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Problem solution · C++

Maximum Sum Queries

Maximum Sum Queries: a C++ solution using binary search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Binary search
Source
walkccc LeetCode Solutions
Length
60 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 Maximum Sum Queries, 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

  • 60 lines of C++ from the credited upstream file 2736.cpp.
  • The implementation visibly relies on sequence storage.
  • 5 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 walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMaximum Sum Queries · C++C++
Use this to learn the idea, then write your own version.
struct Pair {  int x;  int y;}; struct IndexedQuery {  int queryIndex;  int minX;  int minY;}; class Solution { public:  vector<int> maximumSumQueries(vector<int>& nums1, vector<int>& nums2,                                vector<vector<int>>& queries) {    const vector<Pair> pairs = getPairs(nums1, nums2);    vector<int> ans(queries.size());    vector<pair<int, int>> stack;  // [(y, x + y)]     int pairsIndex = 0;    for (const auto& [queryIndex, minX, minY] : getIndexedQueries(queries)) {      while (pairsIndex < pairs.size() && pairs[pairsIndex].x >= minX) {        const auto [x, y] = pairs[pairsIndex++];        // x + y is a better candidate. Given that x is decreasing, the        // condition "x + y >= stack.back().second" suggests that y is        // relatively larger, thereby making it a better candidate.        while (!stack.empty() && x + y >= stack.back().second)          stack.pop_back();        if (stack.empty() || y > stack.back().first)          stack.emplace_back(y, x + y);      }      const auto it = ranges::lower_bound(stack, pair<int, int>{minY, INT_MIN});      ans[queryIndex] = it == stack.end() ? -1 : it->second;    }     return ans;  }  private:  vector<Pair> getPairs(const vector<int>& nums1, const vector<int>& nums2) {    vector<Pair> pairs;    for (int i = 0; i < nums1.size(); ++i)      pairs.push_back({nums1[i], nums2[i]});    ranges::sort(pairs, ranges::greater{},                 [](const Pair& pair) { return pair.x; });    return pairs;  }   vector<IndexedQuery> getIndexedQueries(const vector<vector<int>>& queries) {    vector<IndexedQuery> indexedQueries;    for (int i = 0; i < queries.size(); ++i)      indexedQueries.push_back({i, queries[i][0], queries[i][1]});    ranges::sort(indexedQueries,                 [](const IndexedQuery& a, const IndexedQuery& b) {      return a.minX > b.minX;    });    return indexedQueries;  }};

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