Medium
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 <= 105
s
consists of lowercase English letters.class Solution {
fun repeatLimitedString(s: String, repeatLimit: Int): String {
val result = CharArray(s.length)
val freq = IntArray(128)
var index = 0
for (c in s.toCharArray()) {
freq[c.code]++
}
var max = 'z'
var second = 'y'
while (true) {
while (max >= 'a' && freq[max.code] == 0) {
max--
}
if (max < 'a') {
break
}
second = Math.min(max.code - 1, second.code).toChar()
var count = Math.min(freq[max.code], repeatLimit)
freq[max.code] -= count
while (count-- > 0) {
result[index++] = max
}
if (freq[max.code] == 0) {
max = second--
continue
}
while (second >= 'a' && freq[second.code] == 0) {
second--
}
if (second < 'a') {
break
}
result[index++] = second
freq[second.code]--
}
return String(result, 0, index)
}
}