Easy Approach | Beats 80% in Runtime | Proper Explaination
Intuitive C++ Solution with Frequency Array - Detailed Explanation\n\n## Approach\n\nInstead of actually transforming the string at each step, we can use a freq
# Intuitive C++ Solution with Frequency Array - Detailed Explanation\n\n## Approach\n\nInstead of actually transforming the string at each step, we can use a frequency array to track the count of each character and calculate the length efficiently. The key observation is that during each transformation:\n\n1. Every character except \'z\' becomes the next character in the alphabet\n2. Each \'z\' becomes "ab", increasing the length by 1\n3. We don\'t need the actual string, just the length and character frequencies\n\n\n## Time Complexity: O(n + t)\nwhere n is the length of string and t is the number of transformations\n\n    O(n) for counting initial frequencies\n    O(t * 26) = O(t) for processing transformations\n\n\n## Space Complexity: O(1)\n\n    We use a fixed-size array of 26 elements regardless of input size\n\n\n\n## Step-by-Step Example\nLet\'s take s = "zee", t = 2:\n\n### Initial state:\n\nfreq[] stores: e:2, z:1\nlen = 3\n\n\n### After 1st transformation:\n\nz becomes "ab": len += 1 (now 4)\ne becomes f\nNew frequencies: a:1, b:1, f:2\n\n\n### After 2nd transformation:\n\nNo z\'s, just shift letters\nFinal frequencies: b:1, c:1, g:2\nFinal length = 4\n\n\n\n## Key Points\n- Use frequency array instead of string manipulation\n- Only need to track length changes from \'z\' transformations\n\n- Proper modulo arithmetic to handle large numbers\n\n- Efficient shifting of frequencies for character transformations\n\n\n## Solution\n\n```cpp\nclass Solution {\npublic:\n    static const int MOD = 1000000007;\n    \n    int lengthAfterTransformations(string s, int t) {\n        // Initialize frequency array for each character\n        vector<long long> freq(26, 0);\n        \n        // Count frequency of each character\n        for (char c : s) {\n            freq[c - \'a\']++;\n        }\n        \n        // Initial length of string\n        long long len = s.length();\n        \n        // Process each transformation\n        for (int i = 0; i < t; i++) {\n            // Each \'z\' becomes "ab", so add number of z\'s to length\n            len = (len + freq[25]) % MOD;\n            \n            // Store count of \'z\'s before shifting\n            long long z_count = freq[25];\n            \n            // Shift all frequencies right (each character becomes next in alphabet)\n            for (int j = 25; j > 0; j--) {\n                freq[j] = freq[j - 1];\n            }\n            \n            // Handle \'z\' -> "ab" transformation:\n            // Each \'z\' produces one \'a\' and one \'b\'\n            freq[0] = z_count;  // \'a\' count\n            freq[1] = (freq[1] + z_count) % MOD;  // \'b\' count\n        }\n        \n        return len;\n    }\n};
['C++']