2182. Construct String With Repeat Limit

2182. Construct String With Repeat Limit #

题目 #

You are given a string s and an integer repeatLimit. Construct a new string repeatLimitedString using the characters of s such that no letter appears more than repeatLimit times in a row. You do not have to use all characters from s.

Return the lexicographically largest repeatLimitedString possible.

A string a is lexicographically larger than a string b if in the first position where a and b differ, string a has a letter that appears later in the alphabet than the corresponding letter in b. If the first min(a.length, b.length) characters do not differ, then the longer string is the lexicographically larger one.

Example 1:

Input: s = "cczazcc", repeatLimit = 3
Output: "zzcccac"
Explanation: We use all of the characters from s to construct the repeatLimitedString "zzcccac".
The letter 'a' appears at most 1 time in a row.
The letter 'c' appears at most 3 times in a row.
The letter 'z' appears at most 2 times in a row.
Hence, no letter appears more than repeatLimit times in a row and the string is a valid repeatLimitedString.
The string is the lexicographically largest repeatLimitedString possible so we return "zzcccac".
Note that the string "zzcccca" is lexicographically larger but the letter 'c' appears more than 3 times in a row, so it is not a valid repeatLimitedString.

Example 2:

Input: s = "aababab", repeatLimit = 2
Output: "bbabaa"
Explanation: We use only some of the characters from s to construct the repeatLimitedString "bbabaa".
The letter 'a' appears at most 2 times in a row.
The letter 'b' appears at most 2 times in a row.
Hence, no letter appears more than repeatLimit times in a row and the string is a valid repeatLimitedString.
The string is the lexicographically largest repeatLimitedString possible so we return "bbabaa".
Note that the string "bbabaaa" is lexicographically larger but the letter 'a' appears more than 2 times in a row, so it is not a valid repeatLimitedString.

Constraints:

  • 1 <= repeatLimit <= s.length <= 10^5
  • s consists of lowercase English letters.

题目大意 #

给你一个字符串 s 和一个整数 repeatLimit ,用 s 中的字符构造一个新字符串 repeatLimitedString ,使任何字母 连续 出现的次数都不超过 repeatLimit 次。你不必使用 s 中的全部字符。

返回 字典序最大的 repeatLimitedString 。

如果在字符串 a 和 b 不同的第一个位置,字符串 a 中的字母在字母表中出现时间比字符串 b 对应的字母晚,则认为字符串 a 比字符串 b 字典序更大 。如果字符串中前 min(a.length, b.length) 个字符都相同,那么较长的字符串字典序更大。

解题思路 #

  • 利用贪心的思想,由于题意要求返回字典序最大的字符串,所以先从字典序最大的字母开始选起。然后选择当前字典序最大的字母个数和 limit 的最小值。如果当前字典序最大的字母比较多,多于 limit,不能一直选择它。选完 limit 个以后,需要选一个字典序次大的字母,选完这个字母以后再次选择字典序最大的字母。因为 limit 限制字母不能连续多于 limit 个。如此循环,直到所有的字母都选完。这样的策略排列出来的字母串为最大字典序。

代码 #

package leetcode

func repeatLimitedString(s string, repeatLimit int) string {
	cnt := make([]int, 26)
	for _, c := range s {
		cnt[int(c-'a')]++
	}
	var ns []byte
	for i := 25; i >= 0; {
		k := i - 1
		for cnt[i] > 0 {
			for j := 0; j < min(cnt[i], repeatLimit); j++ {
				ns = append(ns, byte(i)+'a')
			}
			cnt[i] -= repeatLimit
			if cnt[i] > 0 {
				for ; k >= 0 && cnt[k] == 0; k-- {
				}
				if k < 0 {
					break
				} else {
					ns = append(ns, byte(k)+'a')
					cnt[k]--
				}
			}
		}
		i = k
	}
	return string(ns)
}
func min(a, b int) int {
	if a < b {
		return a
	} else {
		return b
	}
}

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