Total Characters in String After Transformations II

HARD

Description

You are given a string s consisting of lowercase English letters, an integer t representing the number of transformations to perform, and an array nums of size 26. In one transformation, every character in s is replaced according to the following rules:

Replace s[i] with the next nums[s[i] - 'a'] consecutive characters in the alphabet. For example, if s[i] = 'a' and nums[0] = 3, the character 'a' transforms into the next 3 consecutive characters ahead of it, which results in "bcd".
The transformation wraps around the alphabet if it exceeds 'z'. For example, if s[i] = 'y' and nums[24] = 3, the character 'y' transforms into the next 3 consecutive characters ahead of it, which results in "zab".

Return the length of the resulting string after exactly t transformations.

Since the answer may be very large, return it modulo 10⁹ + 7.

Example 1:

Input: s = "abcyy", t = 2, nums = [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2]

Output: 7

Explanation:

First Transformation (t = 1):
- 'a' becomes 'b' as nums[0] == 1
- 'b' becomes 'c' as nums[1] == 1
- 'c' becomes 'd' as nums[2] == 1
- 'y' becomes 'z' as nums[24] == 1
- 'y' becomes 'z' as nums[24] == 1
- String after the first transformation: "bcdzz"
Second Transformation (t = 2):
- 'b' becomes 'c' as nums[1] == 1
- 'c' becomes 'd' as nums[2] == 1
- 'd' becomes 'e' as nums[3] == 1
- 'z' becomes 'ab' as nums[25] == 2
- 'z' becomes 'ab' as nums[25] == 2
- String after the second transformation: "cdeabab"
Final Length of the string: The string is "cdeabab", which has 7 characters.

Example 2:

Input: s = "azbk", t = 1, nums = [2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2]

Output: 8

Explanation:

First Transformation (t = 1):
- 'a' becomes 'bc' as nums[0] == 2
- 'z' becomes 'ab' as nums[25] == 2
- 'b' becomes 'cd' as nums[1] == 2
- 'k' becomes 'lm' as nums[10] == 2
- String after the first transformation: "bcabcdlm"
Final Length of the string: The string is "bcabcdlm", which has 8 characters.

Constraints:

1 <= s.length <= 10⁵
s consists only of lowercase English letters.
1 <= t <= 10⁹
nums.length == 26
1 <= nums[i] <= 25

Approaches

Checkout 3 different approaches to solve Total Characters in String After Transformations II. Click on different approaches to view the approach and algorithm in detail.

Brute Force Simulation

This approach directly simulates the transformation process as described in the problem. It starts with the initial string s and, for t times, generates a new string by replacing each character of the current string with its corresponding transformation.

Algorithm

Initialize currentString = s.
Loop t times (from 1 to t): a. Create an empty StringBuilder called nextString. b. For each character c in currentString: i. Get the transformation length k = nums[c - 'a']. ii. For j from 1 to k: - Calculate the next character: nextChar = 'a' + ((c - 'a' + j) % 26). - Append nextChar to nextString. c. Update currentString = nextString.toString().
Return currentString.length().

The algorithm iterates t times. In each iteration, it constructs a new string. It traverses the current string character by character. For each character c, it determines the sequence of new characters based on nums[c - 'a']. The new characters are (c+1)%26, (c+2)%26, etc., wrapping around the alphabet from 'z' to 'a'. These new characters are appended to a temporary string builder. After iterating through all characters of the current string, the temporary string builder's content becomes the new current string for the next iteration. This process is repeated t times. Finally, the length of the resulting string is returned. Due to the potentially massive growth in string length and the large value of t, this approach is extremely slow and memory-intensive. It's not feasible for the given constraints and will result in Time Limit Exceeded (TLE) and Memory Limit Exceeded (MLE) errors.

// This is a conceptual implementation and will not pass due to performance issues.
public int totalCharacters(String s, int t, int[] nums) {
    String currentString = s;
    for (int i = 0; i < t; i++) {
        StringBuilder nextString = new StringBuilder();
        for (char c : currentString.toCharArray()) {
            int len = nums[c - 'a'];
            for (int j = 1; j <= len; j++) {
                char nextChar = (char) ('a' + (c - 'a' + j) % 26);
                nextString.append(nextChar);
            }
        }
        currentString = nextString.toString();
        // The length can exceed memory limits long before t iterations.
    }
    return currentString.length(); // Modulo arithmetic is omitted for clarity of the basic idea.
}

Complexity Analysis

Time Complexity: O(t * L_avg * max(nums)), where `L_avg` is the average length of the string across transformations. Since the length can grow exponentially, this is infeasible.Space Complexity: O(L_max), where `L_max` is the maximum length of the string, which can be huge.

Pros and Cons

Pros:

Simple to understand and implement.
Directly follows the problem description.

Cons:

Extremely inefficient for large t as it requires t iterations.
The string length can grow exponentially, leading to Memory Limit Exceeded.
String concatenations in a loop are slow, leading to Time Limit Exceeded.

Code Solutions

Checking out 3 solutions in different languages for Total Characters in String After Transformations II. Click on different languages to view the code.

class Solution {
private
  final int mod = (int)1 e9 + 7;
public
  int lengthAfterTransformations(String s, int t, List<Integer> nums) {
    final int m = 26;
    int[] cnt = new int[m];
    for (char c : s.toCharArray()) {
      cnt[c - 'a']++;
    }
    int[][] matrix = new int[m][m];
    for (int i = 0; i < m; i++) {
      int num = nums.get(i);
      for (int j = 1; j <= num; j++) {
        matrix[i][(i + j) % m] = 1;
      }
    }
    int[][] factor = matpow(matrix, t, m);
    int[] result = vectorMatrixMultiply(cnt, factor);
    int ans = 0;
    for (int val : result) {
      ans = (ans + val) % mod;
    }
    return ans;
  }
private
  int[][] matmul(int[][] a, int[][] b) {
    int n = a.length;
    int p = b.length;
    int q = b[0].length;
    int[][] res = new int[n][q];
    for (int i = 0; i < n; i++) {
      for (int k = 0; k < p; k++) {
        if (a[i][k] == 0)
          continue;
        for (int j = 0; j < q; j++) {
          res[i][j] = (int)((res[i][j] + 1L * a[i][k] * b[k][j]) % mod);
        }
      }
    }
    return res;
  }
private
  int[][] matpow(int[][] mat, int power, int m) {
    int[][] res = new int[m][m];
    for (int i = 0; i < m; i++) {
      res[i][i] = 1;
    }
    while (power > 0) {
      if ((power & 1) != 0) {
        res = matmul(res, mat);
      }
      mat = matmul(mat, mat);
      power >>= 1;
    }
    return res;
  }
private
  int[] vectorMatrixMultiply(int[] vector, int[][] matrix) {
    int n = matrix.length;
    int[] result = new int[n];
    for (int i = 0; i < n; i++) {
      long sum = 0;
      for (int j = 0; j < n; j++) {
        sum = (sum + 1L * vector[j] * matrix[j][i]) % mod;
      }
      result[i] = (int)sum;
    }
    return result;
  }
}

Video Solution

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Patterns:

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Data Structures:

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