# Get Equal Substrings Within Budget 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 twith 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;
}
};