Simple Approach : )
IntuitionThe problem requires us to find the maximum length of a substring that can be transformed from string 's' to string 't 'with a given maximum cost. The
# Intuition
The problem requires us to find the maximum length of a substring that can be transformed from string 's' to string 't 'with a given maximum cost. The cost is defined as the sum of the absolute differences between the ASCII values of the characters in the two strings at the same positions.

# Approach
Sliding Window Technique: We will use a sliding window approach to keep track of the current substring we are considering.

Cost Calculation: As we iterate through the characters of the strings, we will calculate the cost of transforming the substring from s to t.

Adjusting the Window: If the cost exceeds maxCost, we will move the start of the window forward to reduce the cost until it is within the allowed limit.

Max Length Tracking: Throughout the process, we will keep track of the maximum length of the valid substring found.

# Complexity
- Time complexity:
O(n)), where (n) is the length of the strings. We traverse the strings once.

- Space complexity:
(O(1)), as we are using a constant amount of extra space.

# Code
```python3 []
class Solution:
    def equalSubstring(self, s: str, t: str, maxCost: int) -> int:
        n = 0  # Start of the sliding window
        newCost = 0  # Current cost of the substring
        maxLength = 0  # Maximum length of valid substring found

        # Iterate through each character in the strings
        for i in range(len(s)):
            # Calculate the cost of transforming s[i] to t[i]
            newCost += abs(ord(s[i]) - ord(t[i]))

            # If the cost exceeds maxCost, move the start of the window
            while newCost > maxCost:
                newCost -= abs(ord(s[n]) - ord(t[n]))  # Remove the cost of the character at the start of the window
                n += 1  # Move the start of the window forward

            # Update the maximum length of the valid substring
            maxLength = max(maxLength, i - n + 1)

        return maxLength  # Return the maximum length found

```
['Python3']