Easy C++ Solution | Full Explanation
Intuition\nThe main challenge in this problem is managing the transformations efficiently, especially with potentially large values of \( t \). Replacing charac
### Intuition\nThe main challenge in this problem is managing the transformations efficiently, especially with potentially large values of \\( t \\). Replacing characters repeatedly could lead to exponential growth in the string\'s length, which would be impractical to compute directly. Instead, we can track character counts without directly constructing the transformed string.\n\n### Approach\n1. **Character Frequency Array**: We use a frequency array `freq` of size 26, where `freq[i]` represents the count of the character corresponding to the \\( i \\)-th position in the alphabet (`a` to `z`).\n2. **Initialize Frequency**: We count the initial frequency of each character in the string `s`.\n3. **Transformation Simulation**:\n   - For each transformation:\n     - Store the count of \'z\' characters in a temporary variable `c` (since \'z\' transforms to "ab").\n     - Shift all character counts up by one position, effectively making each character the "next" character.\n     - Update `freq[1]` (representing \'b\') by adding the count of `c` (for the "ab" replacement).\n     - Update `freq[0]` (representing \'a\') with the count of `c`.\n4. **Answer Calculation**: Track the total length of the resulting string after every transformation, using the modulus \\( 10^9 + 7 \\) to avoid overflow.\n\n### Complexity\n- **Time complexity**: \\( O(t) \\), where \\( t \\) is the number of transformations. Each transformation involves shifting values in the frequency array, which is constant in size (26 elements).\n- **Space complexity**: \\( O(1) \\) (constant space), as we only use the frequency array and a few variables.\n\n### Code\n```cpp\ntypedef long long ll;\n\nclass Solution {\npublic:\n    const ll N = 1e9+7;\n\n    int lengthAfterTransformations(string s, int t) {\n        // Initialize frequency array for characters \'a\' to \'z\'\n        vector<ll> freq(26);\n        ll n = s.length();\n\n        // Count the frequency of each character in the initial string\n        for (ll i = 0; i < n; i++) {\n            freq[s[i] - \'a\']++;\n        }\n\n        // Initialize answer with the initial string length\n        ll ans = n;\n\n        // Perform transformations for \'t\' times\n        while (t--) {\n            // Store the count of \'z\' characters (that will turn into "ab")\n            ll c = freq[25];\n            \n            // Shift character frequencies by one position\n            for (ll i = 25; i >= 1; i--) {\n                freq[i] = freq[i - 1];\n            }\n\n            // Update \'a\' and \'b\' positions due to "ab" transformation\n            freq[1] = (freq[1] % N + c % N) % N;  // Adding to \'b\' count\n            freq[0] = c;                           // Setting \'a\' count to \'z\' count\n\n            // Update total length with the newly added characters\n            ans = (ans % N + c % N) % N;\n        }\n\n        return ans;\n    }\n};\n```\n
['C++']