Simple and Clean Java Solution | O(26*t)
Intuition\nThe key observation is that transforming characters incrementally each time results in a predictable shift through the alphabet. Special cases arise 
### Intuition\nThe key observation is that transforming characters incrementally each time results in a predictable shift through the alphabet. Special cases arise only with the letter `z`, which becomes two characters (`ab`). By keeping track of counts of each character in the alphabet, we can avoid creating strings repeatedly, saving on time and memory usage.\n\n### Approach\n1. **Character Frequency Array**: Use an array to store the frequency of each character in the string `s` as they transform over time.\n2. **Transformation Steps**:\n   - For each transformation:\n     - Update the character counts based on the transformation rule:\n       - Each character increments to the next character in the alphabet, except `z`, which becomes `ab`.\n     - This is achieved by shifting the counts to the next index and handling `z` by adding its count to both `a` and `b`.\n3. **Calculate Resulting Length**:\n   - After `t` transformations, sum up all character counts and return the result modulo \\(10^9 + 7\\).\n\n### Complexity Analysis\n- **Time Complexity**: \\(O(26*t)\\), which simplifies to \\(O(t)\\) because we process a constant number of 26 alphabet letters for each of the `t` transformations.\n- **Space Complexity**: \\(O(26) = O(1)\\), since we only store 26 counts.\n\n### Code\n\n```java\nclass Solution {\n    int mod = (int)(1e9) + 7;\n\n    public int lengthAfterTransformations(String s, int t) {\n        // Array to keep count of each letter from \'a\' to \'z\'\n        int[] alpha = new int[26];\n        \n        // Initialize count of characters in the given string\n        for (char ch: s.toCharArray()) {\n            alpha[ch - \'a\']++;\n        }\n\n        // Perform transformations t times\n        for (int i = 0; i < t; i++) {\n            int[] newAlpha = new int[26];\n            \n            // Process each character\'s count\n            for (int j = 0; j < 26; j++) {\n                if (j < 25) {\n                    // Shift the count to the next character in the alphabet\n                    newAlpha[j + 1] = alpha[j];\n                } else {\n                    // For \'z\', add its count to \'a\' and \'b\'\n                    newAlpha[0] = (newAlpha[0] + alpha[j]) % mod;\n                    newAlpha[1] = (newAlpha[1] + alpha[j]) % mod;\n                }\n            }\n            // Update alpha with newAlpha values for next transformation\n            alpha = newAlpha;\n        }\n\n        // Sum the counts to get the total length of the transformed string\n        int len = 0;\n        for (int a : alpha) {\n            len = (len + a) % mod;\n        }\n        return len;\n    }\n}\n```\n\nThis solution efficiently calculates the length of the resulting string without constructing it, which is crucial for handling large values of `t` and long initial strings.
['Java']