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Problem solution · C++

Minimum Possible Maximum Waiting Time

Minimum Possible Maximum Waiting Time: 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
192 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 Minimum Possible Maximum Waiting Time, 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

  • 192 lines of C++ from the credited upstream file minimum-possible-maximum-waiting-time.cpp.
  • The implementation visibly relies on sequence storage, cached states.
  • 12 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 codeMinimum Possible Maximum Waiting Time · C++C++
Use this to learn the idea, then write your own version.
// Time:  O(n * rlogr * (f / 64)) = O(n * rlogr), r = max(demand), f = fuel[0]// Space: O(r * (f / 64)) = O(r) // binary search, dp, bitmasksclass Solution {public:    int minMaxWaitingTime(vector<int>& demand, vector<int>& fuel) {        const auto& binary_search = [](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 left;        };         const auto& low = [](int x) -> uint64_t {            return x >= 63 ? ~uint64_t{0} : x >= 0 ? ((uint64_t{1} << (x + 1)) - 1) : 0;        };         const auto& high = [&](int x) -> uint64_t {            return ~uint64_t{0} ^ low(x - 1);        };         const auto& find_max_served = [&]() -> int {            uint64_t mask = 1;            for (int i = 0, total = 0; i < size(demand); ++i) {                mask = ((mask << demand[i]) & low(fuel[0])) | (mask & high(total + demand[i] - fuel[1]));                if (!mask) {                    return i;                }                total += demand[i];            }            return size(demand);        };         const auto& l = find_max_served();        if (!l) {            return -1;        }            const auto& mx = ranges::max(demand);        const auto& check = [&](int w) {            // dp[last][gap] = bit used0 is set if the state (last, gap, used0) is reachable.            // - last: dispenser serving the previous car            // - gap: remaining busy time of the other dispenser            // - used0: total fuel consumed by dispenser 0            vector<vector<uint64_t>> dp(2, vector<uint64_t>(mx + 1));            dp[0][0] = 1;            for (int i = 0, total = 0; i < l; ++i) {                vector<vector<uint64_t>> new_dp(2, vector<uint64_t>(mx + 1));                const auto& update = [&](int last, int gap, uint64_t mask) {                    if (last == 0) {                        mask = (mask << demand[i]) & low(fuel[0]);                    } else {                        mask &= high(total + demand[i] - fuel[1]);                    }                    new_dp[last][gap] |= mask;                };                 for (int last = 0; last < size(dp); ++last) {                    for (int gap = 0; gap < size(dp[0]); ++gap) {                        if (!dp[last][gap]) {                            continue;                        }                        if ((i - 1 >= 0 ? demand[i - 1] : 0) <= w) {                            update(last, max(gap - (i - 1 >= 0 ? demand[i - 1] : 0), 0), dp[last][gap]);                        }                        if (gap <= w) {                            update(last ^ 1, max((i - 1 >= 0 ? demand[i - 1] : 0) - gap, 0), dp[last][gap]);                        }                    }                }                dp = move(new_dp);                total += demand[i];            }            return ranges::any_of(dp, [](const auto& row) {                return ranges::any_of(row, [](uint64_t mask) {                    return mask != 0;                });            });        };         return binary_search(0, mx, check);    }}; // Time:  O(n * rlogr * f),  r = max(demand), f = fuel[0]// Space: O(r * f)// binary search, dpclass Solution2 {public:    int minMaxWaitingTime(vector<int>& demand, vector<int>& fuel) {        const auto& binary_search = [](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 left;        };         const auto& find_max_served = [&]() -> int {            // dp[used0] = whether the current prefix can be served with state (used0).            vector<bool> dp(fuel[0] + 1);            dp[0] = true;            for (int i = 0, total = 0; i < size(demand); ++i) {                vector<bool> new_dp(fuel[0] + 1);                for (int used0 = 0; used0 <= fuel[0]; ++used0) {                    if (!dp[used0]) {                        continue;                    }                    if (used0 + demand[i] <= fuel[0]) {                        new_dp[used0 + demand[i]] = true;                    }                    if (total - used0 + demand[i] <= fuel[1]) {                        new_dp[used0] = true;                    }                }                if (!ranges::any_of(new_dp, [](bool ok) { return ok; })) {                    return i;                }                dp = move(new_dp);                total += demand[i];            }            return size(demand);        };         const auto& l = find_max_served();        if (!l) {            return -1;        }         const auto& mx = ranges::max(demand);        const auto& check = [&](int w) {            // dp[last][gap][used0] = whether the current prefix can be served with state (last, gap, used0).            // - last: which dispenser serving the previous car            // - gap: remaining busy time of the other dispenser            // - used0: total fuel consumed by dispenser 0            vector<vector<vector<bool>>> dp(2, vector<vector<bool>>(mx + 1, vector<bool>(fuel[0] + 1)));            dp[0][0][0] = true;            for (int i = 0, total = 0; i < l; ++i) {                vector<vector<vector<bool>>> new_dp(2, vector<vector<bool>>(mx + 1, vector<bool>(fuel[0] + 1)));                const auto& update = [&](int last, int gap, int used0) {                    if (last == 0) {                        if (used0 + demand[i] <= fuel[0]) {                            new_dp[last][gap][used0 + demand[i]] = true;                        }                    } else {                        if (total - used0 + demand[i] <= fuel[1]) {                            new_dp[last][gap][used0] = true;                        }                    }                };                    for (int last = 0; last < size(dp); ++last) {                    for (int gap = 0; gap < size(dp[0]); ++gap) {                        for (int used0 = 0; used0 < size(dp[0][0]); ++used0) {                            if (!dp[last][gap][used0]) {                                continue;                            }                            if ((i - 1 >= 0 ? demand[i - 1] : 0) <= w) {                                update(last, max(gap - (i - 1 >= 0 ? demand[i - 1] : 0), 0), used0);                            }                            if (gap <= w) {                                update(last ^ 1, max((i - 1 >= 0 ? demand[i - 1] : 0) - gap, 0), used0);                            }                        }                    }                }                dp = move(new_dp);                total += demand[i];            }            return ranges::any_of(dp, [](const auto& matrix) {                return ranges::any_of(matrix, [](const auto& row) {                    return ranges::any_of(row, [](bool x) {                        return x;                    });                });            });        };         return binary_search(0, mx, check);    }};

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