Easiest Solution 🔥✅ | Recursion + Memoization
Intuition\nTo find the final length of the string after t transformations, we need to track how each character evolves. Most characters simply shift forward in 
# Intuition\nTo find the final length of the string after `t` transformations, we need to track how each character evolves. Most characters simply shift forward in the alphabet, but \'z\' expands to "ab," increasing the length. By using recursion and caching, we can efficiently compute the length after all transformations.\n\n# Approach\nDefine a recursive helper function with caching to compute the length after transformations:\n1. For each character, if adding `t` transformations does not exceed \'z,\' simply add 1 to the count.\n2. If `t` transformations exceed \'z\' \n    - Calculate the remaining transformations needed after reaching \'z\'\n    - Recursively calculate the effect of replacing \'z\' with "ab" until transformations are exhausted.\n3. Return the total length using modulo\n\n# Complexity\n- Time complexity: $$O(n * t)$$, where n is the length of s and t is the number of transformations\n\n- Space complexity: $$O(n * t)$$, due to caching\n\n# Code\n```python3 []\nclass Solution:\n    def lengthAfterTransformations(self, s: str, t: int) -> int:\n        @cache\n        def helper(s, t):\n            ans = 0\n            for c in s:\n                if ord(c) + t <= ord("z"):\n                    ans += 1\n                else:\n                    new_t = t\n                    new_t -= ord("z") + 1 - ord(c)\n                    ans += helper("ab", new_t)\n            return ans\n        return helper(s, t) % ((10 ** 9) + 7)\n```\n```cpp []\n#include <string>\n#include <unordered_map>\n\nclass Solution {\npublic:\n    int lengthAfterTransformations(std::string s, int t) {\n        std::unordered_map<std::string, int> cache;\n        return helper(s, t, cache) % (1000000007);\n    }\n    \nprivate:\n    int helper(const std::string& s, int t, std::unordered_map<std::string, int>& cache) {\n        std::string key = std::to_string(t) + ":" + s;\n        if (cache.find(key) != cache.end()) {\n            return cache[key];\n        }\n\n        int ans = 0;\n        for (char c : s) {\n            if (c + t <= \'z\') {\n                ans += 1;\n            } else {\n                int new_t = t - (\'z\' + 1 - c);\n                ans += helper("ab", new_t, cache);\n                ans %= 1000000007;\n            }\n        }\n\n        cache[key] = ans % 1000000007;\n        return ans;\n    }\n};\n```\n```java []\nimport java.util.HashMap;\n\nclass Solution {\n    private static final int MOD = 1000000007;\n\n    public int lengthAfterTransformations(String s, int t) {\n        HashMap<String, Integer> cache = new HashMap<>();\n        return helper(s, t, cache) % MOD;\n    }\n\n    private int helper(String s, int t, HashMap<String, Integer> cache) {\n        String key = t + ":" + s;\n        if (cache.containsKey(key)) {\n            return cache.get(key);\n        }\n\n        int ans = 0;\n        for (char c : s.toCharArray()) {\n            if (c + t <= \'z\') {\n                ans++;\n            } else {\n                int newT = t - (\'z\' + 1 - c);\n                ans += helper("ab", newT, cache);\n                ans %= MOD;\n            }\n        }\n\n        cache.put(key, ans % MOD);\n        return ans;\n    }\n}\n\n```
['Recursion', 'Memoization', 'C++', 'Java', 'Python3']