content: package codingquestions.leetcode;
  
  /*
  3335. Total Characters in String After Transformations I
  Solved
  Medium
  Topics
  Companies
  Hint
  You are given a string s and an integer t, representing the number of transformations to perform. In one transformation, every character in s is replaced according to the following rules:
  
  If the character is 'z', replace it with the string "ab".
  Otherwise, replace it with the next character in the alphabet. For example, 'a' is replaced with 'b', 'b' is replaced with 'c', and so on.
  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
  
  Output: 7
  
  Explanation:
  
  First Transformation (t = 1):
  'a' becomes 'b'
  'b' becomes 'c'
  'c' becomes 'd'
  'y' becomes 'z'
  'y' becomes 'z'
  String after the first transformation: "bcdzz"
  Second Transformation (t = 2):
  'b' becomes 'c'
  'c' becomes 'd'
  'd' becomes 'e'
  'z' becomes "ab"
  'z' becomes "ab"
  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
  
  Output: 5
  
  Explanation:
  
  First Transformation (t = 1):
  'a' becomes 'b'
  'z' becomes "ab"
  'b' becomes 'c'
  'k' becomes 'l'
  String after the first transformation: "babcl"
  Final Length of the string: The string is "babcl", which has 5 characters.
   
  
  Constraints:
  
  1 <= s.length <= 105
  s consists only of lowercase English letters.
  1 <= t <= 105
  */
  public class TotalCharactersInStringAfterTransformationsI {
      public int lengthAfterTransformations(String s, int t) {
          int[] cnt = new int[26];
          int MOD = 1000000007;
          for (int i = 0; i < s.length(); i++) {
              cnt[s.charAt(i) - 'a']++;
          }
          for (int i = 1; i <= t; i++) {
              int[] temp = new int[26];
              for (int j = 1; j < 26; j++)
                  temp[j] = cnt[j - 1];
  
              temp[0] = (temp[0] + cnt[25]) % MOD;
              temp[1] = (temp[1] + cnt[25]) % MOD;
              cnt = temp;
          }
          int res = 0;
          for (int i = 0; i < 26; i++)
              res = (res + cnt[i]) % MOD;
  
          return res;
      }
  }
  