You are given two strings `s`

and `t`

of the same length. You want to change `s`

to `t`

. Changing the `i`

-th character of `s`

to `i`

-th character of `t`

costs `|s[i] - t[i]|`

that is, the absolute difference between the ASCII values of the characters.

You are also given an integer `maxCost`

.

Return the maximum length of a substring of `s`

that can be changed to be the same as the corresponding substring of `t`

with a cost less than or equal to `maxCost`

.

If there is no substring from `s`

that can be changed to its corresponding substring from `t`

, return `0`

.

**Example 1:**

```
Input: s = "abcd", t = "bcdf", maxCost = 3
Output: 3
Explanation: "abc" of s can change to "bcd". That costs 3, so the maximum length is 3.
```

**Example 2:**

```
Input: s = "abcd", t = "cdef", maxCost = 3
Output: 1
Explanation: Each character in s costs 2 to change to charactor in t, so the maximum length is 1.
```

**Example 3:**

```
Input: s = "abcd", t = "acde", maxCost = 0
Output: 1
Explanation: You can't make any change, so the maximum length is 1.
```

**Constraints:**

`1 <= s.length, t.length <= 10^5`

`0 <= maxCost <= 10^6`

`s`

and`t`

only contain lower case English letters.

**Solution:**

Two pointer.

```
class Solution {
public:
int equalSubstring(string s, string t, int maxCost) {
vector<int> dif(s.size(), 0);
for(int i = 0; i<s.size(); i++){
dif[i] = abs((int)s[i]-t[i]);
}
int a = 0, b=0;
int cur = 0, ans =0 ;
while(b < s.size()){
cur += dif[b];
while(cur > maxCost && a <= b){
cur -= dif[a];
a++;
}
ans = max(ans, b-a+1);
b++;
}
return ans;
}
};
```