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Problem solution · C++

Unique Substrings With Equal Digit Frequency

Unique Substrings With Equal Digit Frequency: 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
63 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 Unique Substrings With Equal Digit Frequency, 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

  • 63 lines of C++ from the credited upstream file 2168.cpp.
  • The implementation visibly relies on sequence storage, hash lookup.
  • 5 loop blocks 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 codeUnique Substrings With Equal Digit Frequency · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int equalDigitFrequency(string s) {    vector<vector<int>> counts;  // counts[i] := the counter map of s[0..i]    vector<int> count(10);    vector<long> pows{1};  // pows[i] := kBase^i % kHash    // hash[i] = the hash of the first i letters of s, where hash[i] =    // (26^(i - 1) * s[0] + 26^(i - 2) * s[1] + ... + s[i - 1]) % kHash    vector<long> hash{0};    unordered_set<int> seen;     for (const char c : s) {      ++count[c - '0'];      counts.push_back(count);      pows.push_back(pows.back() * kBase % kHash);      hash.push_back((hash.back() * kBase + val(c)) % kHash);    }     for (int i = 0; i < s.length(); ++i)      for (int j = i; j < s.length(); ++j)        if (isSameFreq(counts, i, j))          seen.insert(getRollingHash(i, j + 1, hash, pows));     return seen.size();  }  private:  static constexpr int kMax = 1001;  static constexpr int kBase = 11;  static constexpr int kHash = 1'000'000'007;   static constexpr int val(char c) {    return c - '0' + 1;  }   // Returns true if s[i..j] has the same digit frequency.j  bool isSameFreq(const vector<vector<int>>& counts, int i, int j) {    vector<int> count = counts[j];    if (i > 0)      for (int num = 0; num < 10; ++num)        count[num] -= counts[i - 1][num];    return equalFreq(count);  }   bool equalFreq(const vector<int>& count) {    int minfreq = kMax;    int maxfreq = 0;    for (const int freq : count)      if (freq > 0) {        minfreq = min(minfreq, freq);        maxfreq = max(maxfreq, freq);      }    return minfreq == maxfreq;  }   // Returns the rolling hash of s[l..r).  int getRollingHash(int l, int r, const vector<long>& hash,                     const vector<long>& pows) {    const long h = (hash[r] - hash[l] * pows[r - l]) % kHash;    return h < 0 ? h + kHash : h;  }};

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