BEATS 80% SIMPLEST APPROACH WITHOUT DP
Intuition\nThe problem requires transforming a string based on specific character replacement rules. Each transformation can significantly change the length of 
**Intuition**\nThe problem requires transforming a string based on specific character replacement rules. Each transformation can significantly change the length of the string, particularly when characters like \'z\' are replaced with "ab". The key is to efficiently track how these transformations affect character counts without explicitly building the entire string for every transformation.\n\n**Approach**\n1. **Count Character Frequencies**: Use an array to count the occurrences of each character in the input string. This helps in efficiently managing the transformations.\n\n2. **Simulate Transformations**: For each transformation, iterate over the character count array:\n   - If a character is \'z\', it will create two \'a\'s and one \'b\', so update the counts for \'a\' and \'b\' accordingly.\n   - For all other characters, simply increment the count of the next character.\n\n3. **Repeat for t Transformations**: Continue this process for a total of `t` transformations.\n\n4. **Calculate Result**: After completing all transformations, sum up the counts of all characters to get the final length of the transformed string, ensuring to return the result modulo \\(10^9 + 7\\).\n\n**Complexity**\n- **Time complexity**: $$O(t \\cdot 26)$$  \nEach transformation requires processing 26 characters, and since we perform this for `t` transformations, the overall time complexity is linear with respect to the number of transformations and constant with respect to the character set size.\n\n- **Space complexity**: $$O(1)$$  \nThe space used for the character count array is constant (size 26), which does not scale with the input size. Therefore, the space complexity remains constant.\n\n# Code\n```cpp []\ntypedef long long ll;\nconst ll mod = 1e9 + 7;\nll mod_add(ll a, ll b, ll m = mod) {a = a % m; b = b % m; return (((a + b) % m) + m) % m;}\n\nclass Solution {\npublic:\n    int lengthAfterTransformations(string s, int t) {\n        int nums[26] = {0};\n        for (char ch : s) nums[ch - \'a\']++;\n        while (t--) {\n            int cur[26] = {0};\n            for (int j = 0; j < 26; j++) {\n                if (j == 25 && nums[j] > 0) {\n                    cur[0] = mod_add(cur[0], nums[j]);\n                    cur[1] = mod_add(cur[1], nums[j]);\n                } else {\n                    if (j < 25) cur[j + 1] = mod_add(cur[j + 1], nums[j]);\n                }\n            }\n            for (int j = 0; j < 26; j++) nums[j] = cur[j];\n        }\n        ll cnt = 0;\n        for (int i : nums) cnt = mod_add(cnt, i);\n        return (int)cnt;\n    }\n};\n\n```\n```Python []\nclass Solution:\n    MOD = 10**9 + 7\n\n    def mod_add(self, a, b):\n        a %= self.MOD\n        b %= self.MOD\n        return (a + b) % self.MOD\n\n    def lengthAfterTransformations(self, s: str, t: int) -> int:\n        nums = [0] * 26\n        for ch in s:\n            nums[ord(ch) - ord(\'a\')] += 1\n        while t > 0:\n            t -= 1\n            cur = [0] * 26\n            for j in range(26):\n                if j == 25 and nums[j] > 0:\n                    cur[0] = self.mod_add(cur[0], nums[j])\n                    cur[1] = self.mod_add(cur[1], nums[j])\n                elif j < 25:\n                    cur[j + 1] = self.mod_add(cur[j + 1], nums[j])\n            nums = cur\n        return sum(nums) % self.MOD\n\n```\n```Java []\nimport java.util.Arrays;\n\nclass Solution {\n    private static final long MOD = 1000000007;\n\n    private long modAdd(long a, long b, long m) {\n        a = a % m;\n        b = b % m;\n        return (((a + b) % m) + m) % m;\n    }\n\n    public int lengthAfterTransformations(String s, int t) {\n        int[] nums = new int[26];\n        for (char ch : s.toCharArray()) {\n            nums[ch - \'a\']++;\n        }\n        while (t-- > 0) {\n            int[] cur = new int[26];\n            for (int j = 0; j < 26; j++) {\n                if (j == 25 && nums[j] > 0) {\n                    cur[0] = (int) modAdd(cur[0], nums[j], MOD);\n                    cur[1] = (int) modAdd(cur[1], nums[j], MOD);\n                } else {\n                    if (j < 25) {\n                        cur[j + 1] = (int) modAdd(cur[j + 1], nums[j], MOD);\n                    }\n                }\n            }\n            nums = Arrays.copyOf(cur, cur.length);\n        }\n        long cnt = 0;\n        for (int i : nums) {\n            cnt = modAdd(cnt, i, MOD);\n        }\n        return (int) cnt;\n    }\n}\n\n\n\n```\n\n
['Python', 'C++', 'Java']