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Problem solution · C++

Avoid Flood in The City

Avoid Flood in The City: 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
36 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 Avoid Flood in The City, 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

  • 36 lines of C++ from the credited upstream file 1488.cpp.
  • The implementation visibly relies on sequence storage, hash lookup, ordered lookup.
  • 2 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 codeAvoid Flood in The City · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  vector<int> avoidFlood(vector<int>& rains) {    vector<int> ans(rains.size(), -1);    unordered_map<int, int> lakeIdToFullDay;    set<int> emptyDays;  // indices of rains[i] == 0     for (int i = 0; i < rains.size(); ++i) {      const int lakeId = rains[i];      if (lakeId == 0) {        emptyDays.insert(i);        continue;      }      if (const auto itFullDay = lakeIdToFullDay.find(lakeId);          itFullDay != lakeIdToFullDay.cend()) {        // The lake was full in a previous day. Greedily find the closest day        // to make the lake empty.        const auto itEmptyDay = emptyDays.upper_bound(itFullDay->second);        if (itEmptyDay == emptyDays.cend())  // Not found.          return {};        // Empty the lake at this day.        ans[*itEmptyDay] = lakeId;        emptyDays.erase(itEmptyDay);      }      // The lake with `lakeId` becomes full at the day `i`.      lakeIdToFullDay[lakeId] = i;    }     // Empty an arbitrary lake if there are remaining empty days.    for (const int emptyDay : emptyDays)      ans[emptyDay] = 1;     return ans;  }};

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