CPMagical CP Library

Problem solution · C++

Count Beautiful Numbers

Count Beautiful Numbers: a C++ solution using hash-based lookup. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Hash-based lookup
Source
walkccc LeetCode Solutions
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

Hash-based lookup

For Count Beautiful Numbers, the implementation stores previously seen values or frequencies in a hash table for direct membership and lookup operations.

  1. Decide the key that represents the information needed later.
  2. Update its count or stored state while scanning the input.
  3. Use constant-time expected lookups to detect matches or assemble the result.

Code notes

  • 46 lines of C++ from the credited upstream file 3490.cpp.
  • The implementation visibly relies on hash lookup.
  • 1 loop block detected.

Complexity

Expected hash operations are constant time, but the surrounding scan and the number of stored keys determine total work and memory.

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 Beautiful Numbers · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int beautifulNumbers(int l, int r) {    return count(to_string(r), 0, /*tight=*/true, /*isLeadingZero=*/true,                 /*hasZero=*/false, /*sum=*/0, /*prod=*/1, {}) -           count(to_string(l - 1), 0, /*tight=*/true, /*isLeadingZero=*/true,                 /*hasZero=*/false, /*sum=*/0, /*prod=*/1, {});  }  private:  int count(const string& s, int i, bool tight, bool isLeadingZero,            bool hasZero, int sum, int prod, unordered_map<string, int>&& mem) {    if (i == s.length()) {      if (isLeadingZero)        return 0;      return (hasZero || prod % sum == 0) ? 1 : 0;    }    const string key = hash(i, tight, isLeadingZero, hasZero, sum, prod);    if (!isLeadingZero && hasZero && !tight)      return mem[key] = pow(10, s.length() - i);    if (const auto it = mem.find(key); it != mem.end())      return it->second;     int res = 0;    const int maxDigit = tight ? s[i] - '0' : 9;     for (int d = 0; d <= maxDigit; ++d) {      const bool nextTight = tight && (d == maxDigit);      const bool nextIsLeadingZero = isLeadingZero && d == 0;      const bool nextHasZero = !nextIsLeadingZero && d == 0;      const int nextProd = nextIsLeadingZero ? 1 : prod * d;      res += count(s, i + 1, nextTight, nextIsLeadingZero, nextHasZero, sum + d,                   nextProd, std::move(mem));    }     return mem[key] = res;  }   string hash(int i, bool tight, bool isLeadingZero, bool hasZero, int sum,              int prod) {    return to_string(i) + "_" + (tight ? "1" : "0") + "_" +           (isLeadingZero ? "1" : "0") + "_" + (hasZero ? "1" : "0") + "_" +           to_string(sum) + "_" + to_string(prod);  }};

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗