Easy solution 
IntuitionThe key insight is that instead of actually performing string transformations, we can track the count of each character and how these counts change aft
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The key insight is that instead of actually performing string transformations, we can track the count of each character and how these counts change after each transformation. When we see a 'z', it produces two characters ('a' and 'b'), which means the length can grow over time. The trick is to handle these transitions efficiently using a count array.

# Approach
<!-- Describe your approach to solving the problem. -->
1.Create a count array to store the frequency of each character (0 to 25 representing 'a' to 'z')
2.For each transformation:
Create a new count array to store the results of the transformation
For each character position (0 to 25):
If current count is 0, skip it
  If it's 'z' (position 25), add its count to both 'a' and 'b' in the  new array
  Otherwise, move the count to the next character position
Update the original count array with the new counts
3.Finally, sum up all counts to get the final length
4.Handle modulo operations throughout to prevent overflow

# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
 O(t * k) where t is the number of transformations and k is the size of the alphabet (26 in this case). For each transformation, we need to process all possible characters.

- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
O(1) since we use two fixed-size vectors of size 26, regardless of input size. The space used doesn't grow with input size or number of transformations.

# Code
```cpp []
class Solution {
public:
    int lengthAfterTransformations(string s, int t) {
        const int MOD = 1e9 + 7;
        vector<long long> counts(26, 0);  
        
        
        for (char c : s) {
            counts[c - 'a']++;
        }
        
       
        while (t--) {
            vector<long long> newCounts(26, 0);
            
            
            for (int i = 0; i < 26; i++) {
                if (counts[i] == 0) continue;  
                
                if (i == 25) {  
                    newCounts[0] = (newCounts[0] + counts[i]) % MOD;  
                    newCounts[1] = (newCounts[1] + counts[i]) % MOD;  
                } else {
                  
                    newCounts[i + 1] = (newCounts[i + 1] + counts[i]) % MOD;
                }
            }
            
            counts = newCounts;
        }
        
        
        long long totalLength = 0;
        for (long long count : counts) {
            totalLength = (totalLength + count) % MOD;
        }
        
        return totalLength;
    }
};
```
['C++']