README.md

1208. 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 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.

Solutions (Ruby)

1. Two Pointers

# @param {String} s
# @param {String} t
# @param {Integer} max_cost
# @return {Integer}
def equal_substring(s, t, max_cost)
  s = s.bytes
  t = t.bytes
  i = -1
  cost = 0
  ret = 0

  (0...s.size).each do |j|
    cost += (s[j] - t[j]).abs
    while cost > max_cost
      i += 1
      cost -= (s[i] - t[i]).abs
    end
    ret = [ret, j - i].max
  end

  ret
end

Solutions (Rust)

1. Two Pointers

impl Solution {
    pub fn equal_substring(s: String, t: String, max_cost: i32) -> i32 {
        let s = s.as_bytes();
        let t = t.as_bytes();
        let mut i = -1;
        let mut cost = 0;
        let mut ret = 0;

        for j in 0..s.len() {
            cost += (s[j] as i32 - t[j] as i32).abs();
            while cost > max_cost {
                i += 1;
                cost -= (s[i as usize] as i32 - t[i as usize] as i32).abs();
            }
            ret = ret.max(j as i32 - i);
        }

        ret
    }
}