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Problem solution · Java

Total Characters in String After Transformations II

Total Characters in String After Transformations II: a Java 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
60 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 Total Characters in String After Transformations 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

  • 60 lines of Java from the credited upstream file 3337.java.
  • The implementation visibly relies on sequence storage.
  • 9 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 codeTotal Characters in String After Transformations II · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  // Similar to 3335. Total Characters in String After Transformations I  public int lengthAfterTransformations(String s, int t, List<Integer> nums) {    // T[i][j] := the number of ways to transform ('a' + i) to ('a' + j)    int[][] T = getTransformationMatrix(nums);    int[][] poweredT = matrixPow(T, t);    int[] count = new int[26];    // lengths[i] := the total length of ('a' + i) after t transformations    long[] lengths = new long[26];     for (final char c : s.toCharArray())      ++count[c - 'a'];     for (int i = 0; i < 26; ++i)      for (int j = 0; j < 26; ++j) {        lengths[j] += (long) count[i] * poweredT[i][j];        lengths[j] %= MOD;      }     return (int) (Arrays.stream(lengths).sum() % MOD);  }   private static final int MOD = 1_000_000_007;   private int[][] getTransformationMatrix(List<Integer> nums) {    int[][] T = new int[26][26];    for (int i = 0; i < nums.size(); ++i)      for (int step = 1; step <= nums.get(i); ++step)        ++T[i][(i + step) % 26];    return T;  }   private int[][] getIdentityMatrix(int sz) {    int[][] I = new int[sz][sz];    for (int i = 0; i < sz; ++i)      I[i][i] = 1;    return I;  }   // Returns A * B.  private int[][] matrixMult(int[][] A, int[][] B) {    final int sz = A.length;    int[][] C = new int[sz][sz];    for (int i = 0; i < sz; ++i)      for (int j = 0; j < sz; ++j)        for (int k = 0; k < sz; ++k)          C[i][j] = (int) ((C[i][j] + (long) A[i][k] * B[k][j]) % MOD);    return C;  }   // Returns M^n.  private int[][] matrixPow(int[][] M, int n) {    if (n == 0)      return getIdentityMatrix(M.length);    if (n % 2 == 1)      return matrixMult(M, matrixPow(M, n - 1));    return matrixPow(matrixMult(M, M), n / 2);  }}

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