CPMagical CP Library

Problem solution · C++

Maximum Capacity Within Budget

Maximum Capacity Within Budget: a C++ solution using binary search. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Binary search
Source
Kamyu LeetCode Solutions
Length
143 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 Capacity Within Budget, 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

  • 143 lines of C++ from the credited upstream file maximum-capacity-within-budget.cpp.
  • The implementation visibly relies on sequence storage.
  • 12 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 Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMaximum Capacity Within Budget · C++C++
Use this to learn the idea, then write your own version.
// Time:  O(n + b)// Space: O(b) // hash table, prefix sumclass Solution {public:    int maxCapacity(vector<int>& costs, vector<int>& capacity, int budget) {        const auto& mid = (budget - 1) / 2;        vector<int> lookup(budget);        for (int i = 0; i < size(costs); ++i) {            if (costs[i] >= budget) {                continue;            }            lookup[costs[i]] = max(lookup[costs[i]], capacity[i]);        }        for (int i = 0; i + 1 <= mid; ++i) {            lookup[i + 1] = max(lookup[i + 1], lookup[i]);        }        int result = 0, mx = 0;        for (int i = 0; i < size(costs); ++i) {            if (costs[i] > mid) {                continue;            }            result = max(result, mx + capacity[i]);            mx = max(mx, capacity[i]);        }        for (int i = mid + 1; i <= budget - 1; ++i) {            result = max(result, lookup[i] + lookup[(budget - 1) - i]);        }        return result;    }}; // Time:  O(nlogn)// Space: O(n)// sort, mono stackclass Solution2 {public:    int maxCapacity(vector<int>& costs, vector<int>& capacity, int budget) {        vector<int> idxs(size(costs));        iota(begin(idxs), end(idxs), 0);        sort(begin(idxs), end(idxs), [&](const auto& a, const auto& b) {            return costs[a] < costs[b];        });         int result = 0;        vector<pair<int, int>> stk;        for (const auto& i : idxs) {            const auto& cost = costs[i], &cap = capacity[i];            if (cost >= budget) {                break;            }            while (!empty(stk) && stk.back().first + cost >= budget) {                stk.pop_back();            }            result = max(result, (!empty(stk) ? stk.back().second : 0) + cap);            if (empty(stk) || stk.back().second < cap) {                stk.emplace_back(cost, cap);            }        }        return result;    }}; // Time:  O(nlogn)// Space: O(n)// sort, binary searchclass Solution3 {public:    int maxCapacity(vector<int>& costs, vector<int>& capacity, int budget) {        vector<int> idxs(size(costs));        iota(begin(idxs), end(idxs), 0);        sort(begin(idxs), end(idxs), [&](const auto& a, const auto& b) {            return costs[a] < costs[b];        });         vector<int> prefix(size(capacity) + 1);        for (int i = 0; i < size(capacity); ++i) {            prefix[i + 1] = max(prefix[i], capacity[idxs[i]]);        }        int result = 0;        vector<pair<int, int>> stk;        vector<int> sorted_costs;        sorted_costs.reserve(size(costs));        for (const auto& i : idxs) {            sorted_costs.emplace_back(costs[i]);        }        for (int i = 0; i < size(idxs); ++i) {            const auto& cost = costs[idxs[i]], &cap = capacity[idxs[i]];            if (cost >= budget) {                break;            }            const auto& j = distance(cbegin(sorted_costs), lower_bound(cbegin(sorted_costs), cbegin(sorted_costs) + i, budget - cost)) - 1;            result = max(result, prefix[j + 1] + cap);        }        return result;    }}; // Time:  O(nlogn)// Space: O(n)// sort, binary searchclass Solution4 {public:    int maxCapacity(vector<int>& costs, vector<int>& capacity, int budget) {        const auto& binary_search_right = [](int left, int 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 right;        };         vector<int> idxs(size(costs));        iota(begin(idxs), end(idxs), 0);        sort(begin(idxs), end(idxs), [&](const auto& a, const auto& b) {            return costs[a] < costs[b];        });         vector<int> prefix(size(capacity) + 1);        for (int i = 0; i < size(capacity); ++i) {            prefix[i + 1] = max(prefix[i], capacity[idxs[i]]);        }        int result = 0;        vector<pair<int, int>> stk;        for (int i = 0; i < size(idxs); ++i) {            const auto& cost = costs[idxs[i]], &cap = capacity[idxs[i]];            if (cost >= budget) {                break;            }            const auto& j = binary_search_right(0, i - 1, [&](const auto& x) {                return costs[idxs[x]] + cost < budget;            });            result = max(result, prefix[j + 1] + cap);        }        return result;    }};

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 ↗