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Problem solution · C++

Minimum Time to Break Locks II

Minimum Time to Break Locks II: 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
84 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 Time to Break Locks II, 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

  • 84 lines of C++ from the credited upstream file 3385.cpp.
  • The implementation visibly relies on sequence storage.
  • 7 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 codeMinimum Time to Break Locks II · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int findMinimumTime(vector<int>& strength) {    vector<vector<int>> costs;     for (int turn = 1; turn <= strength.size(); ++turn) {      vector<int> cost;      for (const int s : strength)        cost.push_back((s + turn - 1) / turn);      costs.push_back(cost);    }     return hungarian(costs).back();  }  private:  // Updates the currentMinimum if potentialMinimum is smaller and returns true.  bool updateMinimum(int& currentMinimum, const int& potentialMinimum) {    if (potentialMinimum < currentMinimum) {      currentMinimum = potentialMinimum;      return true;    }    return false;  }   // Returns an array `res` of length n (costs.length), with `res[i]` equaling  // the minimum cost to assign the first (i + 1) turns to the first (i + 1)  // locks using Hungarian algorithm, where costs[i][j] is the energy required  // to break j-th lock in i-th turn.  //  // https://en.wikipedia.org/wiki/Hungarian_algorithm  vector<int> hungarian(const vector<vector<int>>& costs) {    const int numLocks = costs.size();    vector<int> res;    vector<int> lockAssignments(numLocks + 1, -1);    vector<int> turnPotentials(numLocks);    vector<int> lockPotentials(numLocks + 1);     for (int currentTurn = 0; currentTurn < numLocks; ++currentTurn) {      int currentLock = numLocks;      lockAssignments[currentLock] = currentTurn;      vector<int> minReducedCosts(numLocks + 1, INT_MAX);      vector<int> previousLockAssignments(numLocks + 1, -1);      vector<bool> locksInOptimalPath(numLocks + 1);       while (lockAssignments[currentLock] != -1) {        locksInOptimalPath[currentLock] = true;        const int assignedTurn = lockAssignments[currentLock];        int minCostDelta = INT_MAX;        int nextLock;         for (int lock = 0; lock < numLocks; ++lock)          if (!locksInOptimalPath[lock]) {            const int reducedCost = costs[assignedTurn][lock] -                                    turnPotentials[assignedTurn] -                                    lockPotentials[lock];            if (updateMinimum(minReducedCosts[lock], reducedCost))              previousLockAssignments[lock] = currentLock;            if (updateMinimum(minCostDelta, minReducedCosts[lock]))              nextLock = lock;          }         for (int lock = 0; lock <= numLocks; ++lock)          if (locksInOptimalPath[lock]) {            turnPotentials[lockAssignments[lock]] += minCostDelta;            lockPotentials[lock] -= minCostDelta;          } else {            minReducedCosts[lock] -= minCostDelta;          }         currentLock = nextLock;      }       for (int lock; currentLock != numLocks; currentLock = lock)        lockAssignments[currentLock] =            lockAssignments[lock = previousLockAssignments[currentLock]];       res.push_back(-lockPotentials[numLocks]);    }     return res;  }};

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