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Problem solution · C++

Count the Number of Ideal Arrays

Count the Number of Ideal Arrays: 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
64 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 Count the Number of Ideal Arrays, 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

  • 64 lines of C++ from the credited upstream file 2338.cpp.
  • The implementation visibly relies on sequence storage, cached states.
  • 9 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 codeCount the Number of Ideal Arrays · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int idealArrays(int n, int maxValue) {    // Since 2^14 > 10^4, the longest strictly increasing array is [1, 2, 4,    // ..., 2^13]    const int maxLength = min(14, n);    const vector<vector<int>> factors = getFactors(maxValue);    // dp[i][j] := the number of strictly increasing ideal arrays of length i    // ending in j    // dp[i][j] := sum(dp[i - 1][k]), where j % k == 0    // dp[i][0] := sum(dp[i][j]) where 1 <= j <= maxValue    vector<vector<long>> dp(maxLength + 1, vector<long>(maxValue + 1));    vector<vector<long>> mem(n, vector<long>(maxLength, -1));    long ans = 0;     for (int j = 1; j <= maxValue; ++j)      dp[1][j] = 1;     for (int i = 2; i <= maxLength; ++i)      for (int j = 1; j <= maxValue; ++j)        for (const int k : factors[j]) {          dp[i][j] += dp[i - 1][k];          dp[i][j] %= kMod;        }     for (int i = 1; i <= maxLength; ++i)      for (int j = 1; j <= maxValue; ++j) {        dp[i][0] += dp[i][j];        dp[i][0] %= kMod;      }     for (int i = 1; i <= maxLength; ++i) {      // nCk(n - 1, i - 1) := the number of ways to create an ideal array of      // length n from a strictly increasing array of length i      ans += dp[i][0] * nCk(n - 1, i - 1, mem);      ans %= kMod;    }     return ans;  }  private:  static constexpr int kMod = 1'000'000'007;   vector<vector<int>> getFactors(int maxValue) {    vector<vector<int>> factors(maxValue + 1);    for (int i = 1; i <= maxValue; ++i)      // Start from i * 2 because of strictly increasing.      for (int j = i * 2; j <= maxValue; j += i)        factors[j].push_back(i);    return factors;  }   long nCk(int n, int k, vector<vector<long>>& mem) {    if (k == 0)      return 1;    if (n == k)      return 1;    if (mem[n][k] != -1)      return mem[n][k];    return mem[n][k] = (nCk(n - 1, k, mem) + nCk(n - 1, k - 1, mem)) % kMod;  }};

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