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Problem solution · C++

Count Numbers with Non-Decreasing Digits

Count Numbers with Non-Decreasing Digits: 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
97 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 Count Numbers with Non-Decreasing Digits, 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

  • 97 lines of C++ from the credited upstream file 3519.cpp.
  • The implementation visibly relies on sequence storage.
  • 8 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.

Open problem ↗Original source ↗
Full codeCount Numbers with Non-Decreasing Digits · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int countNumbers(const string& l, const string& r, const int b) {    const vector<int> rDigits = convertToBaseB(r, b);    vector<int> lDigits = convertToBaseB(l, b);    vector<int> lMinus1Digits = convertToBaseB(decrement(l), b);    padToSameLength(lDigits, rDigits);    padToSameLength(lMinus1Digits, rDigits);    return (countWithMem(rDigits, b) - countWithMem(lMinus1Digits, b) + kMod) %           kMod;  }  private:  static constexpr int kMod = 1'000'000'007;  void padToSameLength(vector<int>& a, const vector<int>& b) {    a.insert(a.begin(), b.size() - a.size(), 0);  }   int countWithMem(const vector<int>& digits, const int b) {    vector<vector<vector<int>>> mem(digits.size(),                                    vector<vector<int>>(2, vector<int>(b, -1)));    return count(digits, 0, 0, true, b, mem);  }   int count(const vector<int>& num, int pos, int lastDigit, bool tight, int b,            vector<vector<vector<int>>>& mem) {    if (pos == num.size())      return 1;     if (mem[pos][tight][lastDigit] != -1)      return mem[pos][tight][lastDigit];     int res = 0;    const int limit = tight ? num[pos] : b - 1;     for (int d = lastDigit; d <= limit; d++) {      const bool newTight = tight && (d == limit);      res = (res + count(num, pos + 1, d, newTight, b, mem)) % kMod;    }     return mem[pos][tight][lastDigit] = res;  }   string decrement(string s) {    for (int i = s.length() - 1; i >= 0; --i) {      if (s[i] > '0') {        --s[i];        break;      } else {        s[i] = '9';      }    }    return s[0] == '0' && s.length() > 1 ? s.substr(1) : s;  }   vector<int> convertToBaseB(const string& numStr, const int b) {    vector<int> digits;    vector<int> currentNum(1, 0);     for (const char c : numStr) {      const int d = c - '0';       int carry = 0;      for (int i = 0; i < currentNum.size(); ++i) {        const long long product = (long long)currentNum[i] * 10 + carry;        currentNum[i] = product % b;        carry = product / b;      }       while (carry > 0) {        currentNum.push_back(carry % b);        carry /= b;      }       carry = d;      for (int i = 0; i < currentNum.size() && carry; ++i) {        const int sum = currentNum[i] + carry;        currentNum[i] = sum % b;        carry = sum / b;      }       while (carry > 0) {        currentNum.push_back(carry % b);        carry /= b;      }    }     for (int i = currentNum.size() - 1; i >= 0; --i)      digits.push_back(currentNum[i]);     if (digits.empty())      digits.push_back(0);     return digits;  }};

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