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Problem solution · C++

Check If Digits Are Equal in String After Operations II

Check If Digits Are Equal in String After Operations II: 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
56 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 Check If Digits Are Equal in String After Operations II, 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

  • 56 lines of C++ from the credited upstream file 3463.cpp.
  • The implementation keeps its working state in language-native values and containers.
  • 3 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 codeCheck If Digits Are Equal in String After Operations II · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  // Same as 3461. Check If Digits Are Equal in String After Operations I  bool hasSameDigits(const string& s) {    const int n = s.length();    int num1 = 0;    int num2 = 0;     for (int i = 0; i + 1 < n; ++i) {      const int coefficient = nCkMod10(n - 2, i);      num1 += (coefficient * (s[i] - '0')) % 10;      num1 %= 10;      num2 += (coefficient * (s[i + 1] - '0')) % 10;      num2 %= 10;    }     return num1 == num2;  }  private:  // Returns (n, k) % 10.  int nCkMod10(int n, int k) {    const int mod2 = lucasTheorem(n, k, 2);    const int mod5 = lucasTheorem(n, k, 5);    static constexpr int lookup[2][5] = {        {0, 6, 2, 8, 4},  // mod2 == 0        {5, 1, 7, 3, 9}   // mod2 == 1    };    return lookup[mod2][mod5];  }   // Returns (n, k) % prime.  int lucasTheorem(int n, int k, int prime) {    int res = 1;    while (n > 0 || k > 0) {      const int nMod = n % prime;      const int kMod = k % prime;      res *= nCk(nMod, kMod);      res %= prime;      n /= prime;      k /= prime;    }    return res;  }   // Returns (n, k).  int nCk(int n, int k) {    int res = 1;    for (int i = 0; i < k; ++i) {      res *= (n - i);      res /= (i + 1);    }    return res;  }};

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