    long long maxTotalValue(vector<int>& nums, int k) {        SparseTable rmq_min(nums, [](const auto& a, const auto& b) {            return a < b ? a : b;        });        SparseTable rmq_max(nums, [](const auto& a, const auto& b) {            return a > b ? a : b;        });        using P = pair<int, pair<int, int>>;        vector<P> arr(size(nums));        for (int i = 0; i < size(nums); ++i) {            arr[i] = {rmq_max.query(i, size(nums) - 1) - rmq_min.query(i, size(nums) - 1), {i, size(nums) - 1}};        }        priority_queue<P> max_heap(cbegin(arr), cend(arr));        int64_t result = 0;        for (int _ = 0; _ < k; ++_) {            const auto [v, i_j] = max_heap.top(); max_heap.pop();            const auto& [i, j] = i_j;            result += v;            if (i <= j - 1) {                max_heap.push({rmq_max.query(i, j - 1) - rmq_min.query(i, j - 1), {i, j - 1}});            }        }        return result;    } private:    // Reference: https://cp-algorithms.com/data_structures/sparse-table.html    class SparseTable {    public:        SparseTable(const vector<int>& arr, function<int (int, int)> fn)         :  fn(fn) {  // Time: O(nlogn) * O(fn) = O(nlogn), Space: O(nlogn)            const int n = size(arr);            const int k = __lg(n);            st.assign(k + 1, vector<int64_t>(n));            st[0].assign(cbegin(arr), cend(arr));            for (int i = 1; i <= k; ++i) {                for (int j = 0; j + (1 << i) <= n; ++j) {                    st[i][j] = fn(st[i - 1][j], st[i - 1][j + (1 << (i - 1))]);                }            }         }         int64_t query(int L, int R) const {            const int i = __lg(R - L + 1);            return fn(st[i][L], st[i][R - (1 << i) + 1]);  // Time: O(fn) = O(1)        }        private:        vector<vector<int64_t>> st;        const function<int (int, int)>& fn;    };}; // Time:  O((n + k) * logn)// Space: O(n)// heap, segment treeconst int NEG_INF = numeric_limits<int>::min();const int POS_INF = numeric_limits<int>::max();class Solution3 {public:    long long maxTotalValue(vector<int>& nums, int k) {        SegmentTreeMin st_min(nums);        SegmentTreeMax st_max(nums);        using P = pair<int, pair<int, int>>;        vector<P> arr(size(nums));        for (int i = 0; i < size(nums); ++i) {            arr[i] = {st_max.query(i, size(nums) - 1) - st_min.query(i, size(nums) - 1), {i, size(nums) - 1}};        }        priority_queue<P> max_heap(cbegin(arr), cend(arr));        int64_t result = 0;        for (int _ = 0; _ < k; ++_) {            const auto [v, i_j] = max_heap.top(); max_heap.pop();            const auto& [i, j] = i_j;            result += v;            if (i <= j - 1) {                max_heap.push({st_max.query(i, j - 1) - st_min.query(i, j - 1), {i, j - 1}});            }        }        return result;    } private:    class SegmentTreeMax {    public:        explicit SegmentTreeMax(const auto& nums)         :  base_(size(nums) > 1 ? 1 << (__lg(size(nums) - 1) + 1) : 1),            tree(size(nums) > 1 ? 1 << (__lg(size(nums) - 1) + 2) : 2, NEG_INF) {            for (int i = base_; i < base_ + size(nums); ++i) {                tree[i] = nums[i - base_];            }            for (int i = base_ - 1; i >= 1; --i) {                tree[i] = max(tree[i << 1], tree[(i << 1) + 1]);            }        }         int query(int L, int R) {            if (L > R) {                return NEG_INF;            }            L += base_;            R += base_;            int left = NEG_INF, right = NEG_INF;            for (; L <= R; L >>= 1, R >>= 1) {                if ((L & 1) == 1) {                    left = max(left, tree[L++]);                }                if ((R & 1) == 0) {                    right = max(tree[R--], right);                }            }            return max(left, right);        }         vector<int> tree;        int base_;    };     class SegmentTreeMin {    public:        explicit SegmentTreeMin(const auto& nums)         :  base_(size(nums) > 1 ? 1 << (__lg(size(nums) - 1) + 1) : 1),            tree(size(nums) > 1 ? 1 << (__lg(size(nums) - 1) + 2) : 2, POS_INF) {            for (int i = base_; i < base_ + size(nums); ++i) {                tree[i] = nums[i - base_];            }            for (int i = base_ - 1; i >= 1; --i) {                tree[i] = min(tree[i << 1], tree[(i << 1) + 1]);            }        }         int query(int L, int R) {            if (L > R) {                return POS_INF;            }            L += base_;            R += base_;            int left = POS_INF, right = POS_INF;            for (; L <= R; L >>= 1, R >>= 1) {                if ((L & 1) == 1) {                    left = min(left, tree[L++]);                }                if ((R & 1) == 0) {                    right = min(tree[R--], right);                }            }            return min(left, right);        }         vector<int> tree;        int base_;    };};

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