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Problem solution · C++

Sell Diminishing-Valued Colored Balls

Sell Diminishing-Valued Colored Balls: a C++ solution using sorting and greedy selection. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Sorting and greedy selection
Source
walkccc LeetCode Solutions
Length
41 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

Sorting and greedy selection

For Sell Diminishing-Valued Colored Balls, the implementation first exposes a useful order, then scans that order while making locally justified choices.

  1. Choose the key that reveals the greedy or grouping structure.
  2. Sort the relevant records by that key.
  3. Scan in order, maintaining the invariant that makes each local choice safe.

Code notes

  • 41 lines of C++ from the credited upstream file 1648.cpp.
  • The implementation visibly relies on sequence storage.
  • 1 loop block detected.

Complexity

Sorting is typically the dominant term unless the subsequent scan uses a more expensive nested operation.

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 codeSell Diminishing-Valued Colored Balls · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int maxProfit(vector<int>& inventory, int orders) {    constexpr int kMod = 1'000'000'007;    long ans = 0;    long largestCount = 1;     ranges::sort(inventory, greater<>());     for (int i = 0; i < inventory.size(); ++i, ++largestCount)      if (i == inventory.size() - 1 || inventory[i] > inventory[i + 1]) {        // If we are at the last inventory, or inventory[i] > inventory[i + 1].        // In either case, we will pick inventory[i - largestCount + 1..i].        const int pick = (i == inventory.size() - 1)                             ? inventory[i]                             : inventory[i] - inventory[i + 1];        if (largestCount * pick >= orders) {          // We have run out of orders, so we need to recalculate the number of          // balls that we actually pick for inventory[i - largestCount + 1..i].          const int actualPick = orders / largestCount;          const int remaining = orders % largestCount;          return (ans +                  largestCount *                      trapezoid(inventory[i], inventory[i] - actualPick + 1) +                  static_cast<long>(remaining) * (inventory[i] - actualPick)) %                 kMod;        }        ans += largestCount * trapezoid(inventory[i], inventory[i] - pick + 1);        ans %= kMod;        orders -= largestCount * pick;      }     throw;  }  private:  long trapezoid(long a, long b) {    return (a + b) * (a - b + 1) / 2;  }};

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