Easy Solution - DP with Recursion (Fastest Execution 12ms)
Approach\nThe simple approach to solve this problem to identify that we can compute result for the string "ab" for a given t recursively.\n\nWe used memoization
# Approach\nThe simple approach to solve this problem to identify that we can compute result for the string "ab" for a given t recursively.\n\nWe used memoization to reduce the time complexity and stored the precomputed results. \n\nHope this helps you. I will answer all your queries. Just ask in the comments\n\n# Complexity\n- Time complexity:\nO(nlogt)\n\n- Space complexity:\nO(n + t)\n\n# Code\n```javascript []\n/**\n * @param {string} s\n * @param {number} t\n * @return {number}\n */\nconst MOD = 1e9 + 7;\n\nconst memo = new Array(1e5 + 1).fill(-1);\nconst computeForAB = (t) => {\n    if (memo[t] > -1) return memo[t];\n    const s = \'ab\';\n    \n    let change = 0;\n    for (let i = 0; i < 2; i++) {\n        const idx = s.charCodeAt(i) - 97;\n        if (idx + t > 25) {\n            let rem = t - (26 - idx);\n            change = (change + computeForAB(rem) + 1) % MOD;\n        }\n    }\n\n    memo[t] = change;\n    return change;\n}\n\nvar lengthAfterTransformations = function(s, t) {\n    const n = s.length;\n    /**\n        z -> ab -> yz -> zab -> abbc -> xyyz -> yzzab -> zababbc -> abbcbccd\n        0.        2.         23          \n        abcyy -> cdeabab -> xyzuwuw -> \n    */\n    let res = n;\n    for (let i = 0; i < n; i++) {\n        const idx = s.charCodeAt(i) - 97;\n        if (idx + t > 25) {\n            let rem = t - (26 - idx);\n            res = (res + computeForAB(rem) + 1) % MOD;\n        }\n    }\n\n    return res;\n};\n```\n\n```java []\npublic class Transformation {\n    private static final long MOD = (long) 1e9 + 7;\n    private static long[] memo = new long[100001];\n\n    static {\n        for (int i = 0; i < memo.length; i++) {\n            memo[i] = -1;\n        }\n    }\n\n    private static long computeForAB(int t) {\n        if (memo[t] > -1) return memo[t];\n        String s = "ab";\n\n        long change = 0;\n        for (int i = 0; i < 2; i++) {\n            int idx = s.charAt(i) - \'a\';\n            if (idx + t > 25) {\n                int rem = t - (26 - idx);\n                change = (change + computeForAB(rem) + 1) % MOD;\n            }\n        }\n\n        memo[t] = change;\n        return change;\n    }\n\n    public static long lengthAfterTransformations(String s, int t) {\n        int n = s.length();\n        long res = n;\n\n        for (int i = 0; i < n; i++) {\n            int idx = s.charAt(i) - \'a\';\n            if (idx + t > 25) {\n                int rem = t - (26 - idx);\n                res = (res + computeForAB(rem) + 1) % MOD;\n            }\n        }\n\n        return res;\n    }\n\n    public static void main(String[] args) {\n        String s = "yourString"; // Replace with your input string\n        int t = yourTransformationValue; // Replace with your t value\n        System.out.println(lengthAfterTransformations(s, t));\n    }\n}\n```\n\n```python []\nMOD = int(1e9 + 7)\nmemo = [-1] * (100001)\n\ndef compute_for_ab(t):\n    if memo[t] > -1:\n        return memo[t]\n    \n    s = \'ab\'\n    change = 0\n\n    for i in range(2):\n        idx = ord(s[i]) - ord(\'a\')\n        if idx + t > 25:\n            rem = t - (26 - idx)\n            change = (change + compute_for_ab(rem) + 1) % MOD\n\n    memo[t] = change\n    return change\n\ndef length_after_transformations(s, t):\n    n = len(s)\n    res = n\n\n    for i in range(n):\n        idx = ord(s[i]) - ord(\'a\')\n        if idx + t > 25:\n            rem = t - (26 - idx)\n            res = (res + compute_for_ab(rem) + 1) % MOD\n\n    return res\n\n# Example usage\ns = "yourString"  # Replace with your input string\nt = your_transformation_value  # Replace with your t value\nprint(length_after_transformations(s, t))\n```\n\n```C++ []\n#include <iostream>\n#include <vector>\n#include <string>\n\nconst long long MOD = 1e9 + 7;\nstd::vector<long long> memo(100001, -1);\n\nlong long computeForAB(int t) {\n    if (memo[t] > -1) return memo[t];\n    std::string s = "ab";\n    \n    long long change = 0;\n    for (int i = 0; i < 2; i++) {\n        int idx = s[i] - \'a\';\n        if (idx + t > 25) {\n            int rem = t - (26 - idx);\n            change = (change + computeForAB(rem) + 1) % MOD;\n        }\n    }\n\n    memo[t] = change;\n    return change;\n}\n\nlong long lengthAfterTransformations(const std::string& s, int t) {\n    int n = s.length();\n    long long res = n;\n\n    for (int i = 0; i < n; i++) {\n        int idx = s[i] - \'a\';\n        if (idx + t > 25) {\n            int rem = t - (26 - idx);\n            res = (res + computeForAB(rem) + 1) % MOD;\n        }\n    }\n\n    return res;\n}\n\nint main() {\n    std::string s = "yourString"; // Replace with your input string\n    int t = yourTransformationValue; // Replace with your t value\n    std::cout << lengthAfterTransformations(s, t) << std::endl;\n    return 0;\n}\n```\n\n![upvote.jpeg](https://assets.leetcode.com/users/images/ea12736d-2147-42b9-aaf7-565bebe53b2b_1730002235.1330783.jpeg)\n\n
['Dynamic Programming', 'Python', 'C++', 'Java', 'JavaScript']