Problem solution · C++

Maximum Profitable Triplets With Increasing Prices II

Maximum Profitable Triplets With Increasing Prices II: a C++ solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
55 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

Direct simulation

For Maximum Profitable Triplets With Increasing Prices II, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 55 lines of C++ from the credited upstream file 2921.cpp.
  • The implementation visibly relies on sequence storage.
  • 3 loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

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.

Full codeMaximum Profitable Triplets With Increasing Prices II · C++C++
Use this to learn the idea, then write your own version.
class FenwickTree { public:  FenwickTree(int n) : vals(n + 1) {}   void maximize(int i, int val) {    while (i < vals.size()) {      vals[i] = max(vals[i], val);      i += lowbit(i);    }  }   int get(int i) const {    int res = 0;    while (i > 0) {      res = max(res, vals[i]);      i -= lowbit(i);    }    return res;  }  private:  vector<int> vals;   static int lowbit(int i) {    return i & -i;  }}; class Solution { public:  // Same as 2907. Maximum Profitable Triplets With Increasing Prices I  int maxProfit(vector<int>& prices, vector<int>& profits) {    const int maxPrice = ranges::max(prices);    int ans = -1;    FenwickTree maxProfitTree1(maxPrice);    FenwickTree maxProfitTree2(maxPrice);     for (int i = 0; i < prices.size(); ++i) {      const int price = prices[i];      const int profit = profits[i];      // max(proftis[i])      const int maxProfit1 = maxProfitTree1.get(price - 1);      // max(proftis[i]) + max(profits[j])      const int maxProfit2 = maxProfitTree2.get(price - 1);      maxProfitTree1.maximize(price, profit);      if (maxProfit1 > 0)        maxProfitTree2.maximize(price, profit + maxProfit1);      if (maxProfit2 > 0)        ans = max(ans, profit + maxProfit2);    }     return ans;  }}; 

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