{ "data": { "question": { "questionId": "3346", "questionFrontendId": "100242", "categoryTitle": "Algorithms", "boundTopicId": 2725542, "title": "Lexicographically Smallest String After Operations With Constraint", "titleSlug": "lexicographically-smallest-string-after-operations-with-constraint", "content": "

You are given a string s and an integer k.

\n\n

Define a function distance(s1, s2) between two strings s1 and s2 of the same length n as:

\n\n\n\n

For example, distance("ab", "cd") == 4, and distance("a", "z") == 1.

\n\n

You can change any letter of s to any other lowercase English letter, any number of times.

\n\n

Return a string denoting the lexicographically smallest string t you can get after some changes, such that distance(s, t) <= k.

\n\n

 

\n

Example 1:

\n\n
\n

Input: s = "zbbz", k = 3

\n\n

Output: "aaaz"

\n\n

Explanation:

\n\n

Change s to "aaaz". The distance between "zbbz" and "aaaz" is equal to k = 3.

\n
\n\n

Example 2:

\n\n
\n

Input: s = "xaxcd", k = 4

\n\n

Output: "aawcd"

\n\n

Explanation:

\n\n

The distance between "xaxcd" and "aawcd" is equal to k = 4.

\n
\n\n

Example 3:

\n\n
\n

Input: s = "lol", k = 0

\n\n

Output: "lol"

\n\n

Explanation:

\n\n

It's impossible to change any character as k = 0.

\n
\n\n

 

\n

Constraints:

\n\n\n", "translatedTitle": "满足距离约束且字典序最小的字符串", "translatedContent": "

给你一个字符串 s 和一个整数 k 。

\n\n

定义函数 distance(s1, s2) ，用于衡量两个长度为 n 的字符串 s1 和 s2 之间的距离，即：

\n\n\n\n

例如，distance(\"ab\", \"cd\") == 4 ，且 distance(\"a\", \"z\") == 1 。

\n\n

你可以对字符串 s 执行 任意次 操作。在每次操作中，可以将 s 中的一个字母 改变 为 任意 其他小写英文字母。

\n\n

返回一个字符串，表示在执行一些操作后你可以得到的 字典序最小 的字符串 t ，且满足 distance(s, t) <= k 。

\n\n

 

\n\n

示例 1：

\n\n
\n输入：s = \"zbbz\", k = 3\n输出：\"aaaz\"\n解释：在这个例子中，可以执行以下操作：\n将 s[0] 改为 'a' ，s 变为 \"abbz\" 。\n将 s[1] 改为 'a' ，s 变为 \"aabz\" 。\n将 s[2] 改为 'a' ，s 变为 \"aaaz\" 。\n\"zbbz\" 和 \"aaaz\" 之间的距离等于 k = 3 。\n可以证明 \"aaaz\" 是在任意次操作后能够得到的字典序最小的字符串。\n因此，答案是 \"aaaz\" 。\n
\n\n

示例 2：

\n\n
\n输入：s = \"xaxcd\", k = 4\n输出：\"aawcd\"\n解释：在这个例子中，可以执行以下操作：\n将 s[0] 改为 'a' ，s 变为 \"aaxcd\" 。\n将 s[2] 改为 'w' ，s 变为 \"aawcd\" 。\n\"xaxcd\" 和 \"aawcd\" 之间的距离等于 k = 4 。\n可以证明 \"aawcd\" 是在任意次操作后能够得到的字典序最小的字符串。\n因此，答案是 \"aawcd\" 。\n
\n\n

示例 3：

\n\n
\n输入：s = \"lol\", k = 0\n输出：\"lol\"\n解释：在这个例子中，k = 0，更改任何字符都会使得距离大于 0 。\n因此，答案是 \"lol\" 。
\n\n

 

\n\n

提示：

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