[LeetCode] 1208. Get Equal Substrings Within Budget
grandyang opened this issue · comments
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
andt
only contain lower case English letters.
这道题给了两个字符串s和t,定义了一种 cost,是将 s[i] 变为 t[i] 的花费就是两个字母的 ASCII 码值的差的绝对值。现在给了一个最大花费额度 maxCost,问最多可以把s串的多少个连续字母转为t的对应字母,从而使得花费不超过给定的最大额度。这里强调了是s的子串,即中间不能有断开,若某个位置 s[i] 和 t[i] 相等,那最好了,花费为0,可以继续延长子串的长度。既然是子串,肯定是要连续的字母,那就一个一个的字母的累加,当花费超过给定值了,就将开头的字母去掉,从而减少总花费。这道 Medium 的题目其实主要就是考察滑动窗口 Sliding Window,维护一个特定长度的窗口,此时若窗口内的总花费小于给定值,则扩大窗口长度,即增加窗口的右边界。而一旦总花费超过给定值了,就要移除窗口最左边的元素,有时可能要连续移除多个,所以需要用一个 while 循环,条件是窗口内的总花费大于给定值,且左边界小于等于右边界,然后移除左边界的元素,并且左边界自增1。同时别忘了每次都用更新后的窗口长度来更新最终结果 res 即可,参见代码如下:
class Solution {
public:
int equalSubstring(string s, string t, int maxCost) {
int res = 0, n = s.size(), cur = 0, start = 0;
for (int i = 0; i < n; ++i) {
cur += abs(s[i] - t[i]);
while (cur > maxCost && start <= i) {
cur -= abs(s[start] - t[start]);
++start;
}
res = max(res, i - start + 1);
}
return res;
}
};
Github 同步地址:
参考资料:
https://leetcode.com/problems/get-equal-substrings-within-budget/