CPMagical CP Library

Problem solution · C++

Maximum Total Beauty of the Gardens

Maximum Total Beauty of the Gardens: 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
56 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 Total Beauty of the Gardens, 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

  • 56 lines of C++ from the credited upstream file 2234.cpp.
  • The implementation visibly relies on sequence storage.
  • 4 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 codeMaximum Total Beauty of the Gardens · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  long long maximumBeauty(vector<int>& flowers, long long newFlowers,                          int target, int full, int partial) {    const int n = flowers.size();     // If a garden is already complete, clamp it to the target.    for (int& flower : flowers)      flower = min(flower, target);    ranges::sort(flowers);     // All gardens are complete, so nothing we can do.    if (flowers[0] == target)      return static_cast<long>(n) * full;     // Having many new flowers maximizes the beauty value.    if (newFlowers >= static_cast<long>(n) * target -                          accumulate(flowers.begin(), flowers.end(), 0L))      return max(static_cast<long>(n) * full,                 (n - 1L) * full + (target - 1L) * partial);     long ans = 0;    long leftFlowers = newFlowers;    // cost[i] := the cost to make flowers[0..i] the same    vector<long> cost(n);     for (int i = 1; i < n; ++i)      // Plant (flowers[i] - flowers[i - 1]) flowers for flowers[0..i - 1].      cost[i] =          cost[i - 1] + static_cast<long>(i) * (flowers[i] - flowers[i - 1]);     int i = n - 1;  // flowers' index (flowers[i + 1..n) are complete)    while (flowers[i] == target)      --i;     for (; leftFlowers >= 0; --i) {      // To maximize the minimum number of incomplete flowers, we find the first      // index j that we can't make flowers[0..j] equal to flowers[j], then we      // know we can make flowers[0..j - 1] equal to flowers[j - 1]. In the      // meantime, evenly increase each of them to seek a bigger minimum value.      const int j = firstGreater(cost, i, leftFlowers);      const long minIncomplete =          flowers[j - 1] + (leftFlowers - cost[j - 1]) / j;      ans = max(ans, (n - 1L - i) * full + minIncomplete * partial);      leftFlowers -= max(0, target - flowers[i]);    }     return ans;  }  private:  int firstGreater(const vector<long>& A, int maxIndex, long target) {    return upper_bound(A.begin(), A.begin() + maxIndex + 1, target) - A.begin();  }};

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 ↗