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2719. Count of Integers 👍

  • Time: $O(|\texttt{num2}| \cdot \texttt{max\_sum} \cdot 2^2 \cdot 10)$
  • Space: $O(|\texttt{num2}| \cdot \texttt{max\_sum} \cdot 2^2)$
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class Solution {
 public:
  int count(string num1, string num2, int min_sum, int max_sum) {
    const string num1WithLeadingZeros =
        string(num2.length() - num1.length(), '0') + num1;
    // dp[i][j][k1][k2] := # of valid integers with size i and at most sum of j.
    // The variables k1 and k2 indicate tight constraints (0/1) for `num1` and
    // `num2` respectively.
    dp.resize(num2.length(),
              vector<vector<vector<int>>>(
                  max_sum + 1, vector<vector<int>>(2, vector<int>(2, -1))));
    return count(num1WithLeadingZeros, num2, 0, max_sum, true, true) -
           count(num1WithLeadingZeros, num2, 0, min_sum - 1, true, true);
  }

 private:
  static constexpr int kMod = 1'000'000'007;
  vector<vector<vector<vector<int>>>> dp;

  int count(const string& num1, const string& num2, int i, int sum,
            bool isTight1, bool isTight2) {
    if (sum < 0)
      return 0;
    if (i == num2.length())
      return 1;
    if (dp[i][sum][isTight1][isTight2] != -1)
      return dp[i][sum][isTight1][isTight2];

    int res = 0;

    const int minDigit = isTight1 ? num1[i] - '0' : 0;
    const int maxDigit = isTight2 ? num2[i] - '0' : 9;
    for (int d = minDigit; d <= maxDigit; ++d) {
      const bool nextIsTight1 = isTight1 && (d == minDigit);
      const bool nextIsTight2 = isTight2 && (d == maxDigit);
      res += count(num1, num2, i + 1, sum - d, nextIsTight1, nextIsTight2);
      res %= kMod;
    }

    return dp[i][sum][isTight1][isTight2] = res;
  }
};