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Intuition\n1. Goal: We want to find out how long a string will become after doing t transformations. Each transformation changes every character:\n\n- If the ch
# Intuition\n1. Goal: We want to find out how long a string will become after doing t transformations. Each transformation changes every character:\n\n- If the character is \'z\', it becomes "ab".\n- Otherwise, it changes to the next letter in the alphabet (e.g., \'a\' to \'b\', \'b\' to \'c\').\n\n2. Challenges: Since each transformation can quickly make the string huge, we can\u2019t just keep transforming the string. Instead, we calculate the length directly using a mathematical approach.\n\n3. Recursive Function (getLength):\n\n- This function calculates the length after transforming a specific character t times.\n- It remembers previous calculations (using a table called dp) so that if we need the length for the same character and transformation count again, it can reuse the answer instead of recalculating.\n- This helps speed up the solution.\n4. Transformation Steps:\n\n- For each character, it checks how many transformations (t) are needed.\n- If t is big enough, it simulates what happens if the character reaches or changes to \'z\' and splits into "ab".\n- Otherwise, it counts the length change for a simple next-letter transformation.\n\n\n\n\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```cpp []\nclass Solution {\npublic:\n    int mod = 1000000007;\n\n    int getLength(int t, char c, vector<vector<int>>& dp) {\n        int ct = \'z\' - c;\n        int len = 0;\n        int ch = (c == \'a\') ? 0 : 1;\n       \n\n        if (dp[t][ch] != -1)\n            return dp[t][ch];\n\n        if (ct < t) {\n            int lenA = getLength(t - ct-1, \'a\', dp) % mod;\n            int lenB = getLength(t - ct-1, \'b\', dp) % mod;\n            len = (lenA + lenB) % mod;\n        } else {\n            len= 1;\n        }\n        \n        return dp[t][ch] = len;\n    }\n\n    int lengthAfterTransformations(string s, int t) {\n        long long len = 0;\n        vector<vector<int>> dp(t + 1, vector<int>(2, -1));\n\n        for (int i = 0; i < s.size(); i++) {\n            int ct = \'z\' - s[i];\n            if (t > ct) {\n                int lenA = getLength(t - ct-1, \'a\', dp) % mod;\n                int lenB = getLength(t - ct-1, \'b\', dp) % mod;\n                len = (len + lenA + lenB) % mod;\n            } else {\n                len = (len + 1) % mod;\n            }\n        }\n        return len;\n    }\n};\n\n```
['Math', 'Dynamic Programming', 'Recursion', 'Memoization', 'C++']