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Problem solution · C++

Maximum Total Reward Using Operations I

Maximum Total Reward Using Operations I: a C++ solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Dynamic programming
Source
walkccc LeetCode Solutions
Length
52 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

Dynamic programming

For Maximum Total Reward Using Operations I, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 52 lines of C++ from the credited upstream file 3180.cpp.
  • The implementation visibly relies on sequence storage, cached states.
  • 2 loop blocks detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

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 Reward Using Operations I · C++C++
Use this to learn the idea, then write your own version.
// According to the constraint rewardValues[i] <= 5 * 10^4, the maximum total// reward < 2 * (5 * 10^4) = 10^5. We can use bitset to record whether each// `rewardValue` is achievable in O(1).//// Let's use `rewardValues = [1, 3, 4]` as an example.//// The maximum reward is 4, so the maximum possible total < 2 * 4 = 8.// Therefore, we can set the size of the bitset to 8 to represent possible// total rewards from 0 to 7.//// Let's define a bitset `dp` to record whether each total reward is// achievable. dp[num] = true if reward `num` is achievable.//// Initially, dp = 0b00000001 := reward 0 is achievable.//// * rewardValues[0] = 1, for each dp[i] = 1, where i + 1 < 10, dp[i + 1] = 1.//   => dp = 0b00000011 := rewards 0 and 1 are achievable.//// * rewardValues[1] = 3, for each dp[i] = 1, where i + 3 < 10, dp[i + 3] = 1.//   => dp = 0b00011011 := rewards 0, 1, 3, and 4 are achievable.//// * rewardValues[2] = 4, for each dp[i] = 1, where i + 4 < 10, dp[i + 4] = 1.//   => dp = 0b10011011 := rewards 0, 1, 3, 4, 5, and 7 are achievable.//// Therefore, the maximum total reward is 7. class Solution { public:  int maxTotalReward(vector<int>& rewardValues) {    constexpr int kPossibleRewards = 100'000;    // dp[num] := true if reward `num` is achievable    bitset<kPossibleRewards> dp;    dp[0] = true;     ranges::sort(rewardValues);     for (const int num : rewardValues) {      bitset<kPossibleRewards> newBits = dp;      // Remove the numbers >= the current number.      newBits <<= kPossibleRewards - num;      newBits >>= kPossibleRewards - num;      dp |= newBits << num;    }     for (int ans = kPossibleRewards - 1; ans >= 0; --ans)      if (dp[ans])        return ans;     throw;  }};

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