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Problem solution · C++

CCC 2017 S2 - High Tide, Low Tide

CCC 2017 S2 - High Tide, Low Tide: a C++ solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to CCCSolutions.

Technique
Dynamic programming
Source
CCCSolutions
Length
46 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 CCC 2017 S2 - High Tide, Low Tide, 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

  • 46 lines of C++ from the credited upstream file ccc17s2.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 CCCSolutions by CCCSolutions contributors · Milliken Mills High School and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCCC 2017 S2 - High Tide, Low Tide · C++C++
Use this to learn the idea, then write your own version.
// CCC 2017 Senior 2: High Tide, Low Tide // by John Liao, Lo-Ellen Park S.S. // To rearrange the numbers alternatingly and increasingly, we sort the numbers from lowest to highest,// then, starting from the center, alternatingly return the small and higher numbers.  // Runtime: O(N logN)   due to sorting// Memory: O(N) #include<iostream>#include<vector>#include<algorithm> using namespace std; int main() {    int N;    cin >> N;      vector<int> data(N);     for (int i = 0; i < N; i++) {        cin >> data[i];    }     sort(data.begin(), data.end());     int i = (N - 1) / 2, j = (N - 1) / 2 + 1;    while (i >= 0) {        cout << data[i] << " ";         if (j < N) {            cout << data[j] << " ";        }         i--;         j++;     }     cout << endl;     return 0;}

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