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Problem solution · C++

Minimum Cost to Merge Sorted Lists

Minimum Cost to Merge Sorted Lists: 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
78 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 Minimum Cost to Merge Sorted Lists, 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

  • 78 lines of C++ from the credited upstream file minimum-cost-to-merge-sorted-lists.cpp.
  • The implementation visibly relies on sequence storage, work queue.
  • 9 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 codeMinimum Cost to Merge Sorted Lists · C++C++
Use this to learn the idea, then write your own version.
// Time:  O(l * nlogn + 2^n * log(n * l) * n * logl + 3^n), n = len(lists), l = max(len(list) for list in lists)// Space: O(n * l + 2^n) // dp, sort, heap, binary search, submask enumerationclass Solution {public:    long long minMergeCost(vector<vector<int>>& lists) {        static const int64_t INF = numeric_limits<int64_t>::max();         const auto& merge = [](const auto& lists) {            using T = tuple<int, int, int>;            vector<T> data;            data.reserve(size(lists));            for (int i = 0; i < size(lists); ++i) {                data.emplace_back(lists[i][0], i, 0);            }            priority_queue<T, vector<T>, greater<T>> min_heap(cbegin(data), cend(data));            vector<int> result;            while (!empty(min_heap)) {                const auto [x, i, j] = min_heap.top(); min_heap.pop();                result.emplace_back(x);                if (j + 1 < size(lists[i])) {                    min_heap.emplace(lists[i][j + 1], i, j + 1);                }            }            return result;        };         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;        };          vector<int> dp1(1 << size(lists));        for (int i = 0; i < size(lists); ++i) {  // Time: O(2^n)            dp1[1 << i] = size(lists[i]);        }        for (int mask = 1; mask < size(dp1); ++mask) {  // Time: O(2^n)            dp1[mask] = dp1[mask ^ (mask & -mask)] + dp1[mask & -mask];        }        auto sorted_vals = merge(lists);  // Time: O((n * l) * logn)        sorted_vals.erase(unique(begin(sorted_vals), end(sorted_vals)), end(sorted_vals));        vector<int> dp2(1 << size(lists));        for (int mask = 1; mask < size(dp2); ++mask) {  // Time: O(2^n * log(n * l) * n * logl)            const auto& check = [&](int x) {                int cnt = 0;                for (int i = 0; i < size(lists); ++i) {                    if ((mask & (1 << i)) == 0) {                        continue;                    }                    cnt += distance(cbegin(lists[i]), upper_bound(cbegin(lists[i]), cend(lists[i]), sorted_vals[x]));                }                return cnt >= (dp1[mask] + 1) / 2;            };             dp2[mask] = sorted_vals[binary_search(0, size(sorted_vals) - 1, check)];        }        vector<int64_t> dp3(1 << size(lists));        for (int mask = 1; mask < size(dp3); ++mask) {  // Time: O(3^n)            if ((mask & (mask - 1)) == 0) {                continue;            }            dp3[mask] = INF;            for (int submask = (mask - 1) & mask; submask > (mask ^ submask); submask = (submask - 1) & mask) {                dp3[mask] = min(dp3[mask], dp3[submask] + dp3[mask ^ submask] + abs(dp2[submask] - dp2[mask ^ submask]) + dp1[mask]);            }        }        return dp3.back();    }};

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