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Problem solution · C++

String Transformation

String Transformation: a C++ solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Dynamic programming
Source
walkccc LeetCode Solutions
Length
79 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

Dynamic programming

For String Transformation, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 79 lines of C++ from the credited upstream file 2851.cpp.
  • The implementation visibly relies on sequence storage, cached states.
  • 3 loop blocks detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

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 codeString Transformation · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  // This dynamic programming table dp[k][i] represents the number of ways to  // rearrange the string s after k steps such that it starts with s[i].  // A string can be rotated from 1 to n - 1 times. The transition rule is  // dp[k][i] = sum(dp[k - 1][j]) for all j != i. For example, when n = 4 and  // k = 3, the table looks like this:  //  // -----------------------------------------------------------  // |       | i = 0 | i = 1 | i = 2 | i = 3 | sum = (n - 1)^k |  // -----------------------------------------------------------  // | k = 0 |   1   |   0   |   0   |   0   |        1        |  // | k = 1 |   0   |   1   |   1   |   1   |        3        |  // | k = 2 |   3   |   2   |   2   |   2   |        9        |  // | k = 3 |   6   |   7   |   7   |   7   |       27        |  // -----------------------------------------------------------  //  // By observation, we have  //   * dp[k][!0] = ((n - 1)^k - (-1)^k) / n  //   * dp[k][0] = dp[k][!0] + (-1)^k  int numberOfWays(string s, string t, long long k) {    const int n = s.length();    const int negOnePowK = (k % 2 == 0 ? 1 : -1);  // (-1)^k    const vector<int> z = zFunction(s + t + t);    const vector<int> indices = getIndices(z, n);    vector<int> dp(2);  // dp[0] := dp[k][0]; dp[1] := dp[k][!0]    dp[1] = (modPow(n - 1, k) - negOnePowK + kMod) % kMod *            modPow(n, kMod - 2) % kMod;    dp[0] = (dp[1] + negOnePowK + kMod) % kMod;    return accumulate(indices.begin(), indices.end(), 0L,                      [&](long acc, int index) {      return (acc + dp[index == 0 ? 0 : 1]) % kMod;    });  }  private:  static constexpr int kMod = 1'000'000'007;   long modPow(long x, long n) {    if (n == 0)      return 1;    if (n % 2 == 1)      return x * modPow(x, n - 1) % kMod;    return modPow(x * x % kMod, n / 2);  }   // Returns the z array, where z[i] is the length of the longest prefix of  // s[i..n) which is also a prefix of s.  //  // https://cp-algorithms.com/string/z-function.html#implementation  vector<int> zFunction(const string& s) {    const int n = s.length();    vector<int> z(n);    int l = 0;    int r = 0;    for (int i = 1; i < n; ++i) {      if (i < r)        z[i] = min(r - i, z[i - l]);      while (i + z[i] < n && s[z[i]] == s[i + z[i]])        ++z[i];      if (i + z[i] > r) {        l = i;        r = i + z[i];      }    }    return z;  }   // Returns the indices in `s` s.t. for each `i` in the returned indices,  // `s[i..n) + s[0..i) = t`.  vector<int> getIndices(const vector<int>& z, int n) {    vector<int> indices;    for (int i = n; i < n + n; ++i)      if (z[i] >= n)        indices.push_back(i - n);    return indices;  }};

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