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.
Solution: Greedy
Adding one letter at a time, find the largest one that can be used.
Time complexity: O(26*n)
Space complexity: O(1)
C++
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// Author: Huahua class Solution { public: string repeatLimitedString(string s, int repeatLimit) { vector<int> m(26); for (char c : s) ++m[c - 'a']; string ans; int count = 0; int last = -1; while (true) { bool found = false; for (int i = 25; i >= 0 && !found; --i) if (m[i] && (count < repeatLimit || last != i)) { ans += 'a' + i; count = (last == i) * count + 1; --m[i]; last = i; found = true; } if (!found) break; } return ans; } }; |